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[ { "cid": "1", "user": "user-1", "content": "链表只是一种数据组织形式，还有很多其它的，比如用数组。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "for() 那行最后多了个分号 ; 前面好像有贴有类似问题的", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "谢谢大佬，现在行了。我还专门去人家博客贴了一个同样的来运行，也是一样的结果就不知所以然了哈哈", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "\"%d%d\" 后面少了个分隔的逗号", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "https://jingyan.baidu.com/article/6d704a134d9c4e28da51ca4d.html", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "https://blog.csdn.net/goldenhawking/article/details/119845838 如果是VC应该要在项目属性设置里配置C++/ include ，Link/Library 等属性。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "https://blog.csdn.net/qq_27898413/article/details/121291412", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "A(*buffer,buff_cnt) {}B(*buffer,count) { A(buffer,count);}", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "#include <stdio.h> int main(int argc, char * argv[]) { int a = 0, b = 0, c = 0, i = 0; scanf(\"%d,%d,%d\",&a,&b,&c); printf (\"\\n%d/%d=%.14lf\\n\",a,b,1.0*a/b); if (b==0) return 0; int array[103]; do { int d = a / b; int e = a % b; array[i] = d; a = e; a *=10; }while (i++ <= c+1); if (array[c+1]>=5) { i = c; do{ ++array[i]; --i; }while (i>=0 && !array[i]); } for (i = 0; i<=c;++i) { printf (\"%d\",array[i]); if (!i) putchar('.'); } return 0; }", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "呵呵请注意 c<=100 double哪儿来的那么高的精度？", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "題目并不是要自己做除法函數，儘管題目寫著c<=100，實際并不要求考慮c是否超double精度的問題。", "referer": "user-2" }, { "cid": "4", "user": "user-2", "content": "题目中的表达不是输出c位小数，而是精确到小数点后c位。", "referer": "user-3" }, { "cid": "5", "user": "user-3", "content": "陷阱，估计说的是（double）", "referer": "user-0" }, { "cid": "6", "user": "user-3", "content": "char fmt[8] = {0}; sprintf(fmt, \"%%.%df\"，c); printf(fmt, (double)a / b);", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "因为scanf看见空格就认为输入结束了。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "那两个白色的a应该就是输入输出吧，程序应该都执行完了。", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "输入 a 运行程序？你的源文件是 1.c，生成的执行程序通常是 1.exe 吧，输入一般是输入 1 或 1.exe 运行它。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "可以去 ask.csdn.net 提问", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "第一个if行末有个多余的分号。", "referer": "user-0" }, { "cid": "3", "user": "user-0", "content": "谢谢", "referer": "user-2" } ]

[ { "cid": "1", "user": "user-1", "content": "异常，被 0 除，可能是 i=0 时作为除数求余引起的吧", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "感谢", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "求值顺序的问题，先求值x---还是x+y未定义", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "请截个图证明一下dev显示是1", "referer": "user-0" }, { "cid": "4", "user": "user-2", "content": "看看我的图,我都有点蒙了", "referer": "user-3" }, { "cid": "5", "user": "user-3", "content": "的确无法解释，懵C中。", "referer": "user-2" }, { "cid": "6", "user": "user-2", "content": "我已经知道答案了,1&gt;(0+0),x到了大于号右边已经变成0了", "referer": "user-3" } ]

[ { "cid": "1", "user": "user-1", "content": "“给定一个小点的输入，完整单步跟踪(同时按Alt+7键查看Call Stack里面从上到下列出的对应从里层到外层的函数调用历史)一遍。”是理解递归函数工作原理的不二法门！递归函数关注以下几个因素·退出条件·参数有哪些·返回值是什么·局部变量有哪些·全局变量有哪些·何时输出·会不会导致堆栈溢出", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "这个挺容易的， 你自己先通关这个 clang.csdn.net", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "大佬写完代码之后报更高级的错。平衡了么，亲？", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "我觉得你想的，和编译器对你源码的理解可能不一样。你是想把 \"YES\" 或 “NO\" 字符串放进 a[i] 里吧，编译器应该是以为你要放的是指针；这样的话，因为只是 ”YES“ 和 \"NO\" 两个内容，可以用简单的数据标记代替它，放入和取出显示时进行转换，只定义 a[n] 就可以了；一定要把字符串复制进去，首先，空间要保证够或进行检查截断，其次恐怕得用字符串复制函数。", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "谢谢您的解答，给我提供了一种编译思路，十分感谢", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-0", "content": "", "referer": "user-0" }, { "cid": "2", "user": "user-1", "content": "typedef struct A{...}B; 引用它时要么用struct A xxx要么用B xxx不要混着来。", "referer": "user-0" }, { "cid": "3", "user": "user-1", "content": "你得注意到第一个你说正常的那个其实从未使用带_t的东西。", "referer": "user-0" }, { "cid": "4", "user": "user-1", "content": "去掉带_t的你称为别名的东西的前面的struct就好。", "referer": "user-0" }, { "cid": "5", "user": "user-0", "content": "按照您的说法改动之后，报错变了", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "因为你缺少了必要的头文件，因而缺少了必要的原型声明，所以函数值被误以为是整型。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "if (*p[j]>*p[j+1])应改为if (strcmp(p[j]>p[j+1])>0)", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "提交编译错误了。。。", "referer": "user-1" }, { "cid": "3", "user": "user-0", "content": "不是应该是strcmp（p[j],p[j+1])吗", "referer": "user-0" }, { "cid": "4", "user": "user-1", "content": "对，我刚才&gt;忘改为,了", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "每个 i 循环，k=0;；min=a[0][k]; 要后移到循环外，同时 y=0; ；break; 应在 if 里 for(i=0;i<m;i++)//筛选开始 { k = 0; for(j=0;j<n;j++)//寻找每行最大值 { if(a[i][k]<a[i][j]) { k=j; a[i][k]=a[i][j]; } max=a[i][k]; } min=a[0][k]; y = 0; for(z=0;z<n;z++) { if(a[z][k]<min) { min=a[z][k]; y=z; } } if(min==max) { printf(\"该数组a[%d][%d]有一个鞍点是%4d\",y,k,a[y][k]); printf(\"\\n\"); o++; break; } }", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "谢谢啦，大佬可以给个关注位置吗？", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "srand() 随机数初始化，time() 返回以秒为单位，这样程序运行时，每次的 srand() 几乎每次都一样，因为循环总共才 9次，跨秒的可能性极小，即使侥幸跨秒，数组里也只会有两不同的数，不会有整体随机的效果（也难个与个都不同）。", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "讲解很详细明白了谢谢啦", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "srand(); 应该挪到循环圈外面吧，即 for(i) 之前", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "谢谢已经解决了 ，不过这个原因是什么呢？烦请解释一下可以吗？", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "通过简单的变换得出，要使两个表达式同时成立，必须是x比z大10。如x=10,y=0,z=0，或者z为负数，但是循环里y和z都是从1开始的，所以永远找不到能使两个表达式同时成立的数值，自然也就不会运行输出语句了。", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "x+y+z==10 3x+2y+ z==30 這2個表達式不可能同時成立。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "abs是处理整数的，返回值也是整数。", "referer": "user-0" }, { "cid": "2", "user": "user-1", "content": "浮点需要fabs", "referer": "user-1" }, { "cid": "3", "user": "user-0", "content": "正常了，谢谢", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "这里的C99特性有两点一是随处定义变量二是用变量定义数组大小现在你愉快地使用了这两个特性，换个开发环境不让用了还不明说，就会很懵", "referer": "user-0" }, { "cid": "2", "user": "user-1", "content": "第一，scanf里面不要带转义字符第二，使用C99特性是有风险的，很可能和你以后考试的环境不符第三，数组下标合法值在0到j-1之间，而且你得写个循环来输入才行", "referer": "user-0" }, { "cid": "3", "user": "user-0", "content": "为什么scanf里不能带\\n，书上都是这么写的", "referer": "user-1" }, { "cid": "4", "user": "user-1", "content": "要不您去找找书？", "referer": "user-0" }, { "cid": "5", "user": "user-2", "content": "你可能是看错了，一般是printf打印里面添加\\n换行符，scanf带\\n会存在有一个数据缓存区的问题，scanf遇到换行或者空格就会停止输入", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "printf(\"Text_Chinese:%s image:%p W:%d h:%d\\n\", lv_btn_image1[i].btn_text_Chinese , lv_btn_image1[i].btn_image, lv_btn_image1[i].btn_witch, lv_btn_image1[i].btn_hietch);", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "于是我陷入沉思，那么一定有问题，问题在哪儿呢？", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "凡是钱能解决的问题都不是问题。——《蜗居》宋思明", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "这道题还是有些难度的， 上面的程序有可能在： 当m<k时， 存在问题。下面程序抛砖引玉，还请大伙指正。 #include <stdio.h> int min(int a, int b) { if(a<b) { return a; } return b; } unsigned long long cnt1(int m, int n, int k) { if(k>n) { return 0; } int i; unsigned long long cnt = 1; int x = min(k,m); for(i=0;i<x;i++) { cnt = cnt * (n-i); } for(i=k;i<m;i++) { cnt = cnt * (n - k + 1); } return cnt; } int main() { int m, n, k; scanf(\"%d %d %d\", &m, &n, &k); unsigned long long sum = cnt1(m,n,k); if (sum==0) { printf(\"Input Error\"); } else { printf(\"%llu\", sum); } return 0; } ```c ```", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "两 for 循环里的 a 或 b 都应该是大写的 A 或 B 吧，除了 int 语句里的： for (int A=a; A != 0; A=A / 10) { e = A % 10; f = f * 10 + e; }", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "a = a / 10", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "0,2,1,15.000000,1.000000,1.500000", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "5,2,1,15.000000,1.500000,0.200000", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "因为你在每次输入完数字之后都会输入一个回车键，而回车键也是一个字符（‘\\n’），会被scanf（“%c”）读取，而‘\\n’ 属于小于'0' || 大于‘9’的范围，所以会输出“error”加getchar函数，或者%*c都是不可取的，得到输出结果应该不是博主想要的建议在第一个if后边加 &&p！=‘\\n’", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "为什么加getchar函数，或者%*c最后结果是我想要的呢", "referer": "user-1" }, { "cid": "3", "user": "user-2", "content": "在每个最后不带\\n的printf后面加fflush(stdout);在每个不想受接收缓冲区旧内容影响的scanf前面加rewind(stdin);另外请检查scanf的返回值。 //请今后要用 char c; scanf(\"%c\",&c); //时，都改为 char s[2]; char c; scanf(\"%1s\",s); c=s[0]; //自动跳过一个或多个空格或Tab字符或回车换行，读取下一个字符。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "我按照你写的能运行啊", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "刚才查了一会，是需要在name后加40，提高它的准确性和安全性，不知道为啥你的可以运行，我的加了这个40后才行", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "你代码里面name后面加了40 啊", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "是在第九行那的name后加", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "你试试在在输入的后面加上长度，比如scanf_s(\"%s\",first_name,40);", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "可以也，这是什么原理？", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "好像是scanf_s就得加这个长度，如果是scanf就不用", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "如此修改，这般输入。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "在输入数据之前进行 X, x （这两个变量名也值得商榷）计算？这，显然不合逻辑啊。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "", "referer": "user-0" }, { "cid": "2", "user": "user-1", "content": "没有注释将就看一下", "referer": "user-1" }, { "cid": "3", "user": "user-0", "content": "谢了", "referer": "user-1" }, { "cid": "4", "user": "user-2", "content": "输入数据没带逗号造成输入错误。代码里要求逗号分隔输入数据的。", "referer": "user-0" }, { "cid": "5", "user": "user-0", "content": "谢谢了，是这样的", "referer": "user-2" } ]

[ { "cid": "1", "user": "user-1", "content": "结束符是'\\0'而不是'/0'所以很难理解为什么能通过编译。", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "修改后后面的乱码消失了，但是输出结果是原s内容，没有显示d之前的字符串", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "编译出错信息都不看的么？ d两边应该是单引号 malloc(strlen(s)+1) 这里和p没啥关系", "referer": "user-2" }, { "cid": "4", "user": "user-2", "content": "好了！感谢大佬", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "fab 函数改为 fabs 函数，没有 fab 函数吧。以后发图片，建议用截图的方式，手机拍照可能会拍不清楚，QQ 截图快捷键：Ctrl + Alt + A", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "fabs吧", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "undefined reference to 'fab'你的fab没有定义", "referer": "user-0" }, { "cid": "4", "user": "user-4", "content": "下面的提示，第一行说 fab 不知道是啥，我看了，也不知道是什么函数", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "OJ是不是不能手工输入？scanf(\"%d%d\",&a,&b)!=EOF", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "谢谢，已解决，穷举法过于费力，已经改为辗转相除法，谢谢你的回答！", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-0", "content": "谢谢各位！", "referer": "user-0" }, { "cid": "2", "user": "user-1", "content": "printf(\"%d+%d=%d\\n\",a,b,c);这样是会输出a的值加b的值等于c的值。如果像单纯显示a，b的值是多少的话，直接printf(\"%d %d\",a,b);", "referer": "user-0" }, { "cid": "3", "user": "user-2", "content": "c也可以不需要，直接用a + b、a - b", "referer": "user-0" }, { "cid": "4", "user": "user-3", "content": "printf(\"%d+%d=%d\\n\",a,b,c); // 这样的吧", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "yday = month==2 ? ((year%400==0||year%4==0&&year%100!=0)+28) : ((int)fabs(month-7.5)%2 ? 30 : 31) ;", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "很有用，感谢作者帮助！", "referer": "user-0" }, { "cid": "3", "user": "user-0", "content": "很简洁的代码，太厉害了", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "国家出台的：终身教育平台里面的教程全部免费，看看教学视频跟着做，做好课堂笔记，不懂的就去网上查资料，实在找不到，论坛这么多大神总会有比较热心的大佬点拨一下的；又不是考试，没必要去找练习册；多在论坛看看别人写的东西，看看别人的评论有时候就会有答案了", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "搜索一下OJ", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "（‘a’+'b）2=(97+98)×2=390ab=6×3.4=20.4c/d=10.0÷5=2.0390+20.4-2.0=408.4double类型", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "困难在哪里？", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "#include <stdio.h> int main (){ int a = 1; int b; scanf (\"%d\",&b); printf (\"%d\",a); while (a<b-1)//先循环后判断 { a=a+2; printf (\" %d\",a); } system (\"pause\"); return 0; }你看看，这样不就行了，没必要判断他的奇数还是偶数", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "好的谢谢，想错了。。。", "referer": "user-1" }, { "cid": "3", "user": "user-2", "content": "if ( m % 2 == 1) 吧，不过没意义；else 里的 while (m <= n) 吧，不能写成 =<", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "应该就是n吧 后面你说的是对的，我后来自己注意到了，改成这样了 #include &lt;stdio.h&gt; int main() { \tint n; \tint m = 1; \tscanf(\"%d\",&n); \tif ( n % 2 == 1){ while(m &lt;= n){ printf(\"%d\",m); m += 2; } \t\t}else{ \t\twhile(m &lt;= n){ \t\t\tprintf(\"%d\",m); \t\t\tm = m + 2; \t\t} \t \t} \treturn 0; } 但五分只得了2分，说我“检查程序输出格式”", "referer": "user-2" }, { "cid": "5", "user": "user-3", "content": "既然两块里面内容一样，那为什么还要写两遍？ 另外你的输出格式不对，没有空格。", "referer": "user-0" }, { "cid": "6", "user": "user-0", "content": "确实内容一样，当时想错了，但我不太知道哪里没有空格", "referer": "user-3" } ]

[ { "cid": "1", "user": "user-1", "content": "scanf(\"%d\",&x);while 循环中也要有 scanfif(x == 34) #include <stdio.h> int main() { int a=34; int x; int n=0; printf(\"请输入猜的数字 \"); scanf(\"%d\",&x); while (x!=34) { n++; if(x<34) { printf(\"小了\\n\"); } else {printf(\"大了\\n\"); } scanf(\"%d\",&x); } if (x==34) { printf(\"正确\\n\"); n++; } printf(\"一共猜测次数为 %d\\n\",n); return 0; }", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "if (x==34)", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "while(x!=34) 等於34時不會進去while", "referer": "user-0" }, { "cid": "4", "user": "user-3", "content": "看錯啦", "referer": "user-3" } ]

[ { "cid": "1", "user": "user-1", "content": "用puts少4个字符~~~", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "#include <stdio.h> int main() { printf( \"****\\n\" \"****\\n\" \"****\\n\" \"\\n\" \"\\n\" \"\\n\" \"&&&&&&\\n\" \" &&&&&&\\n\" \" &&&&&&\\n\" ); return 0; }", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "没懂什么意思", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-0", "content": "错误列表： typeof 在windows下c中是没有的。宏实现最外层有() 会导致解析错误。这个好解决，删掉最外层括号。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "看给谁赋值的。局部变量？ 栈区。char *str=NULL; 和 char *str=\"123\"；。因为要存\"123\"这个常量字符串，编译的固件会比较大。也有可能影响业务逻辑。全局变量? 全局变量初始化为0 会影响这个变量是放到 数据段还是bss段。....", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "没啥影响。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "C语言教程（附带C语言100例） http://download.csdn.net/detail/zhao4zhong1/8830817", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "一般情况下，理论强的人，动手不行。实践强的人，动脑不行。我们要理论和实践相结合。既不动手，也不动脑。 =====就是用起来，理论+实践。去github找点简单的项目，看懂了。先模仿，在超越。？", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "你先推导它的表达式，然后就简单了。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "TSEPro11_Setup.exe 百万军中取上将首级如探囊取物， 千万行里改关键源码在弹指瞬间。 功能超强的程序编辑器！ https://blog.csdn.net/zhao4zhong1/article/details/126596717", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "收藏大佬文章了，大佬支持下我呗，感谢：https://bbs.csdn.net/topics/611387347", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "这是什么情况", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "switch(c) 后面不应该有那个分号 ; 吧", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "你觉得好奇，于是决定租一块地，种四颗double菜尝试一下。地主（系统）租给你一块地，在地头打了个桩子上面写着p_d然后你看看你都干了啥？你种地你收获这都没啥说的。你悄悄把桩子挪了个位置，然后企图把地还给地主的时候，地主揍了你。还觉得冤吗，亲？", "referer": "user-0" }, { "cid": "2", "user": "user-1", "content": "如果你在剧本里对还地的情节避而不谈，当然我们就不会看到你卡住的遭遇，但仅仅是没看到，你懂的。", "referer": "user-1" }, { "cid": "3", "user": "user-0", "content": "哈哈哈哈，很生动", "referer": "user-1" }, { "cid": "4", "user": "user-0", "content": "那如果我只是把桩子挪了个位置，会产生哪些影响呢", "referer": "user-1" }, { "cid": "5", "user": "user-2", "content": "p_d都+=1了，就不能再delete[] p_d了", "referer": "user-0" }, { "cid": "6", "user": "user-0", "content": "可以详细解释下吗，我刚学这个看不懂谢谢", "referer": "user-2" }, { "cid": "7", "user": "user-2", "content": "new返回的内容在delete的时候不能变，您应该另外用一个变量比如p_d_old记住p_d刚刚new返回的值，delete[] p_d_old而不能delete[] p_d，因为你已经使用p_d+=1改动了p_d的值。", "referer": "user-0" }, { "cid": "8", "user": "user-0", "content": "谢谢", "referer": "user-2" } ]

[ { "cid": "1", "user": "user-1", "content": "百度搜“跟我一起写MakeFile”", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "makefile我会 太麻烦 没有cmake方便 现在我都用cmake", "referer": "user-1" }, { "cid": "3", "user": "user-0", "content": "makefile指定目录我会", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "k=0", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "while(scanf(\"%d\",&num))把数字输入完成后，最后再输入一个非数字字符作为结束。 for(j=0;j<n;j++) //不是j<=n { printf(\"% d\",list[j]); sum += list[j] ; }", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "while((scanf(\"%d\",&num)!=EOF)) 显示偶数的循环，应该是 list[j]", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "谢谢！！", "referer": "user-3" }, { "cid": "5", "user": "user-4", "content": "scanf(\"%d\",&num)", "referer": "user-0" }, { "cid": "6", "user": "user-0", "content": "谢谢，但是我试了一下还是循环不了", "referer": "user-4" }, { "cid": "7", "user": "user-5", "content": "有没有一种可能，除此之外还有其他错误？", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "我总是在说：代码不是2nd~ #include <stdio.h> int main() { for (int n, i = (scanf(\"%d\", &n), 1); i <= n; i++, puts(\"\")) { for (int j = 0; j < n - i; j++) printf(\" \"); for (int j = 1; j <= i; j++) printf(\"%3d\", j); } return 0; }", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "看要求的模样，输出数据前，不应该是要输出相应的空格的吗，(10-i) 个 printf(\" \");", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "#include <stdio.h> int main() { int N; int a; scanf(\"%d\", &N); for (int i = 1; i <= N; i++) { a=i; while(a<N){ printf(\" \");//空格的数量根据 \"%3d\"，是3就3个空格，是4就四个空格 a++; } for (int j = 1; j <= i; j++) { printf(\"%3d\", j); } if (i < N) { printf(\"\\n\"); } } return 0; }", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "仅供参考： #pragma warning(disable:4996) /*--------------------------------------- 函数型计算器(VC++6.0,Win32 Console) 功能： 目前提供了10多个常用数学函数: ⑴正弦sin ⑵余弦cos ⑶正切tan ⑷开平方sqrt ⑸反正弦arcsin ⑹反余弦arccos ⑺反正切arctan ⑻常用对数lg ⑼自然对数ln ⑽ｅ指数exp ⑾乘幂函数^ ⑿向上取整ceil ⒀向下取整floor ⒁四舍五入取整round ⒂取符号sign ⒃取绝对值abs 用法： 如果要求2的32次幂，可以打入2^32<回车> 如果要求30度角的正切可键入tan(Pi/6)<回车> 注意不能打入：tan(30)<Enter> 如果要求1.23弧度的正弦，有几种方法都有效： sin(1.23)<Enter> sin 1.23 <Enter> sin1.23 <Enter> 如果验证正余弦的平方和公式,可打入sin(1.23)^2+cos(1.23)^2 <Enter>或sin1.23^2+cos1.23^2 <Enter> 此外两函数表达式连在一起,自动理解为相乘如：sin1.23cos0.77+cos1.23sin0.77就等价于sin(1.23)*cos(0.77)+cos(1.23)*sin(0.77) 当然你还可以依据三角变换，再用sin(1.23+0.77)也即sin2验证一下。 本计算器充分考虑了运算符的优先级因此诸如：2+3*4^2 实际上相当于：2+(3*(4*4)) 另外函数名前面如果是数字,那么自动认为二者相乘. 同理，如果某数的右侧是左括号，则自动认为该数与括弧项之间隐含一乘号。 如：3sin1.2^2+5cos2.1^2 相当于3*sin2(1.2)+5*cos2(2.1) 又如：4(3-2(sqrt5-1)+ln2)+lg5 相当于4*(3-2*(√5 -1)+loge(2))+log10(5) 此外，本计算器提供了圆周率Pi键入字母时不区分大小写,以方便使用。 16进制整数以0x或0X开头。 ----------------------------------------*/ #include <stdlib.h> #include <string.h> #include <ctype.h> #include <math.h> #include <stdio.h> #include <string.h> #include <locale.h> #include <windows.h> const char Tab = 0x9; const int DIGIT = 1; #define MAXLEN 16384 char s[MAXLEN],t[MAXLEN], *endss; int pcs = 15; FILE *fp; double sign(double dVal) { if (dVal>0.0) return 1.0; else if (dVal<0.0) return -1.0; else return 0.0; } double round(double dVal, short iPlaces) {//iPlaces>=0 char tmp[30]; double dRetval; sprintf(tmp, \"%.*lf\", iPlaces, dVal); sscanf(tmp, \"%lf\", &dRetval); return (dRetval); } double fun(double x, char op[], int *iop) { while (op[*iop - 1]<32) //本行使得函数嵌套调用时不必加括号,如 arc sin(sin(1.234)) 只需键入arc sin sin 1.234<Enter> switch (op[*iop - 1]) { case 7: x = sin(x); (*iop)--; break; case 8: x = cos(x); (*iop)--; break; case 9: x = tan(x); (*iop)--; break; case 10: x = sqrt(x); (*iop)--; break; case 11: x = asin(x); (*iop)--; break; case 12: x = acos(x); (*iop)--; break; case 13: x = atan(x); (*iop)--; break; case 14: x = log10(x); (*iop)--; break; case 15: x = log(x); (*iop)--; break; case 16: x = exp(x); (*iop)--; break; case 17: x = ceil(x); (*iop)--; break; case 18: x = floor(x); (*iop)--; break; case 19: x = round(x,0);(*iop)--; break; case 20: x = sign(x); (*iop)--; break; case 21: x = fabs(x); (*iop)--; break; } return x; } int isKMG(char c) { if (c=='k'||c=='K'||c=='m'||c=='M'||c=='g'||c=='G') return 1; return 0; } double calc(char *expr, char **addr) { static int deep; //递归深度 static char *fname[] = { \"sin\",\"cos\",\"tan\",\"sqrt\",\"arcsin\",\"arccos\",\"arctan\",\"lg\",\"ln\",\"exp\",\"ceil\",\"floor\",\"round\",\"sign\",\"abs\",NULL }; double ST[10] = { 0.0 }; //数字栈 char op[10] = { '+' }; //运算符栈 char c, *rexp, *pp, *pf; int ist = 1, iop = 1, last, i, n; __int64 i64; if (!deep) { pp = pf = expr; do { c = *pp++; if (c != ' '&& c != Tab && c != ',')//跳过半角逗号(通常作为千分位分割符) *pf++ = c; } while (c != '\\0'); } pp = expr; if ((c = *pp) == '-' || c == '+') { op[0] = c; pp++; } last = !DIGIT; while ((c = *pp) != '\\0') { if (c == '(') {//左圆括弧 deep++; ST[ist++] = calc(++pp, addr); deep--; ST[ist - 1] = fun(ST[ist - 1], op, &iop); pp = *addr; last = DIGIT; if (*pp == '(' || isalpha(*pp) && strnicmp(pp, \"Pi\", 2) && !isKMG(*pp)) {//目的是：当右圆括弧的右侧为左圆括弧或函数名字时，默认其为乘法 op[iop++] = '*'; last = !DIGIT; c = op[--iop]; goto operate; } } else if (c == ')') {//右圆括弧 pp++; break; } else if (isalpha(c)) { if (!strnicmp(pp, \"Pi\", 2)) { if (last == DIGIT) { fprintf(fp,\"π左侧遇）\\n\"); if (fp!=stdout) fclose(fp); exit(1); } ST[ist++] = 3.14159265358979323846264338328; ST[ist - 1] = fun(ST[ist - 1], op, &iop); pp += 2; last = DIGIT; if (isalpha(pp[0]) && pp[0]!='x' && pp[0]!='X') { fprintf(fp,\"两个π或K/M/G相连\\n\"); if (fp!=stdout) fclose(fp); exit(2); } if (*pp == '(') { fprintf(fp,\"π右侧遇（\\n\"); if (fp!=stdout) fclose(fp); exit(3); } } else if (isKMG(pp[0])) { if (last == DIGIT) { fprintf(fp,\"%c左侧遇）\\n\",pp[0]); if (fp!=stdout) fclose(fp); exit(1); } double value=1.0; switch (pp[0]) { case 'k':value=1000.0 ;break; case 'K':value=1024.0 ;break; case 'm':value=1000000.0 ;break; case 'M':value=1048576.0 ;break; case 'g':value=1000000000.0;break; case 'G':value=1073741824.0;break; } ST[ist++] = value; ST[ist - 1] = fun(ST[ist - 1], op, &iop); pp += 1; last = DIGIT; if (isalpha(pp[0]) && pp[0]!='x' && pp[0]!='X') { fprintf(fp,\"两个π或K/M/G相连\\n\"); if (fp!=stdout) fclose(fp); exit(2); } if (*pp == '(') { fprintf(fp,\"%c右侧遇（\\n\",pp[-1]); if (fp!=stdout) fclose(fp); exit(3); } } else { for (i = 0; (pf = fname[i]) != NULL; i++) if (!strnicmp(pp, pf, strlen(pf))) break; if (pf != NULL) { op[iop++] = 07 + i; pp += strlen(pf); } else { fprintf(fp,\"陌生函数名\\n\"); if (fp!=stdout) fclose(fp); exit(4); } } } else if (c == '+' || c == '-' || c == '*' || c == '/' || c == '%' || c == '^') { char cc; if (last != DIGIT) { fprintf(fp,\"运算符粘连\\n\"); if (fp!=stdout) fclose(fp); exit(5); } pp++; if (c == '+' || c == '-') { do { cc = op[--iop]; --ist; switch (cc) { case '+': ST[ist - 1] += ST[ist]; break; case '-': ST[ist - 1] -= ST[ist]; break; case '*': ST[ist - 1] *= ST[ist]; break; case '/': ST[ist - 1] /= ST[ist]; break; case '%': ST[ist - 1] = fmod(ST[ist - 1], ST[ist]); break; case '^': ST[ist - 1] = pow(ST[ist - 1], ST[ist]); break; } } while (iop); op[iop++] = c; } else if (c == '*' || c == '/' || c == '%') { operate: cc = op[iop - 1]; if (cc == '+' || cc == '-') { op[iop++] = c; } else { --ist; op[iop - 1] = c; switch (cc) { case '*': ST[ist - 1] *= ST[ist]; break; case '/': ST[ist - 1] /= ST[ist]; break; case '%': ST[ist - 1] = fmod(ST[ist - 1], ST[ist]); break; case '^': ST[ist - 1] = pow(ST[ist - 1], ST[ist]); break; } } } else { cc = op[iop - 1]; if (cc == '^') { fprintf(fp,\"乘幂符连用\\n\"); if (fp!=stdout) fclose(fp); exit(6); } op[iop++] = c; } last = !DIGIT; } else { if (last == DIGIT) { fprintf(fp,\"两数字粘连\\n\"); if (fp!=stdout) fclose(fp); exit(7); } if (pp[0] == '0' && (pp[1] == 'x' || pp[1] == 'X')) { sscanf(pp + 2, \"%I64x%n\", &i64, &n); rexp = pp + 2 + n; ST[ist++] = (double)i64; } else ST[ist++] = strtod(pp, &rexp); ST[ist - 1] = fun(ST[ist - 1], op, &iop); if (pp == rexp) { fprintf(fp,\"非法字符\\n\"); if (fp!=stdout) fclose(fp); exit(8); } pp = rexp; last = DIGIT; if (*pp == '(' || isalpha(*pp)) { op[iop++] = '*'; last = !DIGIT; c = op[--iop]; goto operate; } } } *addr = pp; if (iop >= ist) { fprintf(fp,\"表达式有误\\n\"); if (fp!=stdout) fclose(fp); exit(9); } while (iop) { --ist; switch (op[--iop]) { case '+': ST[ist - 1] += ST[ist]; break; case '-': ST[ist - 1] -= ST[ist]; break; case '*': ST[ist - 1] *= ST[ist]; break; case '/': ST[ist - 1] /= ST[ist]; break; case '%': ST[ist - 1] = fmod(ST[ist - 1], ST[ist]); break; case '^': ST[ist - 1] = pow(ST[ist - 1], ST[ist]); break; } } return ST[0]; } void pretreatment() {//预处理 int i,L,n,r; char *p,*q; double v; char D[2]; //x2star将两边不是字母且左边不是非数字或串开头紧跟0的x替换为*,目的是支持用x代替*,且和0x开头的16进制数不冲突 L=strlen(s); for (i=1;i<L;i++) { if (s[i]=='x' && (!isalpha(s[i-1]) || isKMG(s[i-1])) && !isalpha(s[i+1])) { if (!( (i==1 && s[0]=='0') || (i>1 && s[i-1]=='0' && !isdigit(s[i-2])) )) s[i]='*'; } } //将\"数字\\.数字[kmgKMG]\"用()括住 strcpy(t,s); p=&t[0]; q=s; while (1) { r=sscanf(p,\"%lf%n%1[GKMgkm]\",&v,&n,D); if (r==EOF) break;// if (r==0) { L=sprintf(q,\"%c\",*p); q+=L; p++; } else if (r==1) { L=sprintf(q,\"%.*s\",n,p); q+=L; p+=n; } else if (r==2) { L=sprintf(q,\"(%.*s%c)\",n,p,D[0]); q+=L; p+=n+1; } } } int main(int argc, char **argv) { int a; setlocale( LC_ALL,\"chs\"); fp=stdout; if (argc<2) { //if (GetConsoleOutputCP() != 936) system(\"chcp 936>NUL\");//中文代码页 printf(\"计算函数表达式的值。\\n支持(),+,-,*,x,/,%%,^,Pi,sin,cos,tan,sqrt,arcsin,arccos,arctan,lg,ln,exp,ceil,floor,round,sign,abs,k,m,g,K,M,G\\n\"); while (1) { printf(\"请输入表达式：\"); fgets(s,MAXLEN,stdin); if ('\\n' == s[strlen(s)-1]) s[strlen(s) - 1] = 0; if (s[0] == 0) break;// pretreatment();printf(\"%s=%.15lg\\n\",s,calc(s, &endss)); } return 0; } if (argc == 2 && 0 == strcmp(argv[1], \"/?\")) { //if (GetConsoleOutputCP() != 936) system(\"chcp 936>NUL\");//中文代码页 printf( \"计算由≥1个命令行参数给出的函数表达式的值。\\n\" \"最后一个参数是.0～.15表示将计算结果小数点后保留0～15位\\n\" \"最后一个参数是g1～g15表示将计算结果保留有效数字1～15位\\n\" \"最后一个参数是e0～e15表示将计算结果用科学计数法表示，且小数点后保留0～15位\\n\" \"最后一个参数是x表示将计算结果以16进制正整数格式输出\\n\" \"支持(),+,-,*,x,/,%%,^^,Pi,sin,cos,tan,sqrt,arcsin,arccos,arctan,lg,ln,exp,ceil,floor,round,sign,abs,k,m,g,K,M,G\\n\" \"16进制整数以0x或0X开头\\n\" \"忽略表达式中的半角逗号(通常作为千分位分割符)\\n\" \"如果第一个参数是/f且c:\\\\jsresult.txt可写，就将所有输出重定向到该文件\\n\" ); return 0; } if (argc>2 && 0 == strcmp(argv[1], \"/f\")) { fp=fopen(\"c:\\\\jsresult.txt\",\"w\"); if (NULL==fp) fp=stdout; for (a=2;a<argc;a++) argv[a-1]=argv[a]; argc--; } strncpy(s, argv[1], MAXLEN - 1); s[MAXLEN - 1] = 0; if (argc>2) { for (a = 2; a<argc - 1; a++) strncat(s, argv[a], MAXLEN - 1);//将空格间隔的各参数连接到s if (1 == sscanf(argv[a], \".%d\", &pcs) && 0 <= pcs && pcs <= 15) {//最后一个参数是.0～.15表示将计算结果保留小数0～15位 pretreatment();fprintf(fp,\"%.*lf\\n\", pcs, calc(s, &endss)); } else if (1 == sscanf(argv[a], \"g%d\", &pcs) && 1 <= pcs && pcs <= 15) {//最后一个参数是g1～g15表示将计算结果保留有效数字1～15位 pretreatment();fprintf(fp,\"%.*lg\\n\", pcs, calc(s, &endss)); } else if (1 == sscanf(argv[a], \"e%d\", &pcs) && 0 <= pcs && pcs <= 15) {//最后一个参数是e0～e15表示将计算结果用科学计数法表示，且小数点后保留0～15位 pretreatment();fprintf(fp,\"%.*le\\n\", pcs, calc(s, &endss)); } else if (argv[a][0] == 'x' || argv[a][0] == 'X') {//最后一个参数是x表示将计算结果以16进制正整数格式输出 pretreatment();fprintf(fp,\"0x%016I64x\\n\", (__int64)calc(s, &endss)); } else { strncat(s, argv[a], MAXLEN - 1); pretreatment();fprintf(fp,\"%.15lg\\n\", calc(s, &endss)); } } else { pretreatment();fprintf(fp,\"%.15lg\\n\", calc(s, &endss)); } if (fp!=stdout) fclose(fp); return 0; }", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "#include <stdio.h> int main() { int a[] = {12,20,6,4,52,6,77,83,91} ; int arrsize = 8; int left=0,right=arrsize,temp; while(left<right) { if(a[left]%2==0) { while(left<right) { if(a[right]%2==1) { temp = a[left]; a[left]=a[right]; a[right]= temp; right--; break; } right--; } } left++; } for (temp=0;temp<=arrsize;temp++) printf(\"%d \", a[temp]); return 0; }", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "#include <stdio.h>void move(int arr[], int sz){ int left = 0; int right = sz - 1; while (left < right) { while ((left < right) && (arr[left] % 2 == 1)) { left++; } while ((left < right) && (arr[right] % 2 == 0)) { right--; } if (left <right) { int temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; } }}void print(int arr[],int sz){ for (int i = 0; i < sz; i++) { printf(\"%d \", arr[i]); }}int main(){ int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; int sz = sizeof(arr) / sizeof(arr[0]); move(arr, sz); print(arr, sz); return 0;}可以参考一下", "referer": "user-0" }, { "cid": "3", "user": "user-1", "content": "如果不用排序的话：从数组第一个数开始判断，如果是偶数则接着判断数组最后一个数是奇数还是偶数，最后一个是奇数则把其和第一个数交换；如果是偶数则不动，继续判断倒数第二个数，同样处理。如果第一个数是奇数或者第一个数已完成处理，则继续判断第二个数，同上处理。以此类推直到把数组遍历一遍后，数组就符合要求了，这时再把数组输出就行了。", "referer": "user-0" }, { "cid": "4", "user": "user-3", "content": "遍历2次A数组，第一次输出奇数，第二次输出偶数。", "referer": "user-0" }, { "cid": "5", "user": "user-4", "content": "另声明一个数组B，遍历一遍A,将奇数依次放入B,再遍历一遍A,将偶数继续放入B,将B所有元素依次放入A覆盖旧A", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "a[i]=true;", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "谢谢", "referer": "user-1" }, { "cid": "3", "user": "user-2", "content": "脑子临时卡住了", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "scanf中各%X之间应该啥也没有，输入时用空格或Tab键或回车换行间隔各项数据", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "你得输入用空格分隔的两个整数吧，然后才输出它们的和；交互友好计，可以输入前来个提示；输入整数无须用 scanf_s()", "referer": "user-0" }, { "cid": "3", "user": "user-0", "content": "刚上大学的萌新", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "建立一个含有该源文件的项目。", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "怎么建立呀？", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "一行就可以了：for (int c = getchar(); printf(\"%c%c%c\\n\", c - 1, c, c + 1);) break;", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "题目没有让你输入输出在一行，你想多了。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "不要啃故弄玄虚的视频，谢谢。", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "看下面的图，那个_Format: 应该只是提示当前输入域的参数说明，不是要你也输入这么个内容的", "referer": "user-0" }, { "cid": "3", "user": "user-0", "content": "好的，十分感谢！！", "referer": "user-2" } ]

[ { "cid": "1", "user": "user-1", "content": "按照字符串接收输入内容，直接通过查找定位小数点在字符串中的位置，剩下的就是字符串索引的事儿了", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "嗯嗯，谢谢", "referer": "user-1" }, { "cid": "3", "user": "user-2", "content": "不可以的，可以通过读取转换函数转换为内部浮点数值 12.36 然后再行后续处理；如果仅仅是获取整数部分个各小数位，可以通过查询小数点来对字符串进行切分。", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "好的，感谢老哥", "referer": "user-2" } ]

[ { "cid": "1", "user": "user-1", "content": "思路，一个是试差迭代，一个是求反函数；你这里好像求反函数即 di 的公式简单，A=f1(...)*di^2，V=f2(...)di^(2/3) AV=di^(8/3)*f3(...) 这样 di 的公式就出来了。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-0", "content": "Visual Studio 2017 这个软件", "referer": "user-0" }, { "cid": "2", "user": "user-1", "content": "什么编译器？应该不是scanf_s导致的，其他代码的问题", "referer": "user-0" }, { "cid": "3", "user": "user-2", "content": "加了_s以后的参数与原来不同了，不要直接把它当作scanf用。对于%s和%c两个格式符，后面需要有两个参数与之对应，分别是地址和长度。文档还是要读的。", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "地址和长度，这个我没有设定long这个变量啊所以还需要设定一个吗 还是", "referer": "user-2" }, { "cid": "5", "user": "user-2", "content": "形如 scanf_s(\"%s\",buf,128);", "referer": "user-0" }, { "cid": "6", "user": "user-1", "content": "scanf_s(\"%s\", &buf); 这里加&画蛇添足了，直接用buf就好，但并不是错误。", "referer": "user-0" }, { "cid": "7", "user": "user-0", "content": "这个&地址我已经去掉了，但是呢还是会冲突 特别是%s 其他还好", "referer": "user-1" }, { "cid": "8", "user": "user-3", "content": "你把地址运算符去掉试试", "referer": "user-0" }, { "cid": "9", "user": "user-0", "content": "这个我知道但是呢 一样还是有冲突", "referer": "user-3" }, { "cid": "10", "user": "user-4", "content": "感谢", "referer": "user-3" } ]

[ { "cid": "1", "user": "user-0", "content": "在下已经把全部代码都放在网了，还请各位大神帮忙解决 在下在此谢过了", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "并非在下惜墨如金是以此%s为误啊而以此%c%d皆可罢了就以各位所说予以改善", "referer": "user-0" }, { "cid": "3", "user": "user-1", "content": "楼主惜墨如金吾等徒呼奈何", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "_tmp_Data_DateFormat.tmpNowtime有问题，不知道这字符串怎么来的。用ue打开文本文件，可以看到写入的不可见字符是什么。", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "显然在某些情况下，_tmp_Data_DateFormat.tmpNowtime、strdata的内容就是错的，你得先检查这两个数据的写操作，而且这两个数据就是字符串，你还sprintf干嘛呢～", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "进程意外退出或电脑意外掉电都有可能导致文件中多一串看不见的@", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "遇到这样的问题 只能人为去修复这样的数据，让他标准化，是这样吗", "referer": "user-3" }, { "cid": "5", "user": "user-3", "content": "大抵如此", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "scanner是接收输入的所以要输数字然后以一个非数字结束 类似这样", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "DosBox了解一下", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "不存在 BorlandC++3.2 这个产品～", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "bc 到 vc ，改动不小吧；纯 dos ，简单的就是直接上 dosbox 了，什么都不用改", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "bc当年为win3.2提供了支持,所以改动真不大,我的目的是编译运行源代码,和把EXE放到虚拟机里面跑是两个概念.这玩得是编程.不过说到这里又让我感叹了,这年头哪里还有我这样的业余编程爱好者,从CSDN就看的出来,一说到编程就像一群饿红了眼的狼,满脑子就是钱钱钱.代码就是个搞钱的工具,谁还会去研究把玩这些.", "referer": "user-3" } ]

[ { "cid": "1", "user": "user-1", "content": "参考一下BIND ，它大概率就是这么干活的。", "referer": "user-0" }, { "cid": "2", "user": "user-1", "content": "药科学生有个梗，叫做抛开剂量谈毒性都是耍流氓。用在这里就是说，抛开工作量不能探讨是否处理得过来。工作量在一定规模以下，那就处理得过来。BTW: 性能分析与优化这种事相当复杂，也确实值钱。因为它需要具体情况具体分析。", "referer": "user-0" }, { "cid": "3", "user": "user-0", "content": "150个左右客户端不知怎样？", "referer": "user-1" }, { "cid": "4", "user": "user-1", "content": "一条铁路线上两端每天对开两对列车够不够？", "referer": "user-0" }, { "cid": "5", "user": "user-0", "content": "现在正在考虑做这个事情，之前没做过，所以不知从何下手。", "referer": "user-1" }, { "cid": "6", "user": "user-2", "content": "多线程，每个线程处理若干个客户端的数据，如果仍然处理不过来就要用集群了", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-0", "content": "求解", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "活太大，建议去悬赏区或者去网上找个demo改下", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "scanf输入要考虑内存问题", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "仅供参考： #include <stdio.h> char s[]=\"123 ab 4\"; char *p; int v,n,k; void main() { p=s; while (1) { k=sscanf(p,\"%d%n\",&v,&n); printf(\"k,v,n=%d,%d,%d\\n\",k,v,n); if (1==k) { p+=n; } else if (0==k) { printf(\"skip char[%c]\\n\",p[0]); p++; } else {//EOF==k break; } } printf(\"End.\\n\"); } //k,v,n=1,123,3 //k,v,n=0,123,3 //skip char[ ] //k,v,n=0,123,3 //skip char[a] //k,v,n=0,123,3 //skip char[b] //k,v,n=1,4,2 //k,v,n=-1,4,2 //End.", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "%c一次读一个字符%s一次读一串字符，引起内存溢出", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "加油", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "如果想看s值，可以printf打印下s。 如果觉得回答的不错，欢迎关注下我", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "结果，你要什么结果？你让程序怎么结果了？程序如何知道你要什么结果？", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "强制转换+位运算大多单片机不都是这么干的么？", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "因为 a++ 的又被 --a 给抵消了；自运算，前置和后置的区别", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "RA_开头的变量是结构体指针，后面的值却是宏（实际是数值），类型不匹配。", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "那请问一个结构体指针 要把它等于一个宏定义的地址 要怎么写呢 我网上找了很久都没找到！！！ 救命", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "强制类型转换", "referer": "user-0" }, { "cid": "4", "user": "user-1", "content": "亲，关注下我还行？", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "不要使用 while (条件) 更不要使用 while (组合条件) 要使用 while (1) { if (条件1) break; //... if (条件2) continue; //... if (条件3) return; //... } 因为前两种写法在语言表达意思的层面上有二义性，只有第三种才忠实反映了程序流的实际情况。 典型如： 下面两段的语义都是当文件未结束时读字符 while (!feof(f)) { a=fgetc(f); //... b=fgetc(f);//可能此时已经feof了！ //... } 而这样写就没有问题： while (1) { a=fgetc(f); if (feof(f)) break; //... b=fgetc(f); if (feof(f)) break; //... } 类似的例子还可以举很多。", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "谢谢，明白了！", "referer": "user-1" }, { "cid": "3", "user": "user-2", "content": "while(n == 0)， 你想要判断的是n != 0的时候继续计算", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "谢谢！", "referer": "user-2" } ]

[ { "cid": "1", "user": "user-1", "content": "10 的 n 次方，不是 ^ 操作符吧，这个是逻辑异或，可能要用 pow() 函数？其它问题，要么调试，要么关键步上放个 printf() 函数显示内容。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-0", "content": "配置是可以实现功能，我在开发一个网络安全的系统，需要通过代码修改目标ip地址", "referer": "user-0" }, { "cid": "2", "user": "user-1", "content": "不能理解您的意图，打算做什么？您提到了netfilter我觉得就应该是配置问题而不是编程问题。可是您又提到编程代码，这就让我糊涂了。不过您提到的那个函数的返回值是32位的类型，所以显然没有能力处理128位的类型。", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "小文件没有分配到磁盘的数据空间, 而是直接储存在mft, 所以没有这个东西", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "直接储存在mft, 要如何找出小文件的资料区位置?", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "NTFS每个文件都在MFT存储属性, 比如文件名, 时间等, 小文件直接在这个里面有个数据属性, 数据就在这里 数据不是独立的, 是属性的一部分 ntfs系统比较复杂, 如果要读取这个文件的数据, 过程: 1. 从MFT的root根据目标文件路径, 找到文件对应的mft的索引 2. 读取并解释该索引的属性, 其中就有文件的数据", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "这方面是初学者, 请问有没有C CODE范例能参考?", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "运行中做不到，反思一下，就说明这样写是不对的。%c格式符是荤素不计的，任意内容都能满足它----跟是否字母没啥关系。", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "scanf()的返回值是实际输入的参数个数，所以可以用x[a]的值是否是字母来判断。", "referer": "user-0" }, { "cid": "3", "user": "user-3", "content": "不要用带回显的输入函数，用getch()，这个虽然不是标准c库函数，但是几乎所有c库实现都支持", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "排序上有问题吧，简单的冒泡，得两重循环的，一个循环无法实现。", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "哦哦，是吼，谢谢大佬指点", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "“确认删除”后面的那个 else ？如果是的话，可能是你的 scanf() 有问题，格式字符串里 %c 前面竟然有个空格？", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "Add的返回类型是CBall，你return r;肯定不行。可以： CBall& Add(CBall& t) // 注意返回的是引用，不是临时对象 { r = ... return *(CBall *)this; }", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "谢谢啦", "referer": "user-1" }, { "cid": "3", "user": "user-0", "content": "报错是在鼠标光标那行", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "if (*p > n) 这个改成 if(*p > num) 因为要和输入num比大小 *q = n; 这里改成 *q = num，把num插入数组 printf(\"%2d\", *p); 这里改成 printf(\"%2d \", *p);，在2d后面加一个空格，显示的更清楚", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "谢谢谢写大佬，我检查的时候没看见，真的是太感动了，真的十分感谢老哥", "referer": "user-1" }, { "cid": "3", "user": "user-0", "content": "再次感谢老哥，真的是非常感动", "referer": "user-1" }, { "cid": "4", "user": "user-1", "content": "不客气，互相学习", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "谢谢", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "#include<stdio.h>#include<math.h>int Func(int n){ int i,leap=1; for(i=2;i<=sqrt(n);i++) { if(n%i==0) leap=0; } return leap;}int main(){ int n,i,sum=0,a[10],j=0; printf(\"请输入n的值：\"); scanf(\"%d\",&n); for(i=2;i<=n;i++) { if(Func(i)) //判断从2开始到n(包括n)的数是否是质数 { a[j]=i; //是质数的存入数组 sum+=i; j++; } } for(i=0;i<j-1;i++) printf(\"%d+\",a[i]); //最后一个先不打，因为最后一个数后面没有+号 printf(\"%d=%d\",a[i],sum); //打印最后一个数和它们的和。}", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "谢谢", "referer": "user-1" }, { "cid": "5", "user": "user-2", "content": "百度可以搜到的", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "getcher改为getchar", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "cout << \"旧姓名：\" << abs->personArray[ret].m_Name << endl; cout << \"请输入新姓名(输入_保持旧姓名不变)：\" << endl; string name; cin >> name; if (name!=\"_\") abs->personArray[ret].m_Name = name;", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "感谢！！！我还想问一下为什么我修改另一个邮政编码的时候！=报错了", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "单看这段代码似乎没啥问题", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "就是不知道怎么单独修改姓名或者编码这些，就不用再重复输入别的信息了", "referer": "user-1" }, { "cid": "5", "user": "user-1", "content": "修改前，显示旧信息，修改时，提示输入一个下划线\"_\"表示保持原来内容不修改，输入后，判断如果刚输入的是\"_\"就不赋值给原来内容。", "referer": "user-0" }, { "cid": "6", "user": "user-0", "content": "这个应该怎么写呀", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "/添加联系人 void addPerson(Addressbooks * abs) { //判断联系人是否满了 if (abs->m_Size == MAX) { cout << \"通讯录已满，无法添加\" << endl; return; } else { string name; cout << \"请输入姓名：\" << endl; cin >> name; int ret = isExist(abs, name); if (ret != -1) { cout << \"通讯录中已经有姓名为 \"<<name<<\" 的人，无法重复添加\" << endl; return; } abs->personArray[abs->m_Size].m_Name = name; cout << \"请输入邮政编码：\" << endl; int postcode; cin >> postcode; abs->personArray[abs->m_Size].m_Code = postcode; cout << \"请输入联系电话：\" << endl; string phone = \"\"; cin >> phone; abs->personArray[abs->m_Size].m_Phone = phone; cout << \"请输入家庭住址：\" << endl; string address; cin >> address; abs->personArray[abs->m_Size].m_Addr = address; cout << \"请输入E_mail：\" << endl; string e_mail; cin >> e_mail; abs->personArray[abs->m_Size].m_Mail = e_mail; //更新通讯录人数 abs->m_Size++; cout << \"添加成功：\" << endl; system(\"pause\"); system(\"cls\"); } }", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "谢谢你！！！！太厉害了老师", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "Who am I！", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-0", "content": "ue看出来是这样的看不到前面那个源代码不知道哪里改", "referer": "user-0" }, { "cid": "2", "user": "user-1", "content": "这里应该把是上面的地方吧，直接用 211.149... 搜索，若搜不到试试 unicode 模式，还搜不到，就是压缩或加密了；修改 exe ，可能有问题，你这里，看顶楼都有 align 4 语句，可能正好还有 1个字节的空间，但实际 exe 里就要看了，若无空间，就麻烦了。", "referer": "user-0" }, { "cid": "3", "user": "user-2", "content": "用UE改", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "我不懂啊就是ue打开看都看不懂个", "referer": "user-2" }, { "cid": "5", "user": "user-1", "content": "这是源程序，不就是直接改下面一行吗，如 db '192.168.200.100', 0 ；标签不能改的，改了其它地方引用这里的就会找不到了。", "referer": "user-0" }, { "cid": "6", "user": "user-0", "content": "我是用pe反编辑看的但是字串哪里改不了，标签改的了，我要把211 ip改成192 ip的，能私信我帮一下不", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "对read这个函数不熟，我一般喜欢用fread。还有你里边的lseek，也属于已经被废弃的函数名。不如换成fseek，，， open,read,close,lseek，好像就这4个函数吧，建议使用stdio.h标准库里边的函数将你的程序重写一下： open==fopen read==fread close==fclose lseek==fseek 注意有些函数的参数顺序会存在差异，你研究一下应该不难替代。完事再试试看还有没有问题。头文件也不用包含辣么多，windows.h一个足矣。", "referer": "user-0" }, { "cid": "2", "user": "user-2", "content": "fd = open(_pcFullFileName, O_RDWR｜O_BINARY, 0x604);不要把fopen(\"...\",\"...\");fscanf,fprintf,fgets,fgetc,fputc,fclose //读时把\\r\\n替换成\\n，写时把\\n替换成\\r\\n;读到\\x1a就设置EOF;读写的内容当字符看待和fopen(\"...\",\"...b\");fseek,ftell,fread,fwrite,fscanf,fprintf,fgets,fgetc,fputc,fclose //不作以上替换，遇到\\x1a仍继续读;读写的内容当字节看待弄混了", "referer": "user-0" }, { "cid": "3", "user": "user-0", "content": "确实 那个位置的字符是 0x1A...", "referer": "user-2" }, { "cid": "4", "user": "user-2", "content": "Who am I !", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "回复有分吗", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "什么分，哥", "referer": "user-1" } ]

[ { "cid": "1", "user": "user-1", "content": "红字说得很清楚了，有啥不好理解的么？", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "第一个for循环中你写的是p='\\0'，应该改成p=='\\0'，写代码要细心呀", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "OKOK😅", "referer": "user-1" }, { "cid": "3", "user": "user-2", "content": "敲代码时丢掉了一个感叹号。", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "OK😅", "referer": "user-2" } ]

[ { "cid": "1", "user": "user-1", "content": "https://docs.microsoft.com/zh-tw/windows/win32/api/cryptuiapi/nf-cryptuiapi-cryptuiwizimport", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "参数看起来没有问题，怎么会卡住呢，而且是个别电脑上", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "UAC、防火墙、杀毒软件、以管理员身份运行、未打最新补丁、未导入最新证书、……", "referer": "user-0" } ]

[ { "cid": "1", "user": "user-1", "content": "这个函数要是没有return的话，你还写他干嘛？", "referer": "user-0" }, { "cid": "2", "user": "user-0", "content": "不好意思，我没有说清楚问题，现在已将问题补充完整了", "referer": "user-1" }, { "cid": "3", "user": "user-1", "content": "哦，是这样啊。 那么不加return理论上是不对的。 如果运行结果是对的，那都是偶然的恰巧的，毕竟彩票中奖这种事也发生过。", "referer": "user-0" }, { "cid": "4", "user": "user-0", "content": "", "referer": "user-1" } ]