Problem:
Score : 100 points

Problem Statement

Takahashi is distributing N balls to K persons.

If each person has to receive at least one ball, what is the maximum possible difference in the number of balls received between the person with the most balls and the person with the fewest balls?

Constraints

1 \leq K \leq N \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the maximum possible difference in the number of balls received.

Sample Input 1

3 2

Sample Output 1

1

The only way to distribute three balls to two persons so that each of them receives at least one ball is to give one ball to one person and give two balls to the other person.

Thus, the maximum possible difference in the number of balls received is 1.

Sample Input 2

3 1

Sample Output 2

0

We have no choice but to give three balls to the only person, in which case the difference in the number of balls received is 0.

Sample Input 3

8 5

Sample Output 3

3

For example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of balls received between the person with the most balls and the person with the fewest balls would be 3, which is the maximum result.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

void inp_str(int*);

int main(void) {

int input1[150];
inp_str(input1);

int a,b,c;
int N,K;

N=input1[0];
K=input1[1];

if(K==1){
	a=0;
} else {
a=N-K;
}





printf("%d\n",a);

    return 0;

}


void inp_str(int instr[]){
char *ptr;
int z,i,count;
char str[20000]={'\0'};
char *ln;

fgets(str, 20000, stdin); // 入力："123456\n"
ln = strchr(str, '\n'); /* 改行文字を検索 */
if (ln != NULL) { /* 改行が読み取られていたかどうか */
   *ln = '\0';    /* 改行文字を終端文字に置き換える */
}
else {            /* 入力ストリーム上に文字が残ってる場合 */
   while (1) {    /* 改行文字が読み取られるまで空読みする */
       z = getchar();
      if (z == '\n' || z == EOF) break;
   }
}
	count=0;
    ptr = strtok(str, " ");
    i = atoi(ptr);
    instr[count]=i;
    count=count+1;
    // 2回目以降
    while(ptr != NULL) {
        // strtok関数により変更されたNULLのポインタが先頭
        ptr = strtok(NULL, " ");
        // ptrがNULLの場合エラーが発生するので対処
        if(ptr != NULL) {
		    i = atoi(ptr);
    instr[count]=i;
    		count=count+1;
        }
    }
}