Problem:
Score : 200 points

Problem Statement

You are given nonnegative integers a and b (a ≤ b), and a positive integer x.
Among the integers between a and b, inclusive, how many are divisible by x?

Constraints

0 ≤ a ≤ b ≤ 10^{18}

1 ≤ x ≤ 10^{18}

Input

The input is given from Standard Input in the following format:

a b x

Output

Print the number of the integers between a and b, inclusive, that are divisible by x.

Sample Input 1

4 8 2

Sample Output 1

3

There are three integers between 4 and 8, inclusive, that are divisible by 2: 4, 6 and 8.

Sample Input 2

0 5 1

Sample Output 2

6

There are six integers between 0 and 5, inclusive, that are divisible by 1: 0, 1, 2, 3, 4 and 5.

Sample Input 3

9 9 2

Sample Output 3

0

There are no integer between 9 and 9, inclusive, that is divisible by 2.

Sample Input 4

1 1000000000000000000 3

Sample Output 4

333333333333333333

Watch out for integer overflows.

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
typedef long long LL;
#define rep(i,n) for(LL i=0; i<(LL)(n); i++)

//qsort
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

//比較
int max(int a, int b){
	if (a>b) return a;
	else return b;
}

int min(int a,int b) {
	if(a>b) return b;
	else return a;
}


//main関数
int main(){
	LL aa,bb,xx;
	scanf("%lld %lld %lld",&aa,&bb,&xx);
	
	LL subans = 0;
	if( aa == 0){
		if(bb == 0){
			printf("0");
			return 0;
		}
	}
	
	if(bb - aa == 0){
		if(bb % xx == 0){
			printf("1");
		}
		else {
			printf("0");
		}
		return 0;
	}
	if( aa % xx == 0 ){
		subans++;
	}
	/*if( (bb - aa) % xx == 0){
		subans--;
	}*/
	printf("%lld",(bb/xx) - (aa/xx) + subans);
}