CodeForces 597A Divisibility

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 597A

Description

Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.

Input

The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 10¹⁸; - 10¹⁸ ≤ a ≤ b ≤ 10¹⁸).

Output

Print the required number.

Sample Input

Input

1 1 10

Output

10

Input

2 -4 4

Output

5

题意：输出k,a,b,求出区间[a,b]中能别k整除的个数。

数据太大，暴力显然不行。

#include<cstdio>#include<algorithm>using namespace std;int main(){long long k,n,m;while(scanf("%lld %lld %lld",&k,&n,&m)!=EOF){long long ans,i,p1,p2;p1 = abs(m)/k;//求出区间[1,m]内能被k乘除的数的个数（原意自己体会，每k个数分成一组） p2 = abs(n)/k;//同上求[1,n]; if(n >= 0)//如果n，m都为正数 {ans = p1 - p2;if(n % k == 0)ans++;}else if(m <= 0)//如果n,m都为负数，与上种情况相反 {ans = p2 - p1;if(m % k == 0)ans++;}else//如果n为负，m为正 {ans = p1 + p2 + 1;//加上中间一个0 }printf("%lld\n",ans);}}