Codeforces Divisibility【水题】

最新推荐文章于 2024-07-31 16:03:18 发布

转载最新推荐文章于 2024-07-31 16:03:18 发布 · 66 阅读

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原文链接：http://www.cnblogs.com/cniwoq/p/6770805.html

本文介绍了一个算法问题：求解在指定区间[a, b]内能被k整除的整数的数量。通过分析输入数据的正负情况，提供了一种有效的计算方法，并给出了完整的C++实现代码。

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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 thata ≤ 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.

Examples
input
1 1 10
output
10
input
2 -4 4
output
5

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
#define space " "
using namespace std;
typedef long long LL;
//typedef __int64 Int;
typedef pair<int,int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double Pi = acos(-1);
const int MOD = 1e9+5;
const int MAXN = 1e5;
long long k, a, b, ans;
int main() {
    while (scanf("%lld%lld%lld", &k, &a, &b) != EOF) {
        ans = 0; 
        if ((a > 0 && b > 0) || (a < 0 && b < 0)) {
            ans += abs(b/k); ans -= abs(a/k);
            if (ans < 0) ans = -ans;
            if (min(abs(a), abs(b))%k == 0) ans++;
        }
        else {
            ans += abs(b/k); ans += abs(a/k);
            ans++;
        }
        printf("%lld\n", ans);
    }
    return 0;
}