HDU-1695-GCD

ACM模版

描述

题解

莫比乌斯反演入门题，给大家推荐一个写的十分详细的博客，由浅入深，大赞！__proto__’s blog，让我更加清晰的认识了莫比乌斯反演的用处，感谢大佬！

代码

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int MAXN = 1e5 + 7;int a, b, c, d, k, cnt;int pri[MAXN];int miu[MAXN];bool vis[MAXN];void init(){    memset(pri, 0, sizeof(pri));    memset(miu, 0, sizeof(miu));    memset(vis, 0, sizeof(vis));    miu[1] = 1;    cnt = 0;    for (int i = 2; i < MAXN; i++)    {        if (!vis[i])        {            pri[cnt++] = i;            miu[i] = -1;        }        for (int j = 0; j < cnt && i * pri[j] < MAXN; j++)        {            vis[i * pri[j]] = 1;            if (i % pri[j])            {                miu[i * pri[j]] = -miu[i];            }            else            {                miu[i * pri[j]] = 0;                break;            }        }    }}int main(){    init();    int T;    cin >> T;    for (int i = 1; i <= T; i++)    {        printf("Case %d: ", i);        scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);        if (k == 0)        {            puts("0");            continue;        }        b /= k;        d /= k;        int t = min(b, d);        long long ans1 = 0, ans2 = 0;        for (int i = 1; i <= t; i++)        {            ans1 += (long long)miu[i] * (b / i) * (d / i);        }        for (int i = 1; i <= t; i++)        {            ans2 += (long long)miu[i] * (t / i) * (t / i);        }        printf("%lld\n", ans1 - (ans2 >> 1));    }    return 0;}