[bzoj2820]YY的GCD 莫比乌斯反演

来源：互联网发布：算法导论适合初学者吗编辑：程序博客网时间：2024/05/24 02:36

2820: YY的GCD

Time Limit: 10 Sec Memory Limit: 512 MB
Submit: 2118 Solved: 1140
[Submit][Status][Discuss]

Description

神犇YY虐完数论后给傻×kAc出了一题给定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对kAc这种

傻×必然不会了，于是向你来请教……多组输入

Input

第一行一个整数T 表述数据组数接下来T行，每行两个正整数，表示N, M

Output

T行，每行一个整数表示第i组数据的结果

Sample Input

2
10 10
100 100

Sample Output

30
2791

HINT

T = 10000

N, M <= 10000000

Source

具体看黄学长的吧，我不会在这上面打式子，写在本子上的

所以这是水博客数量吗

#include<algorithm>#include<iostream>#include<cstdio>#include<cmath> typedef long long ll;const int N = 10000000 + 5;using namespace std;int T,n,m,cnt; ll ans,f[N];int mul[N],pri[N]; bool isnot[N];void init(){mul[1] = 1;for( int i = 2; i <= N; i++ ){if( !isnot[i] ) pri[++cnt]=i, mul[i]=-1;for( int j = 1; j <= cnt && pri[j]*i <= N; j++ ){isnot[i*pri[j]] = 1;if( i%pri[j] == 0 ){ mul[i*pri[j]]=0; break; }mul[i*pri[j]] = -mul[i];}}for( int i = 1; i <= cnt; i++ ){int p = pri[i];for( int j = 1; j*p <= N; j++ ) f[j*p] += mul[j];}for( int i = 1; i <= N; i++ ) f[i] += f[i-1];}int main(){init();scanf("%d", &T);while(T--){ans = 0;scanf("%d%d", &n, &m);if( n > m ) swap(n,m);for( int i = 1, j; i <= n; i = j+1 ){j = min(n/(n/i),m/(m/i));ans += (f[j]-f[i-1])*(n/i)*(m/i);}printf("%lld\n", ans);}return 0;}

阅读全文

1 0