bzoj1101 [POI2007]Zap

Description

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　FGD正在破解一段密码，他需要回答很多类似的问题：对于给定的整数a,b和d，有多少正整数对x,y，满足x<=a
，y<=b，并且gcd(x,y)=d。作为FGD的同学，FGD希望得到你的帮助。

1<=n<= 50000
1<=d<=a,b<=50000

Solution

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sb题，现在我也写到去年66wei写下的题目了，感慨万千

$$ans=\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}[gcd(i,j)=1]=\sum_{x=1}^{\lfloor\frac{n}{d}\rfloor}\mu(x)\lfloor\frac{n}{dx}\rfloor\lfloor\frac{m}{dx}\rfloor$$

Code

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#include <stdio.h>
#include <algorithm>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)

const int N=500005;

int prime[N],mu[N];

bool not_prime[N];

void pre_work(int n) {
    mu[1]=1;
    for (int i=2;i<=n;i++) {
        if (!not_prime[i]) {
            prime[++prime[0]]=i;
            mu[i]=-1;
        }
        for (int j=1;i*prime[j]<=n&&j<=prime[0];j++) {
            not_prime[i*prime[j]]=1;
            if (i%prime[j]==0) {
                mu[i*prime[j]]=0;
                break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    for (int i=1;i<=n;i++) mu[i]+=mu[i-1];
}

void solve(int n,int m) {
    if (n>m) std:: swap(n,m);
    int ans=0;
    for (int i=1,j;i<=n;i=j+1) {
        j=std:: min(n/(n/i),m/(m/i));
        ans+=(mu[j]-mu[i-1])*(n/i)*(m/i);
    }
    printf("%d\n", ans);
}

int main(void) {
    pre_work(N-5);
    int T; scanf("%d",&T);
    while (T--) {
        int n,m,d; scanf("%d%d%d",&n,&m,&d);
        solve(n/d,m/d);
    }
    return 0;
}