【bzoj1101】[POI2007]Zap

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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。  

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <cstdio>
#include <cstring>
using namespace std;

typedef long long LL;

inline int read(){
    int x = 0,f = 1; char ch = getchar();
    while(ch < '0'||ch > '9'){if(ch == '-')f=-1;ch = getchar();}
    while(ch >= '0'&&ch <= '9'){x = x * 10 + ch -'0';ch = getchar();}
    return x*f;
}

//

/*
算法：Mobius + 分块
题目：
  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
  (1<=d<=a,b<=50000)

*/

const int MAXN = 50000 + 10;
int tot;
int mu[MAXN+1],sum[MAXN+1],pri[MAXN+1];
bool mark[MAXN];

void get(){
   mu[1] = 1;
   for(int i = 2;i <= MAXN;++i){
      if(!mark[i])pri[tot++] = i,mu[i] = -1;
      for(int j = 0;j < tot&&i*pri[j] <= MAXN;++j){
        mark[i*pri[j]] = 1;
        if(i % pri[j]==0){mu[i*pri[j]] = 0; break;}
        else mu[i*pri[j]] = -mu[i];
      }
   }

   for(int i = 1;i <= MAXN;++i) //预处理前缀
    sum[i] = sum[i-1] + mu[i];
}

int cal(int n,int m){
    if(n > m) swap(n,m);
    int ans = 0,pos;
    for(int i = 1;i <= n;i = pos + 1){
        pos = min(n/(n/i),m/(m/i));       //分块
        ans += (sum[pos] - sum[i-1]) * (n/i) * (m/i);    
    }
    return ans;
}

int main()
{
    get();
    int T = read();
    while(T--){
        int a = read(),b = read(),d = read();
        printf("%d\n",cal(a/d,b/d));
    }
    return 0;
}