题目

1101: [POI2007]Zap

Time Limit: 10 Sec  Memory Limit: 162 MB

Description

Input

Output

Sample Input

Sample Output

HINT

题解

由数论推知->

且只会有种取值，所以我们暴力就好了！

代码

/*Author:WNJXYK*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
using namespace std;#define LL long long
#define Inf 2147483647
#define InfL 10000000000LLinline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;}
inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;}
inline int remin(int a,int b){if (a<b) return a;return b;}
inline int remax(int a,int b){if (a>b) return a;return b;}
inline LL remin(LL a,LL b){if (a<b) return a;return b;}
inline LL remax(LL a,LL b){if (a>b) return a;return b;}
inline void read(int &x){x=0;int 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();}x=x*f;}
inline void read(LL &x){x=0;LL 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();}x=x*f;}
inline void read(int &x,int &y){read(x);read(y);}
inline void read(LL &x,LL &y){read(x);read(y);}
inline void read(int &x,int &y,int &z){read(x,y);read(z);}
inline void read(int &x,int &y,int &n,int &m){read(x,y);read(n,m);}
inline void read(LL &x,LL &y,LL &z){read(x,y);read(z);}
inline void read(LL &x,LL &y,LL &n,LL &m){read(x,y);read(n,m);}const int Maxn=50000;
LL miu[Maxn+10];
inline void getMiu(){for (int i=1;i<=Maxn;i++){int target=i==1?1:0;int delta=target-miu[i];miu[i]=delta;for (int j=i+i;j<=Maxn;j+=i) miu[j]+=delta;}  for (int i=2;i<=Maxn;i++) miu[i]+=miu[i-1];
}inline LL getAns(int n,int m){if (n>m) swap(n,m);LL Ans=0;int pos;for (int i=1;i<=n;i=pos+1){pos=remin(n/(n/i),m/(m/i));Ans+=(miu[pos]-miu[i-1])*(n/i)*(m/i);} return Ans;
}int T;
LL a,b,d;
int main(){read(T);getMiu();for (;T--;){read(a,b,d);printf("%lld\n",getAns(a/d,b/d));}return 0;
}