BZOJ 2301: [HAOI2011]Problem b


题目


2301: [HAOI2011]Problem b

Time Limit: 50 Sec  Memory Limit: 256 MB

Description

对于给出的n个询问,每次求有多少个数对(x,y),满足a≤x≤b,c≤y≤d,且gcd(x,y) = k,gcd(x,y)函数为x和y的最大公约数。



Input

第一行一个整数n,接下来n行每行五个整数,分别表示a、b、c、d、k

 

Output

共n行,每行一个整数表示满足要求的数对(x,y)的个数

 

Sample Input

2

2 5 1 5 1

1 5 1 5 2



Sample Output


14

3



HINT



100%的数据满足:1≤n≤50000,1≤a≤b≤50000,1≤c≤d≤50000,1≤k≤50000


题解


这道题目,我自己推得公式是sigma{miu[p]*(b/kp-(a-1)/kp)*(d/kp-(c-1)/kp)},这样是O(nd)的复杂度。还是TLE了,我不明觉厉,果然还是太弱!然后看网上的题解真是ORZ啊!


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 10000000000LL

inline 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;}

const int Maxn=50000;
int 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(m,n);
	LL last;
	LL ret=0;
	for (LL i=1;i<=n;i=last+1){
		last=remin(n/(n/i),(m/(m/i)));
		ret+=((LL)miu[last]-(LL)miu[i-1])*(m/i)*(n/i);
	}
	return ret;
}

int T;
int a,b,c,d,k;
LL Ans=0;
int main(){
	getMiu();
	scanf("%d",&T);
	for (;T--;){
		scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
		a-=1;c-=1;
		a/=k;b/=k;c/=k;d/=k;
		Ans=getAns(b,d)+getAns(a,c)-getAns(a,d)-getAns(b,c);
		printf("%lld\n",Ans);
	}
	return 0;
}



posted @ 2014-10-28 19:31  WNJXYK  阅读(...)  评论(...编辑  收藏

WNJXYK-我今年一定要努力!