杭电1695 GCD（莫比乌斯反演）

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本文介绍了一种解决特定GCD问题的有效算法。该算法利用线性筛法预处理莫比乌斯函数，并通过巧妙的数据结构优化，实现了快速查找在给定区间内具有特定最大公约数的整数对数量。

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GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6247 Accepted Submission(s): 2289

Problem Description

Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number pairs.
Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.

Yoiu can assume that a = c = 1 in all test cases.

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.
Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.

Output

For each test case, print the number of choices. Use the format in the example.

Sample Input

  
   2
1 3 1 5 1
1 11014 1 14409 9

Sample Output

  
   Case 1: 9
Case 2: 736427


   
    
     Hint
    For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).

Source

2008 “Sunline Cup” National Invitational Contest

/*
31ms比容斥原理快多了，，，之前用容斥做过，两天没白搞，，不过代码还是猥琐的看的bin神的代码。。。。。加油！！！
Time:2014-12-28 0:26
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX=100000+10;
bool vis[MAX+10];
int mu[MAX+10],prime[MAX];
typedef long long LL;
void mobius(){//线性筛法求顺便mu
    mu[1]=1;int tot=0;
    memset(vis,0,sizeof(vis));
    for(int i=2;i<=MAX;i++){
        if(!vis[i]){
            prime[tot++]=i;
            mu[i]=-1;
        }
        for(int j=0;j<tot;j++){
            if(i*prime[j]>MAX) break;
                vis[i*prime[j]]=true;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                break;
            }else{
                mu[i*prime[j]]=-mu[i];
            }
        }
    }
}
int main(){
    int T;
    mobius();//打表
    int a,b,c,d,k;
    int nCase=0;
    scanf("%d",&T);
    while(T--){
        nCase++;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);//求1---b与1---d内最大公约数是k的个数,相当于求1---b/k与1---d/k之间互质的个数
        if(k==0){
            printf("Case %d: 0\n",nCase);
            continue;
        }
        b/=k;d/=k;
        if(b>d)swap(b,d);
        LL ans=0;
        for(int i=1;i<=b;i++)//1--b与1--d互质的个数
            ans+=(LL)mu[i]*(b/i)*(d/i);
        LL more=0;
        for(int i=1;i<=b;i++)//1--b与1--b互质的个数
            more+=(LL)mu[i]*(b/i)*(b/i);
        ans-=(more>>1);//题目要求5 7 和7 5是一组，刚好减去一半即可
        printf("Case %d: %lld\n",nCase,ans);
    }
return 0;
}