Hdu1695 GCD 2017-06-27 22:19 30人阅读 评论(0) 收藏

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 11177    Accepted Submission(s): 4236

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

 

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——————————————————————————————————

题目的意思是给出两个区间，问两个区间有多少对数GCD=k，(1,2)和(2,1)算一种，区间从1开始

思路：先把区间每个数除以k，然后说白了就是求[1,m/k],[1,n/k]有多少对数互质 可以枚举m/k,在n/k中找有多少互质的数，可以先把枚举的m/k质因数的分解，然后利用容斥定理计数，防止重复计数枚举时可以计算i与区间[i,n/k]的互质数；

防止超时分解的质因数可以先打个表

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
#define LL long long
const int inf=0x3f3f3f3f;

vector<LL>p[100010];
int pr[100010];
int a,b,x,y,k;
LL ans;

void init()
{
    memset(pr,0,sizeof pr);
    for(int i=1;i<=100005;i++) p[i].clear();
    for(int i=2;i<=100005;i+=2)
        p[i].push_back(2);
    for(int i=3;i<=100005;i+=2)
    {
        if(!pr[i])
        {
            for(int j=i;j<=100005;j+=i)
            {
                pr[j]=1;
                p[j].push_back(i);
            }

        }
    }
}

void dfs(int no,int pos,int xx,int cnt)
{
    if(pos>=p[no].size())
    {
        if(cnt==0)
            return;
        if(cnt%2)
        {
            ans+=y/xx;
            ans-=(no-1)/xx;
        }
        else
        {
            ans-=y/xx;
            ans+=(no-1)/xx;
        }
        return;
    }
    dfs(no,pos+1,xx*p[no][pos],cnt+1);
    dfs(no,pos+1,xx,cnt);
}

int main()
{
    init();
    int T;
    int q=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d%d%d",&a,&x,&b,&y,&k);
          if (k == 0) {
            printf("Case %d: 0\n", q++);
            continue;
        }
        x/=k,y/=k;
        if(x>y) swap(x,y);
        LL ans2=0;
        for(int i=1; i<=x; i++)
        {
            ans=0;
            dfs(i,0,1,0);
            ans2+=y-i+1-ans;
        }
        printf("Case %d: %lld\n",q++,ans2);
    }
    return 0;
}