HDU 1695 GCD（容斥定理）

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7529    Accepted Submission(s): 2773

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

    题意：输入五个整数a,b,c,d,k。要求从区间[a,b]取出一个x，从区间[c,d]取出一个y，使得GCD(x,y) == k求出有多少种情况，只是注意的是GCD(5,7)与GCD(7,5)是一种。

    思路：将x,y同一时候除以k。就转变成求x,y互质，就能用容斥定理做了。

点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<vector>
#include<queue>
#include<stack>
#include<map>

#define N 101000

using namespace std;

vector<int>q[N];
int num[N];
int a,b,c,d,k;

void init(){
    for(int i=0;i<=N;i++){
        q[i].clear();
    }
    for(int i=1;i<=100000;i++){
        int p = i;
        int pi = sqrt(p);
        for(int j=2;j<=pi;j++){
            if(p%j == 0){
                q[i].push_back(j);
                while(p%j == 0){
                    p = p/j;
                }
            }
        }
        if(p!=1){
            q[i].push_back(p);
        }
    }
}

__int64 IEP(int ii,int pn){
    int pt = 0;
    __int64 s = 0;
    num[pt++] = -1;
    for(int i=0;i<q[ii].size();i++){
        int l = pt;
        for(int j=0;j<l;j++){
            num[pt++] = num[j]*q[ii][i]*(-1);
        }
    }
    for(int i=1;i<pt;i++){
        s += pn/num[i];
    }
    return s;
}

int main(){
    int T;
    init();
    int kk = 0;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        if(b>d){
            int e = b;
            b = d;
            d = e;
        }
        if(k == 0){
            printf("Case %d: 0
",++kk);
            continue;
        }
        b = b/k;
        c = b+1;
        d = d/k;
        __int64 sum = 0;
        for(int i=1;i<=b;i++){
            sum += b - IEP(i,b);
        }
        sum = (sum+1)/2;
        for(int i=1;i<=b;i++){
            sum += d - c + 1 - IEP(i,d) + IEP(i,c-1);
        }
        printf("Case %d: %I64d
",++kk,sum);
    }
    return 0;
}