font hint cup show lan orm required stdio.h test case

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

HintFor 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



題意：輸入五個整數a,b,c,d,k。要求從區間[a,b]取出一個x，從區間[c,d]取出一個y，使得GCD(x,y) == k求出有多少種情況，只是註意的是GCD(5,7)與GCD(7,5)是一種。

思路：將x,y同一時候除以k。就轉變成求x,y互質，就能用容斥定理做了。

點擊打開鏈接

#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\n",++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\n",++kk,sum);
    }
    return 0;
}

HDU 1695 GCD（容斥定理）