杭电1695 GCD（容斥原理三题）

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GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6245 Accepted Submission(s): 2288

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

21 3 1 5 11 11014 1 14409 9

Sample Output

Case 1: 9Case 2: 736427Hint
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

/*校赛的时候都犯了一次低级错误，没用总个数减去他的倍数，没搞出来，，，今天又是这种低级错误。。。。。。几个小时、、、、我到底怎么回事呢？？？我就纳闷了。。。。是好运气将要来临的标志吗？？嘿嘿，还得加油啊！！Time:2014-12-24 16:46*///题中假设所有测试数据中a=c=1//相当于求1--b/k中与1--d/k中互质的个数，并且2---3与3---2算一种，用欧拉函数求和求出来，比b小的那一部分，然后再用容斥原理求b+1--d的那一部分就不用考虑重复了//f=x(1-1/p1)(1-1/p2)#include<cstdio>#include<cstring>#include<vector>#include<queue>using namespace std;typedef long long LL;const int MAX= 100000+10;vector<int>prime[MAX];bool vis[MAX];LL eular[MAX];void Prime(){    memset(eular,0,sizeof(eular));        eular[1]=1;    for(int i=0;i<MAX;i++)if(prime[i].size())prime[i].clear();    for(int i=2;i<MAX;i++){        if(eular[i]==0){            for(int j=i;j<MAX;j+=i){                if(eular[j]==0)eular[j]=j;                eular[j]=eular[j]*(i-1)/i;                prime[j].push_back(i);            }        }        eular[i]+=eular[i-1];    }}LL DFS(int pos,int b,int n){    //pos表示素数位置,b表示1--b的个数，与n互质    LL ret=0;    int k=prime[n].size();    for(int i=pos;i<k;i++){        int temp=b/prime[n][i];        //这儿求出与第一个质因子互质的个数        //再除以下一个时,就是与第一个和第二个都互质的个数        ret+=temp-DFS(i+1,temp,n);//容斥原理    }    return ret;}int main(){    int T;    Prime();    int a,b,c,d,k;    int nCase=1;    scanf("%d",&T);    while(T--){        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);        printf("Case %d: ",nCase++);        if(k==0){            printf("0\n");            continue;        }        b=b/k;d=d/k;        if(b>d){swap(b,d);}        LL ans=eular[b];        /*        for(int i=1;i<=10;i++){            printf("%d ",eular[i]);        }        printf("ans=%lld\n",ans);        */        for(int i=b+1;i<=d;i++){            ans+=b-DFS(0,b,i);        }        printf("%lld\n",ans);    }return 0;}

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