杭电1695GCD【欧拉函数】【容斥定理】

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10255    Accepted Submission(s): 3868

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: 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).
题意：x∈[a,b],y∈[c,d],x=i*k,y=j*k,则i,j互质，且i∈[1,b/k=m],j∈[1,d/k=n],求存在多少对的i,j,(i,j)与(j,i)相同,(假设i>j)存在多少对(i,j).假设m>n,则分为两个部分求
①i∈[1,n],j∈[1,n],i,j互质,用欧拉函数求
②i∈[n+1,m],j∈[1,n],枚举i,求[1,n]中多少个数与i互质，用容斥定理
#include<stdio.h>#include<string.h>#define LL long longLL n,m,num;LL p[15],o[100010];void eulur(){LL i,j;memset(o,0,sizeof(o));o[1]=1;for(i=2;i<100010;i++){    if(!o[i])    {    for(j=i;j<100010;j+=i)    {    if(!o[j])    o[j]=j;    o[j]=o[j]*(i-1)/i;}}o[i]+=o[i-1];}}void get(LL k){for(LL i=2;i*i<=k;i++){if(k%i==0)p[num++]=i;while(k%i==0)k/=i;}if(k>1)p[num++]=k;}LL solve(LL k){LL i,j,ans=0;for(i=1;i<(1LL<<num);i++){    LL s=0;    LL l=1;for(j=0;j<num;j++){if((i>>j)&1){s++;l*=p[j];}}if(s&1)ans+=k/l;elseans-=k/l;}return ans;}int main(){LL a,b,c,d,k,t,l,sum;scanf("%lld",&t);eulur();for(l=1;l<=t;l++){scanf("%lld%lld%lld%lld%lld",&a,&b,&c,&d,&k);if(k==0){printf("Case %d: 0\n",l);continue;}m=b/k>d/k?b/k:d/k;n=b/k<d/k?b/k:d/k;sum=o[n];for(LL i=n+1;i<=m;i++){num=0;get(i);sum+=n-solve(n);}printf("Case %lld: %lld\n",l,sum);}return 0;}