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.
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.
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).
求（1，b）区间和（1，d）区间里面gcd(x, y) = k的数的对数（1<=x<=b , 1<= y <= d）。
b和d分别除以k之后的区间里面，只需要求gcd(x, y) = 1就可以了，这样子求出的数的对数不变。
这道题目还要求1-3 和 3-1 这种情况算成一种，因此只需要限制x<y就可以了
只需要枚举x，然后确定另一个区间里面有多少个y就可以了。因此问题转化成为区间（1， d）里面与x互素的数的个数
先求出x的所有质因数，因此（1，d）区间里面是x的质因数倍数的数都不会与x互素，因此，只需要求出这些数的个数，减掉就可以了。
如果w是x的素因子，则（1，d）中是w倍数的数共有d/w个。
容斥原理：
所有不与x互素的数的个数= 1个因子倍数的个数 - 2个因子乘积的倍数的个数 + 3个……-……
答案很大，用long long。
所有数的素因子，预先处理保存一下，不然会超时的。
#include<iostream> using namespace std; const int Max=100005; __int64 elur[Max];//存放每个数的欧拉函数值 int num[Max];//存放数的素因子个数 int p[Max][20];//存放数的素因子 void init()//筛选法得到数的素因子及每个数的欧拉函数值 { elur[1]=1; for(int i=2;i<Max;i++) { if(!elur[i]) { for(int j=i;j<Max;j+=i) { if(!elur[j]) elur[j]=j; elur[j]=elur[j]*(i-1)/i; p[j][num[j]++]=i; } } elur[i]+=elur[i-1]; //进行累加（法里数列长度） } } int dfs(int idx,int b,int now)//求不大于b的数中,与now不互质的数的个数; { //dfs()写的容斥原理 int ans=0; for(int i=idx;i<num[now];i++)//容斥原理来求A1并A2并A3.....并Ak的元素的数的个数. ans += b/p[now][i]-dfs(i+1,b/p[now][i],now); return ans; } int main() { int t,a,b,c,d,k; init(); scanf("%d",&t); for(int ca=1;ca<=t;ca++) { scanf("%d%d%d%d%d",&a,&b,&c,&d,&k); printf("Case %d: ",ca); if(k==0) { printf("0 "); continue; } if(b>d) swap(b,d); b/=k; d/=k; __int64 ans=elur[b]; for(int i=b+1;i<=d;i++) ans+=b-dfs(0,b,i);//求不大于b的数中,与i不互质的数的个数 printf("%I64d ",ans); } return 0; }