Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2789 Accepted Submission(s): 1074
Problem Description
Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N. Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.
Input
The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 1015) and (1 <=N <= 109).
Output
For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.
Sample Input
2
1 10 2
3 15 5
Sample Output
Case #1: 5 Case #2: 10
View Code
Hint
In the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.1.容斥初涉
2.找到交叉情况,即有条理的分析题目
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<cstdio> #include<string.h> __int64 a[1200],num; void gett(__int64 n) { __int64 i; num=0; for(i=2;i*i<=n;i++) { if(n%i==0) { a[num++]=i; while(n%i==0) n=n/i; } } if (n>1) a[num++]=n; } __int64 rc(__int64 m) { __int64 q[10000],i,j,k,t=0,sum=0; q[t++]=-1; for(i=0;i<num;i++) { k=t; for(j=0;j<k;j++) q[t++]=q[j]*a[i]*(-1); } for(i=1;i<t;i++) sum=sum+m/q[i]; return sum; } int main() { __int64 T,x,y,n,sum; while (scanf("%I64d",&T)!=EOF) { for (int j=1;j<=T;j++) { scanf("%I64d %I64d %I64d",&x,&y,&n); gett(n); sum=y-rc(y)-(x-1-rc(x-1)); printf("Case #%d: ",j); printf("%I64d ",sum); } } }