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HDU4135(容斥原理)
Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4090 Accepted Submission(s): 1619
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.
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
思路：求1~m中与n互素的数的个数。
#include <cstdio>#include <vector>using namespace std;typedef long long LL;LL sieve(LL m,LL n){ vector<LL> divisor; for(LL i=2;i*i<=n;i++) { if(n%i==0) { divisor.push_back(i); while(n%i==0) n/=i; } } if(n>1) divisor.push_back(n); LL ans=0; for(LL mark=1;mark<(1<<divisor.size());mark++) { LL odd=0; LL mul=1; for(LL i=0;i<divisor.size();i++) { if(mark&(1<<i)) { mul*=divisor[i]; odd++; } } LL cnt=m/mul; if(odd&1) ans+=cnt; else ans-=cnt; } return m-ans;}LL a,b,n;int main(){ int T; scanf("%d",&T); for(int cas=1;cas<=T;cas++) { scanf("%lld%lld%lld",&a,&b,&n); LL res=sieve(b,n)-sieve(a-1,n); printf("Case #%d: ",cas); printf("%lld\n",res); } return 0;}
HDU4135(容斥原理)
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