标签：ios class space 选中 you 没有 ati case tween
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.
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这道题分几步：
1、先把n用算术基本原理拆分成不同的质数，如果a~b中的数是这些质数的倍数那其肯定不与n互质，a~b的总个数减去不与n互质的数剩下来就是与n互质的数啦。
2、先计算1~a-1中与n互质的个数，再计算1~b中与n互质的个数，相减。
#include<iostream> using namespace std; typedef long long ll; ll a[10000]; int main() { ll t; cin>>t; for(ll ii=1;ii<=t;ii++) { ll x,y,n; cin>>x>>y>>n; ll num=1,sum=1,all=0,times; for(int i=2;i*i<=n;i++) //向a数组里塞n的质因数 { if(n%i==0) { num++; a[num]=i; } while(n%i==0) { n=n/i; } } if(n>1) a[++num]=n; for(int i=1;i<(1<<num);i++) { times=0,sum=1; for(int j=0;j<num;j++) //num个因数,乘了几个 { if(1&(i>>j)) //判断有没有选中第j个因数 { sum*=a[j+1]; times++; } } if(sum==1||sum==0) continue; if(times%2==1) { all+=y/sum; all-=(x-1)/sum; } else{ all-=y/sum; all+=(x-1)/sum; } } printf("Case #%lld: %lld\n",ii,y-x+1-all); } return 0; }
以上。
标签：ios class space 选中 you 没有 ati case tween
原文地址：https://www.cnblogs.com/zjydeoneday/p/11361000.html
