题：http://acm.hdu.edu.cn/showproblem.php?pid=4135
题意：求[A,B]与N互质的数的个数
#include<iostream> #include<algorithm> #include<cstring> #include<cstdio> using namespace std; typedef long long ll; const int M=1e5+5; ll a[M]; ll A,B,N; int tot; void init(){ tot=0; ll NN=N; for(int i=2;i*i<=NN;i++){ if(N%i==0){ a[tot++]=i; while(N%i==0) N/=i; } } if(N>1) a[tot++]=N; } ll solve(ll x){ ll ans=0; for(ll i=1;i<((ll)1<<tot);i++){///枚举N的因子相乘子集 ll sum=1; int num=0; for(ll j=0;j<tot;j++){ if(i&((ll)1<<j)) sum*=a[j],num++; } ll tmp=x/sum; ///奇加偶减（容斥原理） ///ans为这些因子相乘能组成的数的个数（小于x）之和 if(num&1) ans+=tmp; else ans-=tmp; } return x-ans; } int main(){ int t; scanf("%d",&t); for(int i=1;i<=t;i++){ scanf("%lld%lld%lld",&A,&B,&N); init(); printf("Case #%d: %lld ",i,solve(B)-solve(A-1ll)); } return 0; }
ecf：https://codeforces.com/contest/1295/problem/D
思路：https://www.cnblogs.com/heyuhhh/p/12243444.html
#include<iostream> #include<algorithm> #include<cstring> #include<cstdio> using namespace std; typedef long long ll; const int M=1e5+5; ll a[M]; ll A,B; int tot; void init(ll N){ tot=0; ll NN=N; for(ll i=2;i*i<=NN;i++){ if(N%i==0){ a[tot++]=i; while(N%i==0) N/=i; } if(i>N) break; } if(N>1) a[tot++]=N; } ll solve(ll x){ ll ans=0; for(ll i=1;i<((ll)1<<tot);i++){ ll sum=1; int num=0; for(ll j=0;j<tot;j++){ if(i&((ll)1<<j)) sum*=a[j],num++; } ll tmp=x/sum; if(num&1) ans+=tmp; else ans-=tmp; } return x-ans; } int main(){ int t; scanf("%d",&t); while(t--){ scanf("%lld%lld",&A,&B); ll g=__gcd(A,B); A/=g; B/=g; init(B); printf("%lld ",solve(B+A-1)-solve(A-1)); } return 0; }