hdu 1796 How many integers can you find 容斥原理

abc=a+b+c-ab-ac-bc+abc

abcd=a+b+c+d-ab-ac-ad-bc-bd-cd+abc+abd+bcd-abcd

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dfs枚举，传参保证当前是加还是减，注意输入检查以及<n

#include <stdio.h>#include <vector>#include <algorithm>#include <iostream>using namespace std;typedef long long ll;ll t[20];ll n,ans;int m;ll gcd(ll a,ll b){    return a%b==0?b:gcd(b,a%b);}void dfs(int p,ll num,int flag){    if(p) ans+=flag*n/num;    for(int i=p+1;i<=m;i++)    {        dfs(i,t[i]/gcd(t[i],num)*num,-flag);    }}int main(){    while(cin>>n>>m)    {        for(int i=1;i<=m;i++)        {            cin>>t[i];            if(t[i]<=0||t[i]>=n) i--,m--;        }        n--;        ans=0;        dfs(0,1,-1);        cout<<ans<<endl;    }    return 0;}