1284 2 3 5 7的倍数 容斥原理

题目链接:https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1284
题意：找出小于等于n，且不为2 3 5 7倍数的数。
思路：容斥原理。

#include<cstdio>#include<queue>#include<iostream>#include<vector>#include<map>#include<cstring>#include<string>#include<set>#include<stack>#include<algorithm>#define cle(a) memset(a,0,sizeof(a))#define inf(a) memset(a,0x3f,sizeof(a))#define ll long long#define Rep(i,a,n) for(int i=a;i<=n;i++)using namespace std;const int INF = ( 2e9 ) + 2;//const int maxn =int main(){    ll n;    scanf("%lld",&n);    ll n1=n/7*1LL;      //7 的倍数的个数      ll n2=n/5*1LL;    ll n3=n/3*1LL;    ll n4=n/2*1LL;    ll n5=n/6*1LL;    ll n6=n/10*1LL;    ll n7=n/14*1LL;    ll n8=n/15*1LL;    ll n9=n/21*1LL;    ll n10=n/35*1LL;    ll n11=n/30*1LL;    ll n12=n/42*1LL;    ll n13=n/70*1LL;    ll n14=n/105*1LL;    ll n15=n/210*1LL;    printf("%lld\n",n-n1-n2-n3-n4+n5+n6+n7+n8+n9+n10-n11-n12-n13-n14+n15);}