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); }