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Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6001    Accepted Submission(s): 1722 
题解:wa
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
vector<int>p;
typedef long long LL;
void rc(int x){
	LL sum=0;
	for(int i=1;i<(1<<p.size());i++){
		LL num=0,cur=1;
		for(int j=0;j<p.size();j++){
			if(i&(1<<j)){
				num++;
				cur*=p[j];
			}
		}
		if(num&1)sum+=x/cur-1;
		else sum-=x/cur-1;
	}
	printf("%lld\n",sum);
}
int main(){
	int N,M,x;
	while(~scanf("%d%d",&N,&M)){
		p.clear();
		for(int i=0;i<M;i++){
			scanf("%d",&x);
			if(x==0)continue;
			p.push_back(x);
		}
		rc(N);
	}
	return 0;
}
How many integers can you find(容斥)
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原文地址:http://www.cnblogs.com/handsomecui/p/4989977.html