首页 > 代码库 > How many integers can you find
How many integers can you find
How many integers can you find
Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Problem Description
Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.
Input
There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.
Output
For each case, output the number.
Sample Input
12 22 3
Sample Output
7
分析：容斥原理，注意long long；
代码：
#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <bitset>#include <map>#include <queue>#include <stack>#include <vector>#define rep(i,m,n) for(i=m;i<=n;i++)#define mod 1000000007#define inf 0x3f3f3f3f#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define ll long long#define pi acos(-1.0)#define pii pair<int,int>#define sys system("pause")const int maxn=1e5+10;using namespace std;inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}inline void umax(ll &p,ll q){if(p<q)p=q;}inline void umin(ll &p,ll q){if(p>q)p=q;}inline ll read(){ ll x=0;int f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f;}int n,m,k,t,fac[20],all;int main(){ int i,j; while(~scanf("%d%d",&m,&n)) { --m; all=0; rep(i,0,n-1) { scanf("%d",&j); if(j)fac[all++]=j; } ll ret=0; rep(i,1,(1<<all)-1) { ll now=1,cnt=0; rep(j,0,all-1) { if(i&(1<<j)) { cnt++; now=now*fac[j]/gcd(now,fac[j]); } } if(cnt&1)ret+=m/now; else ret-=m/now; } printf("%lld\n",ret); } return 0;}
How many integers can you find
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。