ZJU 2836

来源：互联网发布：cpa题库软件编辑：程序博客网时间：2024/05/05 17:18

题意：给你n ， m 和 n 个数a，找出一个不大于m数且能被任一一个数a [i] 整除，求这样的数有多少个。

注意：

1<=n<=10,1=<a[i]<=10,1=<m<=200000000

利用a[i]的范围和质数的性质（2,3,5,7），删选a[i]个数；

算出交并集的个数；

#include<cstdio>#include<stdlib.h>#include<string.h>#include<string>#include<map>#include<cmath>#include<iostream>#include <queue>#include <stack>#include<algorithm>#include<set>using namespace std;#define INF 1e8#define inf -0x3f3f3f3f#define eps 1e-8#define ll long long#define maxn 250001*3#define mol 100007int gcd(int a,int b){if(b==0) return a;return gcd(b,a%b);}int lcm(int a,int b){return a*b/gcd(a,b);}int main(){int n,m;int a[11],num[10];while(scanf("%d%d",&n,&m)!=EOF){int i,j,k;for(i=0;i<n;i++)cin>>a[i];int t=0;memset(num,0,sizeof(num));for(i=0;i<n;i++){    if(a[i]==1)t=1;else if(a[i]%2==0){if(a[i]==2)num[1]=2;if(num[1]==0) num[1]=a[i];    else num[1]=min(num[1],a[i]);}else if(a[i]%3==0){if(a[i]==3)num[2]=3;if(num[2]==0) num[2]=a[i];    else num[2]=min(num[2],a[i]);}else if(a[i]%5==0){if(a[i]==5)num[3]=5;else if(num[3]==0) num[3]=a[i];    else num[3]=min(num[3],a[i]);}else if(a[i]%7==0){if(a[i]==7)num[4]=7;else if(num[4]==0) num[4]=a[i];    else num[4]=min(num[4],a[i]);}}if(t){printf("%d\n",m);continue;}int ans=0;for(i=1;i<=4;i++)if(num[i]!=0)    ans+=m/num[i];for(i=1;i<=4;i++){for(j=i+1;j<=4;j++)if(num[i]!=0&&num[j]!=0)   ans-=m/(lcm(num[i],num[j]));}for(i=1;i<=4;i++){for(j=i+1;j<=4;j++){for(k=j+1;k<=4;k++){if(num[i]!=0&&num[j]!=0&&num[k]!=0)    ans+=m/lcm(num[k],lcm(num[i],num[j]));}}}for(i=1;i<=4;i++){for(j=i+1;j<=4;j++){for(k=j+1;k<=4;k++){for(int l=k+1;l<=4;l++)if(num[i]!=0&&num[j]!=0&&num[k]!=0&&num[l]!=0)         ans-=m/lcm(num[l],lcm(num[k],lcm(num[i],num[j])));}}}printf("%d\n",ans);}return 0;}

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