bzoj1441: Min

a1*x1+a2*x2+…+an*xn=s,s的最小值为gcd(a1,a2,…,an).

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;int main(){    int n;    cin>>n;    int ans,x;    cin>>ans;    ans=abs(ans);    for(int i=2;i<=n;i++)    {        scanf("%d",&x);        ans=__gcd(ans,abs(x));    }    cout<<ans;}