BZOJ5074 小B的数字

　　对b_i取log，则相当于Σb_i<=min{b_i*a_i}。注意到值域很小，那么如果有解，使其成立的最小的Σb_i不会很大，大胆猜想不超过Σa_i。然而一点也不会(xiang)证。暴力枚举就好了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int T,n,a[];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5074.in","r",stdin);
    freopen("bzoj5074.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read();
    while (T--)
    {
        n=read();memset(a,,sizeof(a));
        int m=;
        for (int i=;i<=n;i++)
        {
            int x=read();
            a[x]++;m+=x;
        }
        bool flag=;
        for (int i=;i<=m;i++)
        {
            ll cnt=;
            for (int j=;j<=;j++)
            {
                cnt+=1ll*a[j]*((i-)/j+);
                if (cnt>i) break;
            }
            if (cnt<=i) {flag=;break;}
        }
        if (flag) puts("YES");
        else puts("NO");
    }
    return ;
}