《2020百度之星-初赛2》

Car：

思路：暴搜，对于每一个尾数，显然只会在5天中的某一天被禁止。

那么5^10.不会爆

但是如果用字符串cin输入字符串，就会因为输入过慢而TLE，那么这里直接把字符输入变成整形，然后取最后一位即可。

Code:

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef long double ld;
typedef pair<int,int> pii;
const int N = 1e5+5;
const int M = 1e6+6;
const LL Mod = 2505;
#define pi acos(-1)
#define INF 1e8
#define INM INT_MIN
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read()
{
    int x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
int aa[15],cal[15],cnt,n,ans;
void dfs(int num,int a,int b,int c,int d,int e)
{
    if(num == cnt+1)
    {
        int sum = 0;
        sum = max(sum,n-a);
        sum = max(sum,n-b);
        sum = max(sum,n-c);
        sum = max(sum,n-d);
        sum = max(sum,n-e);
        ans = min(ans,sum);
        return ;
    }
    dfs(num+1,a+cal[num],b,c,d,e);
    dfs(num+1,a,b+cal[num],c,d,e);
    dfs(num+1,a,b,c+cal[num],d,e);
    dfs(num+1,a,b,c,d+cal[num],e);
    dfs(num+1,a,b,c,d,e+cal[num]);
}
int main()
{
    int ca;ca = read();
    while(ca--)
    {
        n = read();
        memset(aa,0,sizeof(aa));
        for(int i = 1;i <= n;++i)
        {
            int x;x = read();
            aa[x%10]++;
        }
        cnt = 0,ans = INF;
        for(int i = 0;i <= 9;++i) if(aa[i] != 0) cal[++cnt] = aa[i];
        dfs(1,0,0,0,0,0);
        printf("%d\n",ans);
    }
    system("pause");
    return 0;
}