CDOJ 1273 God Qing's circuital law

暴力枚举+idea。做的时候mod写错了，写成了1000000009，找了两个多小时才发现......

a[1],a[2],a[3]....a[N]

b[1],b[2],b[3]....b[N]

首先需要枚举b[1]...b[N]与a[1]进行组合。

然后对a[2]...a[N]从小到大排序

对b[1],b[2],b[3]....b[N] 除当前与a[1]组合的以外，剩下的从大到小排序

然后找出每一个a[i]在不破坏a[0]最大值的情况下最大能与哪一个b[i]配对。

然后从第N个人开始往第2个人开始计算，先算N有几种取法，然后算N-1有几种。。。一直算到第2个人有几种

然后把这些数字乘起来就是当前这一次枚举的答案，最后把所有答案加起来就是了

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;

const int maxn =  + ;
const long long MOD = ;
long long tmpa[maxn];
long long tmpb[maxn];
long long a[maxn];
long long b[maxn];
long long flag[maxn];
int n;

bool cmp(const long long&a, const long long&b)
{
    return a>b;
}

int main()
{
    while (~scanf("%d", &n))
    {
        for (int i = ; i <= n; i++) scanf("%lld", &tmpa[i]);
        for (int i = ; i <= n; i++) scanf("%lld", &tmpb[i]);

        long long ans = ;
        for (int t = ; t <= n; t++)
        {
            long long top = tmpa[] * tmpb[t];

            for (int i = ; i <= n - ; i++) a[i] = tmpa[i + ];

            int u = ;
            for (int i = ; i <= n; i++)
            {
                if (i == t) continue;
                else b[u++] = tmpb[i];
            }

            int tot = n - ;
            sort(a + , a +  + tot);
            sort(b + , b +  + tot, cmp);

            for (int i = ; i <= n; i++) flag[i] = (long long);
            for (int i = ; i <= tot; i++)
            {
                for (int j = ; j <= tot; j++)
                {
                    if (a[i] * b[j] >= top){}
                    else
                    {
                        flag[i] = (long long)j;
                        break;
                    }
                }
            }

            long long ans_tmp = ;
            bool fail = ;
            for (int i = ; i <= tot; i++)
            if (flag[i] > (long long)i) { ans_tmp = ; fail = ; break; }

            if (fail == )
            {
                long long now = ;
                long long newpos = (long long)(tot + );
                for (int i = tot; i >= ; i--)
                {
                    long long newz = ;
                    if (flag[i] < newpos)
                    {
                        newz = newpos-flag[i];
                        newpos = flag[i];
                    }
                    now = now + newz;
                    ans_tmp = (ans_tmp*now) % MOD;
                    now--;
                }
            }
            ans = (ans + ans_tmp) % MOD;
        }
        printf("%lld\n", ans);
    }
    return ;
}