ICode9

精准搜索请尝试: 精确搜索
首页 > 其他分享> 文章详细

The 2019 China Collegiate Programming Contest Harbin Site I. Interesting Permutation

2019-11-05 13:02:32  阅读:74  来源: 互联网

标签:Contest int Interesting permutation Programming flag each dp MOD



链接:

https://codeforces.com/gym/102394/problem/I

题意:

DreamGrid has an interesting permutation of 1,2,…,n denoted by a1,a2,…,an. He generates three sequences f, g and h, all of length n, according to the permutation a in the way described below:

For each 1≤i≤n, fi=max{a1,a2,…,ai};
For each 1≤i≤n, gi=min{a1,a2,…,ai};
For each 1≤i≤n, hi=fi−gi.
BaoBao has just found the sequence h DreamGrid generates and decides to restore the original permutation. Given the sequence h, please help BaoBao calculate the number of different permutations that can generate the sequence h. As the answer may be quite large, print the answer modulo 109+7.

思路:

考虑,某个位置为n或者h[i] > h[i+1] ,和第一个位置不为0时,答案都为0.
其他情况,h[i] == h[i+1],考虑前面的空位数, h[i] < h[i+1],可以插最小值,也可以插最大值, 直接×2(没太懂)。

代码:

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
 
const int MAXN = 1e5+10;
const int MOD = 1e9+7;
 
LL dp[MAXN];
int a[MAXN];
int n;
 
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        memset(dp, 0, sizeof(dp));
        scanf("%d", &n);
        a[0] = 0;
        bool flag = true;
        for (int i = 1;i <= n;i++)
        {
            scanf("%d", &a[i]);
            if (a[i] < a[i-1] || a[i] >= n || (i != 1 && a[i] == 0))
                flag = false;
        }
        if (a[1] != 0)
            flag = false;
        if (!flag)
        {
            puts("0");
            continue;
        }
        dp[1] = 1;
        for (int i = 2;i <= n;i++)
        {
            if (a[i] > a[i-1])
                dp[i] = (dp[i-1]*2)%MOD;
            else
            {
                LL sp = a[i]-i+2;
                dp[i] = (dp[i-1]*sp)%MOD;
            }
        }
        printf("%d\n", (int)(dp[n]%MOD));
    }
 
    return 0;
}


标签:Contest,int,Interesting,permutation,Programming,flag,each,dp,MOD
来源: https://www.cnblogs.com/YDDDD/p/11797745.html

专注分享技术,共同学习,共同进步。侵权联系[admin#icode9.com]

Copyright (C)ICode9.com, All Rights Reserved.

ICode9版权所有