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时间：2022-03-22 21:31:33
Rikka with Nash Equilibrium

Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)

Total Submission(s):     Accepted Submission(s): 

Problem Description

Nash Equilibrium is an important concept in game theory.

Rikka and Yuta are playing a simple matrix game. At the beginning of the game, Rikka shows an n×m integer matrix A. And then Yuta needs to choose an integer in [,n], Rikka needs to choose an integer in [,m]. Let i be Yuta's number and j be Rikka's number, the final score of the game is Ai,j. 

In the remaining part of this statement, we use (i,j) to denote the strategy of Yuta and Rikka.

For example, when n=m= and matrix A is

⎡⎣⎢⎢⎤⎦⎥⎥

If the strategy is (,), the score will be ; if the strategy is (,), the score will be .

A pure strategy Nash equilibrium of this game is a strategy (x,y) which satisfies neither Rikka nor Yuta can make the score higher by changing his(her) strategy unilaterally. Formally, (x,y) is a Nash equilibrium if and only if:

{Ax,y≥Ai,y  ∀i∈[,n]Ax,y≥Ax,j  ∀j∈[,m]

In the previous example, there are two pure strategy Nash equilibriums: (,) and (,).

To make the game more interesting, Rikka wants to construct a matrix A for this game which satisfies the following conditions:

. Each integer in [,nm] occurs exactly once in A.

. The game has at most one pure strategy Nash equilibriums. 

Now, Rikka wants you to count the number of matrixes with size n×m which satisfy the conditions.

Input

The first line contains a single integer t(≤t≤), the number of the testcases.

The first line of each testcase contains three numbers n,m and K(≤n,m≤,≤K≤).

The input guarantees that there are at most  testcases with max(n,m)>.

Output

For each testcase, output a single line with a single number: the answer modulo K.

Sample Input

Sample Output

Source

 Multi-University Training Contest 

Recommend

chendu

从大到小填。每次填进去的数都是要在被管住的里面。
#include<iostream>

#include<cstdio>

#include<string.h>

using namespace std;

#define ll long long

ll dp[][][*];

ll  n,m,mod;

ll dfs(ll  x,ll  y,ll z)

{

    if(dp[x][y][z]!=-)

        return dp[x][y][z];

    ll temp=;

    if(x<n)

        temp=(temp+(y*(n-x)%mod)*dfs(x+,y,z+))%mod;

    if(y<m)

        temp=(temp+(x*(m-y)%mod)*dfs(x,y+,z+))%mod;

    if(x*y>z)

        temp=(temp+(x*y-z)%mod*dfs(x,y,z+))%mod;

    return dp[x][y][z]=temp;

}

int main()

{

    int T;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%lld%lld%lld",&n,&m,&mod);

        memset(dp,-,sizeof dp);

        dp[n][m][n*m]=;

        ll ans=((n*m)%mod*dfs(,,)%mod);

        printf("%lld\n",ans);

    }

    return ;

}


找规律看https://www.cnblogs.com/solvit/p/9507207.html


 


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