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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