1233

JSZKC is the captain of the lala team.

There are

JSZKC has to arrange them as an array from left to right and let $h_ihi be the height of the i^{th}ith girl counting from the left. After that, he can calculate the sum of the inversion pairs. A inversion pair counts if h_i>h_jhi>hj with i<ji<j.$

Now JSZKC wants to know how many different arranging plans with the sum of the inversion pairs equaling $h_ihi different in these two plans.$

As the answer may be very large, you should output the answer mod

Input Format

The input file contains several test cases, each of them as described below.

The first line of the input contains two integers $\le N \le 5000,0 \le K \le 5000)(1≤N≤5000,0≤K≤5000), giving the number of girls and the pairs that JSZKC asked.$

There are no more than

Output Format

An integer in one line for each test case, which is the number of the plans mod

样例输入

3 2
3 3

样例输出

2
1

题目来源

The 2018 ACM-ICPC China JiangSu Provincial Programming Contest

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <vector>
 6 #include <queue>
 7 #include <stack>
 8 #include <cstdlib>
 9 #include <iomanip>
10 #include <cmath>
11 #include <cassert>
12 #include <ctime>
13 #include <map>
14 #include <set>
15 using namespace std;
16 int n,m,k;
17 int a[15][2];
18 const int N=2311;
19 int vis[N][N];
20 int x,y,val;
21 const int  inf=0x3f3f3f3f;
22 int   main()
23 {
24     while(~scanf("%d%d",&n,&m)){
25         memset(vis,inf,sizeof(vis));
26         val=-1;
27         scanf("%d",&k);
28         for(int i=0;i<k;i++){
29             scanf("%d%d",&a[i][0],&a[i][1]);
30         }
31         for(int i=1;i<=n;i++){
32             for(int j=1;j<=m;j++){
33                 for(int ii=0;ii<k;ii++){
34                     vis[i][j]=min(vis[i][j],abs(i-a[ii][0])+abs(j-a[ii][1]));
35                 }
36                 if(vis[i][j]>val){
37                     val=vis[i][j];
38                     x=i;
39                     y=j;
40                 }
41                 else if(vis[i][j]==val){
42                     if(i<x){
43                         x=i;
44                         y=j;
45                     }
46                     else if(i==x) y=min(y,j);
47                 }
48             }
49         }
50         printf("%d %d\n",x,y);
51     }
52     return  0;
53 }