链接:https://www.nowcoder.com/acm/contest/139/A
题目描述
Count the number of n x m matrices A satisfying the following condition modulo (109+7).
* Ai, j ∈ {0, 1, 2} for all 1 ≤ i ≤ n, 1 ≤ j ≤ m.
* Ai, j ≤ Ai + 1, j for all 1 ≤ i < n, 1 ≤ j ≤ m.
* Ai, j ≤ Ai, j + 1 for all 1 ≤ i ≤ n, 1 ≤ j < m.
输入描述:
The input consists of several test cases and is terminated by end-of-file. Each test case contains two integers n and m.
输出描述:
For each test case, print an integer which denotes the result.
示例1
输入
1 2 2 2 1000 1000
输出
6 20 540949876
备注:
* 1 ≤ n, m ≤ 103
* The number of test cases does not exceed 105.
分析:
考虑n*m中01的分界线和12的分界线,一开始两条路径的起点都是(1,1)终点都是(n,m)
然后考虑他们相交的情况,相交包括重叠,所以将01的分界线往右上角移动,起点变成(2,0)终点变成(n+1,m-1)
在交点处交换他们后面的边使01的分界线起点变成(2,0)终点变成(n,m)。
12的分界线,起点变成(1,1),终点变成(n+1,m-1)
Cm+n,n(2)-Cm+n,n-1*Cm+n,m-1
代码:
#include <bits/stdc++.h>
const int MOD = 1e9 + 7;
const int N = 1005;
int dp[N][N];
void update(int& x, int a)
{
x += a;
if (x >= MOD) {
x -= MOD;
}
}
int sqr(int x)
{
return 1LL * x * x % MOD;
}
int main()
{
dp[0][0] = 1;
for (int i = 0; i < N; ++ i) {
for (int j = 0; j < N; ++ j) {
if (i) {
update(dp[i][j], dp[i - 1][j]);
}
if (j) {
update(dp[i][j], dp[i][j - 1]);
}
}
}
int n, m;
while (scanf("%d%d", &n, &m) == 2) {
printf("%d\n", ((sqr(dp[n][m]) + MOD - 1LL * dp[n - 1][m + 1] * dp[n + 1][m - 1] % MOD) % MOD));
}
}