题解 | 二维斐波那契数列

#include<bits/stdc++.h>
using namespace std;
int a[2004][2004];
int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (i == 1 && j == 1) {
                a[i][j] = 1;
            } else if (i >= 2 && j == 1) {
                a[i][j] = a[i - 1][j];
            } else if (j >= 2 && i == 1) {
                a[i][j] = a[i][j - 1];
            } else if (i >= 2) {
                a[i][j] = (a[i - 1][j] + a[i][j - 1])%1000000007;
            }
        }
    }
cout << a[n][m];
return 0;
}