题解 | 二维斐波那契数列

#include <iostream>
#include <vector>
#include <bits/stdc++.h>
using namespace std;
const int MODE = 1000000007;
long long dp[1001][1001];
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) dp[i][j] = 1;
            else if(i == 1) dp[i][j] = dp[i][j-1];
            else if(j == 1) dp[i][j] = dp[i-1][j];
            else dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % MODE;
        }
    }
    cout << dp[n][m] << endl;
    return 0;
}
// 64 位输出请用 printf("%lld")