题解 | 放苹果

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int m, n;
    cin >> m >> n;
    vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
    for(int i = 0; i <= m; ++i) { //1个盘子放所有苹果
        dp[i][1] = 1;
    }
    for(int j = 1; j <= n; ++j) { //0个苹果
        dp[0][j] = 1;
    }
    for(int i = 1; i <= m; ++i) {
        for(int j = 1; j <= n; ++j) {
            if(j > i) dp[i][j] = dp[i][i]; //盘子多于苹果 必定有盘子为空
            //苹果多于盘子分为两种情况：盘子全放满：dp[i-j][j] 和 至少有1个盘子为空：dp[i][j-1]
            if(i >= j) dp[i][j] = dp[i-j][j] + dp[i][j-1]; 

        }
    }
    cout << dp[m][n] << endl;
    return 0;
}