题解 | 二维斐波那契数列

//noob 40
//二级指针模拟实现二维数组

#include <stdio.h>

#define MOD 1000000007
int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    //创建二维数组存储结果
    long long** dp = (long long**)malloc((n + 1) * sizeof(long long*));
    if (dp == NULL) {
        perror("malloc fail");
        exit(1);
    }
    for (int i = 0; i <= n; i++) {
        dp[i] = (long long*)malloc((m + 1) * sizeof(long long));
        if (dp[i] == NULL) {
            perror("malloc fail");
            exit(1);
        }
    }
    // 初始化第一行和第一列
    for (int i = 1; i <= n; i++) {
        dp[i][1] = 1;
    }
    for (int j = 1; j <= m; j++) {
        dp[1][j] = 1;
    }

    // 填充dp数组
    for (int i = 2; i <= n; i++) {
        for (int j = 2; j <= m; j++) {
            dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % MOD;
        }
    }

    printf("%lld\n", dp[n][m]);

    // 释放动态分配的内存（避免内存泄漏）
    for (int i = 0; i <= n; i++) {
        free(dp[i]);  // 先释放每行
    }
    free(dp);  //
    return 0;
}