C语言揭秘：青蛙跳上第n级台阶且在第m级停留的跳法总数

#include <stdio.h> int frogJump(int n) { if (n == 0) { return 1; } if (n == 1) { return 1; } if (n == 2) { return 2; } int dp[n + 1]; dp[0] = 1; dp[1] = 1; dp[2] = 2; for (int i = 3; i <= n; i++) { dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3]; } return dp[n]; } int main() { int n, m; printf("请输入第n级台阶和停留在第m级台阶的数值："); scanf("%d %d", &n, &m); int result = frogJump(n); printf("跳上第%d级台阶且在第%d级台阶停留过的总跳法数为：%d\n", n, m, result); return 0; } ######[AI写代码神器 | 396点数解答 | 2024-06-14 10:13:28]