求数塔问题的源代码和伪代码
时间: 2024-05-23 22:09:10 浏览: 15
好的,以下是求解数塔问题的伪代码和源代码:
伪代码:
1. 读入数塔的行数 n
2. 定义数组 a 存储数塔
3. for i = 1 to n do
4. for j = 1 to i do
5. 读入 a[i][j]
6. end for
7. end for
8. for j = 1 to n do
9. dp[n][j] = a[n][j] // 初始化最后一行
10. end for
11. for i = n-1 to 1 do // 从倒数第二行往上递推
12. for j = 1 to i do
13. dp[i][j] = max(dp[i+1][j], dp[i+1][j+1]) + a[i][j] // 状态转移方程
14. end for
15. end for
16. 输出 dp[1][1]
源代码:
#include <iostream>
#include <cstring>
using namespace std;
const int N = 110;
int a[N][N], dp[N][N];
int main()
{
int n;
cin >> n;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= i; j++)
cin >> a[i][j];
for (int j = 1; j <= n; j++)
dp[n][j] = a[n][j];
for (int i = n-1; i >= 1; i--)
for (int j = 1; j <= i; j++)
dp[i][j] = max(dp[i+1][j], dp[i+1][j+1]) + a[i][j];
cout << dp[1][1] << endl;
return 0;
}
请注意,以上代码是伪代码和源代码,只是演示了数塔问题的解法,具体实现中可能还需要进行一些改动和优化。