Dp随笔

Posted 2022-05-21 csyzfraction

**最大子段和**
很简单的一道Dp题，用$dp[i]$表示$1-i$中的最大子段和
能够得出状态转移方程：$dp[i] = max(dp[i], dp[i]+dp[i-1])$
初始化$dp[i]=num[i]$
代码

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int n, ans = -20000;
 5 int num[200005], dp[200005];
 6 
 7 int init() 
 8     scanf("%d", &n);
 9     for(int i = 1; i <= n; i++)scanf("%d", &num[i]), dp[i] = num[i];
10     return 0;
11 
12 
13 int solve() 
14     for(int i = 2; i <= n; i++) dp[i] = max(dp[i], dp[i-1]+dp[i]);
15     for(int i = 1; i <= n; i++) ans = max(ans, dp[i]);
16     printf("%d", ans);
17     return 0;
18 
19 
20 int main() 
21     init();
22     solve();
23     return 0;
24 

最大子段和