HDU-1003 Max Sum 简单区间dp求最大子区间和

借鉴博客： https://blog.csdn.net/hollay/article/details/89433928

　　　　　 https://blog.csdn.net/Alibaba_lhl/article/details/79914061

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

 

Sample Input

2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1: 14 1 4 Case 2: 7 1 6

 

解题思路：

根据简单的dp思想可以很快求出最大值，所以只要根据变化来保存一下他的最大的区间的位置就可以了

dp方程： dp[i] = max（ a[i], dp[i-1]+a[i] ）

 

代码如下：

 1 #include<bits/stdc++.h>
 2 #define mem(a) memset(a,0,sizeof(a))
 3 #define forn(i,n) for(int i=0;i<n;++i)
 4 #define for1(i,n) for(int i=1;i<=n;++i)
 5 #define IO std::ios::sync_with_stdio(false); std::cin.tie(0)
 6 using namespace std;
 7 typedef long long ll;
 8 const int maxn=1e5+5;
 9 const int mod=10007;
10 const int inf=0x3f3f3f3f;
11 
12 int n,m,t,k;
13 int a[maxn];
14 int dp[maxn];
15 
16 
17 int main()
18 {
19     IO;
20     cin>>m;
21     int k=1;
22     while(k<=m)
23     {
24         mem(a);
25         mem(dp);
26         if(k!=1)
27             cout<<endl;
28         cin>>n;
29         forn(i,n)
30         {
31             cin>>dp[i];
32         }
33         int maxs=dp[0];
34         int l=0,r=0,head=0;
35         for(int i=1;i<n;++i)
36         {
37             if(dp[i-1]>=0)
38             {
39                 dp[i]=dp[i]+dp[i-1];
40             }
41             else
42             {
43                 head=i;
44             }
45             if(dp[i]>maxs)
46             {
47                 maxs=dp[i];
48                 l=head;
49                 r=i;
50             }
51         }
52         cout<<"Case "<<k<<":"<<endl;
53         cout<<maxs<<" "<<l+1<<" "<<r+1<<endl;
54         k++;
55     }
56     return 0;
57 }