HDU - 1003 Max Sum

Posted 2020-12-26 daybreaking

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

DP问题，dp[i-1]代表的是前i-1项最大的连续和，dp[i]=max(dp[i-1],con[i])。
如果dp[i-1]比con[i]小，则dp[i]为con[i],舍弃dp[i-1]。

这样就保证了状态设计的正确性，关键是如何想出来要这样设计状态转移方程的？利用了负数的"切断性"，该种性质导致了数列在某一步会减小，使得不如舍弃前几项的和，从新的地方另立门户。

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 #include<math.h>
 5 #include<algorithm>
 6 #include<queue>
 7 #include<stack>
 8 #include<deque>
 9 #include<iostream>
10 using namespace std;
11 typedef long long  LL;
12 int con[100009];
13 int dp[100009];
14 int check[100009];
15 int main()
16 {
17     int i,p,j,n,t;
18     int head,pen;
19     scanf("%d",&t);
20     for(i=1;i<=t;i++)
21     {
22         scanf("%d",&n);
23         for(j=1;j<=n;j++)
24             scanf("%d",&con[j]);
25         head=-1001;
26         pen=0;
27         memset(dp,0,sizeof(dp));    //初值应该赋为什么？？？？？？？
28         memset(check,0,sizeof(check));
29         for(j=1;j<=n;j++)
30         {
31             dp[j]=max(dp[j-1]+con[j],con[j]);
32             if(con[j]>dp[j-1]+con[j])
33                 check[j]=1;
34             if(dp[j]>head)
35             {
36                 head=dp[j];
37                 pen=j;
38             }
39         }
40         for(j=pen;j>=2;j--)
41             if(check[j]==1)
42                 break;
43 
44         printf("Case %d:
",i);
45         printf("%d %d %d
",head,j,pen);
46         if(i<t)
47             putchar(‘
‘);
48     }
49     return 0;
50 }

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