HDU

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。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#include<iostream>
using namespace std;
typedef long long  LL;
int con[100009];
int dp[100009];
int check[100009];
int main()
{
    int i,p,j,n,t;
    int head,pen;
    scanf("%d",&t);
    for(i=1;i<=t;i++)
    {
        scanf("%d",&n);
        for(j=1;j<=n;j++)
            scanf("%d",&con[j]);
        head=-1001;
        pen=0;
        memset(dp,0,sizeof(dp));    //初值应该赋为什么？？？？？？？
        memset(check,0,sizeof(check));
        for(j=1;j<=n;j++)
        {
            dp[j]=max(dp[j-1]+con[j],con[j]);
            if(con[j]>dp[j-1]+con[j])
                check[j]=1;
            if(dp[j]>head)
            {
                head=dp[j];
                pen=j;
            }
        }
        for(j=pen;j>=2;j--)
            if(check[j]==1)
                break;

        printf("Case %d:
",i);
        printf("%d %d %d
",head,j,pen);
        if(i<t)
            putchar('
');
    }
    return 0;
}