HDU 1003

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

 

题意：

给你一个序列，求最大连续子列和。

 

题解：

定义：dp[i]为从a[1~i]中以a[i]为结尾的最大连续子序列和。

那么状态转移方程：dp[i] = (dp[i-1]>=0) ? (dp[i-1]+a[i]) : (a[i]));

最后遍历dp数组，找到最大的那个值就是ans，那个位置就是子序列的末尾位置；

另外为了记录子序列的开始位置，不妨把dp数组定义成结构体，里面定义一个sum用来存最大的和，另外定义一个st来储存开始位置.

 

AC代码： 

#include<bits/stdc++.h>
using namespace std;
const int maxn = 100000+10;

int a[maxn];
struct DP{
    int val;
    int st;
}dp[maxn];

int main()
{
    int t,n;
    scanf("%d",&t);
    for(int kase=1;kase<=t;kase++)
    {
        scanf("%d",&n);

        int maxi=-10000,maxi_idx;
        dp[0].val=-1, dp[0].st=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            dp[i].val = dp[i-1].val>=0 ? (dp[i-1].val+a[i]) : a[i];
            dp[i].st = dp[i-1].val>=0 ? (dp[i-1].st) : i;

            if(maxi<dp[i].val)
            {
                maxi = dp[i].val;
                maxi_idx = i;
            }
        }

        printf("%sCase %d:
",kase>1?"
":"",kase);
        printf("%d %d %d
",maxi,dp[maxi_idx].st,maxi_idx);
    }
}

注：状态转移方程　dp[i] = (dp[i-1]>=0) ? (dp[i-1]+a[i]) : (a[i]));　和　dp[i] = (dp[i-1]>=0) ? (dp[i-1]+a[i]) : (a[i]));　是有区别的。