HDU-1003-Max Sum(dp经典问题-最大连续子序列和)

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 196696    Accepted Submission(s): 45933

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

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

 

Author

Ignatius.L

 

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觉得烦的时候就来A题，这句话说的一点也不错...只是最近烦的次数有点多，A题之心也不如以前那般狂热。

好！闲话少说，这题是动态规划DP的经典问题，题意就是求一段数列的最大连续子序列和，然后让你输出三个数字，

分别是该段数列的最大连续子序列和的值，最大连续子序列第一个数的下标（下标从1开始），最大连续子序列最后一个数的下标。

思路很简单，关键是找出状态转移方程。

1.把数列第一个数存入dp[0].

2.首先你要明确最大连续子序列和是什么意思，就是从中取两个数之间相加最大！是吧，它的特点一定要抓住，就是这个和一定比

   

    它的子序列中任何一个数要大，所以就有了判断条件。

3.状态转移方程，max( dp[i-1] + a[i] , a[i] )

#include <iostream>#include <cstdio>using namespace std;int a[100001],dp[100001];int main(){    int T,n,i=0;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        dp[0] = a[0];        int start = 0,end = 0,max = -1001;        int first = 0,last = 0;        for(int i=0;i<n;i++)        {            if(dp[i-1]+a[i]>=a[i])         //判断条件            {                dp[i] = dp[i-1]+a[i];                end = i;            }            else            {                dp[i] = a[i];                start = end = i;            }            if(max<dp[i])            {                max = dp[i];                first = start;                last = end;            }        }        printf("Case %d:\n%d %d %d\n",++i,max,first+1,last+1);        if(T!=0)        {            printf("\n");        }    }    return 0;}