HDU 1003 DP 问题

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 126465 Accepted Submission(s): 29302

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

题目大意：给出以一组数现在要求求出最大子序列数，并且给出序列的初始位置和终端位置

解题方法：简单DP 主要是找到状态方程 sum[i] = max{ sum[i-1] + a[i], a[i] } 再开始使用这个的意思是什么呢如果前面和是大于零的那么继续加

如果不是那么舍弃前面的从现在开始计数现在就是开始位置之后往下加数

这道题最要注意的就是在加和过程中会产生最大值可能最后加上一个负数，那么整个的最大值就会减小的因此要用max 来标记最大

源代码：

#include<iostream>using  namespace  std;#define  MAX 100000int main(){    //题目大意：就是求最大子序列数, 并且要求出这个序列的位置    //主要利用状态转移方程 sum[i] = max{sum[i-1] +a[i], a[i]}    int t , i;    cin>>t ;    int m;    int count = 1;    while(t--)    {        int n;        cin>>n;        int s ,e;        int a, b;        int sum = -1001;        int max = -1001;        for(int i = 1; i <=n; i++)        {            cin>>m;            if(sum + m < m)            {//根据状态方程                sum = m ;                s = i;                e = i;            }            else            {//这个步骤往下加了一个数                sum += m;                e++;            }            //下面进行动态调试            if(max < sum)            {//为什么要这步 因为之间可能会产生最大值                max = sum ;                a = s;                b = e;            }        }        cout<<"Case "<<count<<":"<<endl;        cout<<max<<" "<<a<<" "<<b<<endl;        count++;    }    return 0;}

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