1001 of dp

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Problem A

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 258 Accepted Submission(s) : 50

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.<br>

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).<br>

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.<br>

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

题目要求：在一串数字中找到最加和最大的子序列。

解题思路：若在第i的数字结束的最大子序列为d[i],则在第i+1个数字结束的最大子序列为d[i+1]=max[a[i+1],d[i+1]+a[i+1]]

由状态转移方程可以得到的每位的最大子序列，再由另为两个数组保存到该数时所取的开头和结尾数列，输出最大的的d[i]和开始终点位置即可。

#include<iostream>#include<cstdio>using namespace std;int main(){    //freopen("right.txt","r",stdin);    int i,t,n,b,e,s,x,max,before;    cin>>t;   for(int j=1;j<=t;j++)    {        cin>>n;        for(i=1;i<=n;i++)        {//cout<<b<<endl;           cin>>s;           if(i==1)           {               max=before=s;               x=b=e=1;           }           else {                if(before<0)                {                    before=s;x=i;                }                else before+=s;                if(before>max)                {max=before;b=x;e=i;}           }        }        if(j!=1)cout<<endl;           printf("Case %d:\n%d %d %d\n",j,max,b,e);    }}

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