标签：
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
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);
}
}
1001 of dp
标签：
原文地址：http://blog.csdn.net/toy_block/article/details/51338119
