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hdu1003 Max Sum--DP


原题链接:​​http://acm.hdu.edu.cn/showproblem.php?pid=1003​​


一:原题内容



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


二:分析理解

一个简单动态规划。。。。


三:AC代码

#include<iostream>
#include<string.h>
#include<algorithm>

using namespace std;

int a[100010];

int main()
{
int T;
scanf("%d", &T);

for (int i = 1; i <= T; i++)
{
scanf("%d", &a[0]);

for (int j = 1; j <= a[0]; j++)
scanf("%d", &a[j]);

int left = 1;
int right = 1;
int cnt = a[1];
int flag = 1;
int max = a[1];

for (int k = 2; k <= a[0]; k++)
{
if (cnt < 0)
{
flag = k;
cnt = a[k];
}
else
cnt += a[k];

if (cnt > max)
{
left = flag;
right = k;
max = cnt;
}
}

printf("Case %d:\n%d %d %d\n", i, max, left, right);
if (i != T)
printf("\n");
}

return 0;
}



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