动态规划入门1

问题：

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.

输入：

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).

输出：

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

题意：求最大和的最长序列，输出这些数的总和和序列的起始点和终点

代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int main()
{
int nase,n,num,start,end,temp,i,j,a[100010],sum,max;


scanf("%d",&nase);


for(i=1;i<=nase;i++)
{
max=-1111;
sum=0;
start=end=temp=1;
scanf("%d",&n);
for(j=1;j<=n;j++)
{
scanf("%d",&num);
sum+=num;
if(sum>max)
{
max=sum;
start=temp;
end=j;
}
if(sum<0)
{
sum=0;
temp=j+1;
}
}
printf("Case %d:\n",i);
printf("%d %d %d\n",max,start,end);
if(i!=nase)
printf("\n");
}
return 0;
}