hdu1003MAX SUM

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 148685 Accepted Submission(s): 34759

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

代码：

#include<stdio.h>
#include<iostream>
#define INF 9999
using namespace std;
int main()
{
int T,n,i,j,a[100000],sum,max;
int pos_1,pos_2,pos;//最大和序列起始位置，最大和序列结束位置，当前位置
cin>>T;
for(j=1;j<=T;++j)
{
max=-INF;//因为一个数a 是-1000~1000的，所以这里相当于变成最小值
pos_1=pos_2=pos=1;
sum=0;
cin>>n;
for(i=1;i<=n;++i)
{
cin>>a[i];
if(sum<0)
{
sum=a[i];
pos=i;//记录当前的位置i,i即是应当更新的起点
}
else sum+=a[i];
if(sum>max)
{
max=sum;
pos_1=pos;
pos_2=i;
}
}
printf("Case %d:\n%d %d %d\n",j,max,pos_1,pos_2);
if(j!=T)
printf("\n");
}
return 0;
}
/*
状态方程 sum[i] = sum[i - 1] > 0 ？ sum[i - 1] + a[i] : a[i]
只有当 sum 处于增长状态时，才能得到最大子序列，
当 sum 处于减小状态时，应当更新起点
*/