HDU 1003 Max Sum(最大连续子序列和)

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 63276 Accepted Submission(s): 14433

Problem Description

Given a sequence a[1],a[2],a[3]......a
, 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

Author

Ignatius.L

解题报告:典型的动态规划，就是求最大连续子序列和的并记录起始位置！

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
const int N = 100010;
int a
,pos
,f
,ans;//pos[]数组记录起始位置，f[]数组记录和
int main()
{
int t,n,i,k,j,ans,begin,end;
//freopen("1001.txt","r",stdin);
scanf("%d",&t);
k=0;
for(j=1;j<=t;j++)
{
scanf("%d",&n);
for(i=1;i<=n;i++)//初始化
{
scanf("%d",&a[i]);
pos[i]=i;
}
f[1]=a[1];
for(i=2;i<=n;i++)
{
if(a[i]+f[i-1]>=a[i])
{
pos[i]=pos[i-1];//关键！起始位置和前一个相同
f[i]=a[i]+f[i-1];
}
else
{
f[i]=a[i];
}
//printf("%d %d\n",i,f[i]);
}
ans=-9999;
for(i=1;i<=n;i++)//遍历一回求最大
{
if(ans<f[i])
{
ans=f[i];
begin=pos[i];
end=i;
}
}
printf("Case %d:\n",++k);
printf("%d %d %d\n",ans,begin,end);
if(k!=t)//最后一行没有空格！
{
printf("\n");
}
}
return 0;
}