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[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

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[N],pos[N],f[N],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;
}