HDU 1003 Max Sum

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

 

题目大意：求最大字段和

n<=100000,t<=20，可知只能使用线性复杂度的算法。

设f[i]表示结尾为i的最大子段和，f[i]=max{f[i-1]+a[i],a[i]}

ans=max{f[i]}

 

#include<iostream>
#include<cstring>
#include<climits>
#include<cstdio>
using namespace std;
const int N=100005;
int T,n,t,a[N],f[N],res,x,y,pth[N];
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        printf("Case %d:
",++t);
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        res=f[0]=INT_MIN;
        for(int i=1;i<=n;i++)
        {
            if(f[i-1]<0)
            {
                f[i]=a[i];
                pth[i]=i;
            }
            else
            {
                f[i]=a[i]+f[i-1];
                pth[i]=pth[i-1];
            }
            if(f[i]>res)
            {
                res=f[i];
                x=pth[i],y=i;
            }
        }
        printf("%d %d %d
",res,x,y);
        if(T)
            printf("
");
    }
    return 0;
}