hdu 1003(最大子段和)

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 204780    Accepted Submission(s): 47877

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

 

模板题:

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
const int N = ;

int a[N];
int dp[N];
int main()
{
    int tcase;
    scanf("%d",&tcase);
    int k =;
    while(tcase--){
        int n;
        scanf("%d",&n);
        for(int i=;i<=n;i++){
            scanf("%d",&a[i]);
        }
        int start=,endd=,t=;
        int mx = dp[] = a[];
        for(int i=;i<=n;i++){
            if(dp[i-]>=){
                dp[i] = dp[i-]+a[i];
            }else{
                t = i;
                dp[i] = a[i];
            }
            if(dp[i]>mx){
                mx = dp[i];
                start = t;
                endd = i;
            }
        }
        printf("Case %d:\n%d %d %d\n",k++,mx,start,endd);
        if(tcase!=) printf("\n");
    }
    return ;
}