hdu 1003

Max Sum

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

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

题意不说 和hdu1231很相似  刚刚写了1231，但被这题坑了  maxsum我开始初值设了0  一直wa
后面测试数据才发现自己错了

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
typedef long long ll;
#define N 100005
using namespace std;
int a[N];
int b[N];
ll dp[N];
int main()
{
    int t;
    int n;
    int i,j;
    int first,next;
    scanf("%d",&t);
    for(j=1;j<=t;j++)
    {
        ll maxsum=-99999999;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(dp,0,sizeof(dp));
        first=1;
        for(i=1;i<=n;i++)
        {
            if(dp[i-1]+a[i]>=a[i])
                dp[i]=dp[i-1]+a[i];
            else
            {
                dp[i]=a[i];
                first=i;
            }
            b[i]=first;
            maxsum=max(maxsum,dp[i]);
        }
        for(i=1;i<=n;i++)
        {
            if(maxsum==dp[i])
            {
                next=i;
                //break;
            }
        }
        printf("Case %d:\n",j);
        if(j!=t)
            printf("%I64d %d %d\n\n",maxsum,b[next],next);
        else
            printf("%I64d %d %d\n",maxsum,b[next],next);
    }
    return 0;
}