HDU 1003

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

 

【题意】 给n个数，求最大连续元素的和；并输出起点和终点。

【分析】

　　设d[i]为以第i个元素为终点的最大连续元素和。则

　　状态转移方程：d[i] = d[i-1] > 0 ? d[i-1]+a[i] : a[i];

 

　　思路应该比较好理解，如果d[i-1]<0， 那么无论a[i]为何值，其和总不如单独的a[i]大；反之如果d[i-1]>0， 那么无论a[i]为何值，其和总大于单独的a[i]；

【代码】　　

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
const int maxn = 100010;
int n, a[maxn];
int d[maxn];
void dp()
{
    memset(d, 0, sizeof(d));
    d[0] = a[0];
    for(int i = 1; i < n; i++)
    {
        if(d[i-1] > 0) d[i] = d[i-1]+a[i];
        else d[i] = a[i];
    }
    //cout << endl;
    int st = 0, en = 0;
    int max_ = d[0];
    for(int i = 1; i < n; i++)
    {
        if(d[i]>max_)
        {
            max_ = d[i];
            en = i;
        }
    }
    st = en;
    for(int j = en-1; j>=0; j--)
    {
        if(d[j] < 0) break;
        else st = j;
    }
    printf("%d %d %d
", max_, st+1, en+1);
}

int main()
{
    int T; scanf("%d", &T);
    for(int kase = 0; kase < T; kase++)
    {
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        if(kase) printf("
");
        printf("Case %d:
", kase+1);
        dp();

    }
    return 0;
}