HDU 1003 - Max Sum（难度：*）

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 = ;
 int n, a[maxn];
 int d[maxn];
 void dp()
 {
     memset(d, , sizeof(d));
     d[] = a[];
     for(int i = ; i < n; i++)
     {
         if(d[i-] > ) d[i] = d[i-]+a[i];
         else d[i] = a[i];
     }
     //cout << endl;
     int st = , en = ;
     int max_ = d[];
     for(int i = ; i < n; i++)
     {
         if(d[i]>max_)
         {
             max_ = d[i];
             en = i;
         }
     }
     st = en;
     for(int j = en-; j>=; j--)
     {
         if(d[j] < ) break;
         else st = j;
     }
     printf("%d %d %d\n", max_, st+, en+);
 }

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

     }
     return ;
 }