动态规划——A 最大子段和

A - 最大子段和

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

 

解题思路：

题目意思：最大子段和是要找出由数组成的一维数组中和最大的连续子序列。比如{5,-3,4,2}的最大子序列就是 {5,-3,4,2}，它的和是8,达到最大；而 {5,-6,4,2}的最大子序列是{4,2}，它的和是6。看的出来了，找最大子序列的方法很简单，只要前i项的和还没有小于0那么子序列就一直向后扩展，否则丢弃之前的子序列开始新的子序列，同时我们要记下各个子序列的和，最后找到和最大的子序列

 

程序代码：

 #include <cstdio>
 using namespace std;
 const int N=;
 int a[N],n,b[N],num[N];
 long long sum;
 int q;
 int main()
 {
     int t,Case=;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         for(int i=;i<=n;i++)
                 scanf("%d",&a[i]);
         b[]=a[];num[]=;
         for(int i=;i<=n;i++)
         {
             if(b[i-]>=)
                {
                   b[i]=b[i-]+a[i];
                   num[i]=num[i-];
                }
              else
              {
                  b[i]=a[i];
                  num[i]=i;
              }
         }
         sum=b[];q=;
         for(int i=;i<=n;i++)
         {
             if(b[i]>sum)
             {
                 sum=b[i];
                 q=i;
             }
         }
         printf("Case %d:\n",++Case);
         printf("%lld %d %d\n",sum,num[q],q);
         if(t) printf("\n");
     }
     return ;
 }