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最大子段和

时间:2015-08-11 09:52:04      阅读:111      评论:0      收藏:0      [点我收藏+]

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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:求最大的子段和,和它的起始位置和终止位置。
 
 
 
解题思路:
     在输入的时候,就可以开始求最大子段和了。当前子段和的值为负值时,就没必要再往下进行,而应将下一位数作为子段的第一位。也就是说,当处理第i个数时,如果以第i-1个数为结尾的子段的和为正数,则不必将第i个数作为新子段的首位进行枚举,因为新子段加上前面子段和所得的正数,一定能得到更大的子段和。然后在标记下位置就好了....
 
 
代码如下:
 
 
 1 #include<stdio.h>
 2 int a[100005];
 3 int main()
 4 {
 5     int T;
 6     int N=0;
 7     scanf("%d",&T);
 8     while(T--)
 9     {
10         N++;
11         int n,max=-1010,l=0,r=0,sum=0,temp=1;
12         scanf("%d",&n);
13         for(int i=1; i<=n; i++)
14         {
15             scanf("%d",&a[i]);
16             sum+=a[i];
17             if(sum>max)
18             {
19                 max=sum;
20                 l=temp;
21                 r=i;
22             }
23             if(sum<0)
24             {
25                 sum=0;
26                 temp=i+1;
27             }
28         }
29         printf("Case %d:\n%d %d %d\n",N,max,l,r);
30         if(T)
31             printf("\n");
32     }
33 }

 

最大子段和

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原文地址:http://www.cnblogs.com/huangguodong/p/4719993.html

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