最大子段和问题(C/C++)

Description

给定有n个整数(可能为负整数)组成的序列a1,a2,...,an,求该序列连续的子段和的最大值。 如果该子段的所有元素和是负整数时定义其最大子段和为0。

Input

第一行有一个正整数n(n<1000)，后面跟n个整数,绝对值都小于10000。直到文件结束。

Output

输出它的最大子段和。

Sample Input

6 -2 11 -4 13 -5 -2

Sample Output

20

参考: https://blog.csdn.net/niteip/article/details/7444973#

1. 穷举法 时间复杂度：$O(n^3)$

   /*O(n^3)*/
   #include <stdio.h>
   #include <stdlib.h>
   
   int main()
   {
       int n;
       int num[1001];
       int i, j, k;
       int sum, max;
       while (~scanf("%d", &n))
       {
           for (i = 0; i < n; i++)
               scanf("%d", &num[i]);
           max = 0;
           for (i = 0; i < n; i++)
           {
               for (j = i; j < n; j++)
               {
                   sum = 0;
                   for (k = i; k <= j; k++)
                       sum += num[k];
                   if (sum > max)
                       max = sum;
               }
           }
           printf("%d\n", max);
       }
       return 0;
   }
   

2. 穷举法

   时间复杂度：$O(n^2)$

   /*O(n^2)*/
   #include <stdio.h>
   #include <stdlib.h>
   
   int main()
   {
       int n;
       int num[1001];
       int i, j, k;
       int sum, max;
       while (~scanf("%d", &n))
       {
           for (i = 0; i < n; i++)
               scanf("%d", &num[i]);
           max = 0;
           for (i = 0; i < n; i++)
           {
               sum = 0;
               for (j = i; j < n; j++)
               {
                   sum += num[j];
                   if (sum > max)
                       max = sum;
               }
           }
           printf("%d\n", max);
       }
       return 0;
   }
   

3. 分治法

   时间复杂度：$O(nlog_2n)$

   /*O(nlogn)分治法*/
   #include <iostream>
   #include <cstdio>
   #include <cstdlib>
   
   int maxSubSegSum(int num[], int left, int right)
   {
       int mid = (left + right) / 2;
   
       if (left == right)
           return num[left] > 0 ? num[left] : 0;
   
       int leftMaxSum = maxSubSegSum(num, left, mid);
       int rightMaxSum = maxSubSegSum(num, mid + 1, right);
   
       int sum = 0;
       int leftSum = 0;
       for (int i = mid; i >= left; i--)
       {
           sum += num[i];
           if (sum > leftSum)
               leftSum = sum;
       }
   
       sum = 0;
       int rightSum = 0;
       for (int i = mid + 1; i <= right; i++)
       {
           sum += num[i];
           if (sum > rightSum)
               rightSum = sum;
       }
   
       int retSum = leftSum + rightSum;
       if (retSum < leftMaxSum)
           retSum = leftMaxSum;
       if (retSum < rightMaxSum)
           retSum = rightMaxSum;
       return retSum;
   }
   
   int main()
   {
       int n;
       int num[1001];
   
       while (~scanf("%d", &n))
       {
           for (int i = 0; i < n; i++)
               scanf("%d", &num[i]);
           printf("%d\n", maxSubSegSum(num, 0, n - 1));
       }
       return 0;
   }
   

4. 动态规划

   时间复杂度：$O(n)$

   /*O(n) 动态规划*/
   #include <stdio.h>
   #include <stdlib.h>
   #include <memory.h>
   
   int main()
   {
       int n;
       int num[1001];
       int f[1001 + 1];
       int i;
       int max;
       while (~scanf("%d", &n))
       {
           for (i = 1; i <= n; i++)
               scanf("%d", &num[i]);
           max = 0;
           memset(f, 0, sizeof(f));
           for (i = 1; i <= n; i++)
           {
               if (f[i - 1] > 0)
                   f[i] = f[i - 1] + num[i];
               else
                   f[i] = num[i];
               if (f[i] > max)
                   max = f[i];
           }
           printf("%d\n", max);
       }
       return 0;
   }
   

   或者

   /*O(n) 动态规划*/
   #include <stdio.h>
   #include <stdlib.h>
   #include <memory.h>
   
   int main()
   {
       int n;
       int num[1001];
       int b;
       int i;
       int max;
       while (~scanf("%d", &n))
       {
           for (i = 1; i <= n; i++)
               scanf("%d", &num[i]);
           max = 0;
           b = 0;
           for (i = 1; i <= n; i++)
           {
               b = b > 0 ? b + num[i] : num[i];
               max = b > max ? b : max;
           }
           printf("%d\n", max);
       }
       return 0;
   }