patest_1007_Maximum Subsequence Sum_(dp)(思维)

1007. Maximum Subsequence Sum (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given a sequence of K integers { N₁, N₂, ..., N_K }. A continuous subsequence is defined to be { N_i, N_i+1, ..., N_j } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

---

这题做了好几次了。。。这次依然没有自己想到，但其实挺简单的。

思路：仔细一想也是动态规划的思想，从num[1]-->num[n]，维护一个连续区间最大值{sum,l,r}。

#include<bits/stdc++.h>
using namespace std;
#define N 10005

int num[N];

int main()
{
  int n;
  scanf("%d",&n);
  for(int i=;i<n;i++)
    scanf("%d",&num[i]);
  int l=,r=,resl,resr,res=-,tmp=;
  for(int i=;i<n;i++)
  {
    tmp+=num[i];
    r=i;
    if(tmp<) //如果求和的值tmp<0,那么弃掉前面的求和，在继续递推
    {
      l=r=i+;
      tmp=;
    }
    else   //如果tmp>0，那么后面可能找到更大的求和
    {
      if(tmp>res)
      {
        res=tmp;
        resl=l;
        resr=r;
      }
    }
  }
  if(res==-)
    printf("%d %d %d\n",,num[],num[n-]);
  else
    printf("%d %d %d\n",res,num[resl],num[resr]);
  return ;
}