PAT甲级1007

Posted 2022-08-30 wx630c98f24f6b8

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

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 10010;
int N[maxn];
int d[maxn];
int main()

  int K;
  scanf("%d", &K);
  for (int i = 0; i < K; i++)
  
    scanf("%d", &N[i]);
  
  d[0] = N[0];//边界
  for (int i = 1; i < K; i++)
  
    d[i] = max(N[i], d[i - 1]+N[i]);//状态转移方程
  
  int k=0;//由于是d[i]表示下标为i的数为尾的最大连续子序列的和，也就是说N[i]必须在其中
  //那么d[K-1]不一定是最大的，需要对d[]进行遍历找最大的，并记录下标
  for (int i = 0; i < K; i++)
  
    if (d[i] > d[k])
    
      k = i;
    
  
  int sum = d[k];
  int j = k;//j用来求最大子序列和的第一个数
  if (sum != 0)//这个地方得注意下，若sum就等于0，j不需要自增
  
    while (sum != 0)
    
      sum -= N[j];
      j--;
    
    j++;
  
  if (d[k] < 0)
    printf("0 %d %d\\n", N[0], N[K-1]);
  else
    printf("%d %d %d\\n", d[k], N[j], N[k]);//注意题目说的是序列中的元素而不是下标
  return 0;