Given a sequence of K integers N 1 , N 2 , . . . , N K { N_1, N_2, ..., N_K } N1,N2,...,NK. A continuous subsequence is defined to be N i , N i + 1 , . . . , N j { N_i, N_{i+1}, ..., N_j } Ni,Ni+1,...,Nj where 1 ≤ i ≤ j ≤ K 1≤i≤j≤K 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 { -2, 11, -4, 13, -5, -2 } −2,11,−4,13,−5,−2, its maximum subsequence is 11 , − 4 , 13 { 11, -4, 13 } 11,−4,13 with the largest sum being 20 20 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 ) K (≤10000) K(≤10000). The second line contains K K 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 i i and j j j (as shown by the sample case). If all the K K 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
题解:
最大子列求和问题,只需进行简单改编即可,注意题目中对于最大和为负情况的处理;
附:子序列问题参考
#include <iostream>
using namespace std;
int main()
{
int k;
cin >> k;
int *arr = new int[k];
int ThisSum = 0, MaxSum = -1; // 最小值为 0,故初始值应为负
int first, last;
int c = 0;
for (int i = 0; i < k; i++)
cin >> arr[i];
for (int i = 0; i < k; i++)
{
ThisSum += arr[i];
c++;
if (ThisSum > MaxSum)
{
MaxSum = ThisSum;
first = arr[i - c + 1];
last = arr[i];
}
else if (ThisSum < 0)
{
ThisSum = 0;
c = 0;
}
}
if (MaxSum == -1)
cout << "0" << " " << arr[0] << " " << arr[k-1] << endl;
else
cout << MaxSum << " " << first << " " << last << endl;
}