Problem Description
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } 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
Solution
这里使用第二种算法，因为需要确定最大子列和的首尾元素，也算是在线处理的思想。
因为存在只有负数和零的情况，MaxSum 不能再初始化为 0 ，否则无法特判出该情况。
#include <stdio.h>
int main()
{
int i, j, N, A[10001], ThisSum, MaxSum, left, right;
scanf("%d", &N);
for (i = 0; i < N; i++) scanf("%d", &A[i]);
MaxSum = -1; // initialize MaxSum to -1, because there are only negative numbers and zeros situations
for (i = 0; i < N; i++) {
ThisSum = 0;
for (j = i; j < N; j++) {
ThisSum += A[j];
if (ThisSum > MaxSum) {
MaxSum = ThisSum;
left = A[i];
right = A[j];
}
}
}
if (MaxSum < 0) printf("%d %d %d", 0, A[0], A[N - 1]); // if there are only negative numbers and zeros, MaxSum will always be -1
else printf("%d %d %d", MaxSum, left, right);
}
