PTA---01-复杂度2 Maximum Subsequence Sum (25分)

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

---

利用在线处理的方法，T(N) = O(N)。

完整代码

#include<stdio.h>
#define size 10000

int main(){
	int n, i, a[size];
	while(scanf("%d", &n) != EOF){  //while语句可测试多组数据
		int left = 0, right = 0, temp = 0;//left为左下标，right为右下标，temp为临时左下标
		int thisSum = 0, maxSum = -1;	//thisSum当前累加和，maxSum最大子列和，设置为负数便于区分后续两种结果的情况
		for( i = 0; i < n; i++ ){//录入数据
			scanf("%d", &a[i]);
		}
		for( i = 0; i < n; i++ ){//在线处理
			thisSum += a[i];
			if(thisSum < 0){
				thisSum = 0;
				temp = i+1;//临时左下标
			}else if(thisSum > maxSum){
			maxSum = thisSum;
			left = temp;//更新左下标，若不更新，即使最大子列和已经找到，随着之后循环的进行，可能会出现thisSum<0的情况，使得temp改变
			right = i;	//更新右下标
			}
		}
		if(maxSum < 0) printf("0 %d %d\n", a[0], a[n-1]);
		else printf("%d %d %d\n", maxSum, a[left], a[right]);
	}
	return 0;
}