pat PAT (Advanced Level) Practise 1007. Maximum Subsequence Sum (25)

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

在线处理，用一个ts保存临时的最大子列和区间的左端点，s为确定的最终左端点，e为右端点

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

const int maxn = 10000, INF = 0x7fffffff;
int a[maxn + 10];

int main()
{
	int K;
	cin >> K;
	bool check = false;
	for (int i = 0; i < K; i++) {
		scanf("%d", &a[i]);
		if (a[i] >= 0) check = true;
	}
	if (!check) {
		printf("0 %d %d\n", a[0], a[K - 1]);
		return 0;
	}
	int s = a[0], e = a[0], ts = a[0];
	int tsum = 0, maxx = -INF;
	bool flag = false;
	for (int i = 0; i < K; i++) {
		//flag == true时说明上一次迭代的时候tsum < 0了，要重新计算最大子列和s
		if (flag) {
			ts = a[i]; //计算新的子列和，取新的左端点
			flag = false; //注意解除标记
		}
		tsum += a[i];
		if (maxx < tsum) {
			s = ts; //更新最终的左端点s
			e = a[i]; //更新右端点e
			maxx = tsum;
		}
		if (tsum < 0) { //debug 不要放到if(maxx < tsum)上面
			flag = true;
			tsum = 0;
		}
	}
	printf("%d %d %d\n", maxx, s, e);
	return 0;
}