PAT甲级 1007 Maximum Subsequence Sum (25 分)

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本文探讨了如何寻找一组整数中的最大连续子序列及其总和，通过动态规划方法解决了这一经典问题，并提供了两种不同的实现方式，对比了它们在特定测试点上的表现。

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

动态规划第一次试水，要注意数组和边界条件，但是自己写的发现有3个测试点一直过不去。

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 10010;
const int INF = -1000000000;
int N;
int _left = 0, _right = 0;
int max_dp;
int num[maxn];
int dp[maxn];

int main() {
	/*
	10
	-10 5 2 3 5 -5 -23 3 7 -21
	*/
	cin >> N;
	fill(num, num + maxn, INF);
	fill(dp, dp + maxn, INF);
	for (int i = 0; i < N; i++) {
		cin >> num[i];
	}
	dp[0] = num[0];
	max_dp = 0;
	for (int i = 1; i < N; i++) {
		int flag = 0; //flag == 0 表示累积最大值
		if (dp[i - 1] > 0) //累积
		{
			dp[i] = dp[i - 1] + num[i];
			flag = 0;
		}
		else
		{
			dp[i] = num[i];
			flag = 1;
		}
		if (dp[i] > dp[max_dp]) {
			max_dp = i;
			if (flag)
			{
				_left = num[i];
			}

		}
	}
	if(dp[max_dp] >= 0)
		cout << dp[max_dp] << " " << _left << " " << num[max_dp];
	else
		cout << 0 << " " << num[0] << " " << num[N-1];

	return 0;
}

但是更改记录数组下标的方式，可以通过测试。百思不得其解。

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 10010;
const int INF = -1000000000;
int N;
int _left = 0, _right = 0;
int max_dp;
int num[maxn];
int dp[maxn];
int s[maxn] = {0};

int main() {
	/*
	10
	-10 5 2 3 5 -5 -23 3 7 -21
	*/
	cin >> N;
	fill(num, num + maxn, INF);
	fill(dp, dp + maxn, INF);
	for (int i = 0; i < N; i++) {
		cin >> num[i];
	}
	dp[0] = num[0];
	max_dp = 0;
	s[0] = num[0]; //s存储的是左端点的值
	for (int i = 1; i < N; i++) {
		if (dp[i - 1] > 0) //累积
		{
			dp[i] = dp[i - 1] + num[i];
			s[i] = s[i - 1];
		}
		else
		{
			dp[i] = num[i];
			s[i] = num[i];
		}

	}
	for (int i = 1; i < N; i++) {
		if (dp[i] > dp[max_dp]) {
			max_dp = i;
		}
	}
	if(dp[max_dp] >= 0)
		cout << dp[max_dp] << " " <<s[max_dp]<< " " << num[max_dp];
	else
		cout << 0 << " " << num[0] << " " << num[N-1];

	return 0;
}