PAT甲级 -- 1007 Maximum Subsequence Sum (25 分)

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博客围绕最大子序列问题展开，给定整数序列，需找出最大子序列的和以及首尾数字。介绍了输入输出规范，还提到运行代码得9分但不知错误原因，看了博主答案也未发现差别。

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 Knumbers, 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

9分:

跑出来结果也没问题，PAT不给测试样例，不知道自己出错在哪...

#include <iostream>
#include <cmath>
using namespace std;

const int maxn = 10010;
int k; //k个元素
int a[maxn]; //存放序列
int dp[maxn]; //以a[i]结尾的最大和

struct node
{
	int start, end;
}index[maxn];

int count_negative = 0, count_0 = 0;


int main()
{
	scanf("%d", &k);
	for(int i = 0; i < k; i++)
	{
		scanf("%d", &a[i]);
		if(a[i] < 0) count_negative++;
		else if(a[i] == 0) count_0++;
	}

	dp[0] = a[0];
	index[0].start = 0, index[0].end = 0;
	for(int i = 1; i < k; i++)
	{
		if (dp[i-1] + a[i] >= a[i])
		{
			dp[i] = dp[i-1] + a[i];
			index[i].start = index[i-1].start, index[i].end = i;
		}else
		{
			dp[i] = a[i];
			index[i].start = i, index[i].end = i;
		}
	}
	int j = 0;
	for(int i = 1; i < k; i++)
	{
		if(dp[i] > dp[j])
		{
			j = i;
		}else if(dp[j] == dp[i])
		{
			if(index[j].start == index[i].start)
			{
				if(index[j].end > index[i].end)
				{
					j = i;
				}
			}else if(index[j].start > index[i].start)
			{
				j = i;
			}
		}
	}

	if(dp[j] <= 0)
	{
		if(count_negative == k)
		{
			printf("0 %d %d", a[0], a[k-1]);
		}else if(count_negative + count_0 == k)
		{
			printf("0 0 0");
		}
	}else
	{
		printf("%d %d %d", dp[j], index[j].start, index[j].end);
	}

	
	return 0;
}

正确code：

看了博主的答案，还是没看出来差别在哪，求告知啊啊啊啊！！！

#include<iostream>
using namespace std;
int main()
{
	int N;
	cin>>N;
	int in[N];
	int result[N];
	int dp[N];
	for(int i=0;i<N;++i)
		cin>>in[i];
	dp[0]=max(in[0],0);         //动态规划，初始状态定义
	for(int i=1;i<N;++i){        
		if(dp[i-1]+in[i] < 0)
			dp[i]=max(in[i],0);
		else
			dp[i]=dp[i-1]+in[i];
	}
    //至此，dp（dynamic programming）数组结束
 
	int maxid=0;
	for(int i=0;i<N;++i)   //这个循环：找最大的序列和
	{
		if(dp[i]>dp[maxid])
			maxid=i;
	}
	if(dp[maxid]>0){
		int temp=maxid;
		int i=0;
		while(dp[temp]>0){
			result[i]=in[temp];
			--temp;
			++i;
		}
		cout<<dp[maxid]<<" "<<result[i-1]<<" "<<result[0]<<endl;
	}
	else{
		int j;
		for(j=0;j<N;++j)
		{
			if(in[j]==0)	
				break;
		}
		if(j<N-1||in[N-1]==0)
			cout<<"0 0 0"<<endl;
		else
			cout<<"0 "<<in[0]<<" "<<in[N-1]<<endl;
	}
	return 0;
}
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作者：天啊野 
来源：优快云 
原文：https://blog.youkuaiyun.com/weixin_38097576/article/details/82715413 
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