PAT 1007 Maximum Subsequence Sum

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最新推荐文章于 2024-12-16 14:46:31 发布

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分类专栏： PAT甲级文章标签：最大子序列和

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本文探讨了最大子序列和问题，这是一种常见的编程面试题，旨在寻找连续子序列中元素和最大的部分。通过分析给定整数序列的例子，如{-2,11,-4,13,-5,-2}

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1007 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 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

#include <iostream>
using namespace std;
int main(){
	int n,ans=0;
	cin>>n;
	int *k=new int[n];
	for(int i=0; i<n; ++i){
		cin>>k[i];
		if(k[i]>=0){
			ans=1;
		}
	}
	if(!ans){
		cout<<"0 "<<k[0]<<" "<<k[n-1];
	}
	else{
		//temp相当于一个先行哨兵，当当前序列已经可以被舍弃时先记录另外一个正序列的开端
		//从这里开始试着计算后面的子序列和，如果出现了大于当前最大子序列和的子串就把start更新为
		//哨兵的值，然后更新最大子序列和的记录值，end也随之更新 
		int maxn = -1,sum = 0,start = 0,temp = 0,end = 0;  //这个地方有必要让maxn=-1,不然对于-1 0 -1这样的序列就不行了，无法更新start和end的值 
		for(int i=0; i<n; ++i){
			sum+=k[i];
			if(sum<0){
				sum=0;
				temp = i+1;
			}
			else if(sum>maxn){
				maxn = sum;
				end = i;
				start = temp;
			}
		}
		cout<<maxn<<" "<<k[start]<<" "<<k[end];
	}
	return 0;
}