[PAT-Advanced] A1007 Maximum Subsequence Sum (25)

原创已于 2022-10-06 16:02:59 修改 · 255 阅读

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于 2019-12-27 10:53:00 首次发布

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本文详细解析了PAT甲级A1007题目——寻找连续子序列中和最大的算法实现。通过动态规划方法，文章提供了C++代码示例，展示了如何找到具有最大和的连续子序列及其边界元素。

文章目录

总题解目录
A1007 Maximum Subsequence Sum (25point(s))
Input Specification:
Output Specification:
Sample Input:
Sample Output:
Analysis
C++ Code

总题解目录

[PAT- Advanced Level] 甲级题解目录（Advanced Level）

A1007 Maximum Subsequence Sum (25point(s))

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

Analysis

求最长子序列和。如果输入全为负数，则结果要求输出的和为0，并输出整个数组的首尾数据。
使用DP思想。只要所有dp的值都是负数，就可以肯定整个数组中的数据都是负数。
注意读题，最后输出的是对应位置数据的值，而不是位置!

C++ Code

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <string.h>
#define MAX_K 100000

using namespace std;


int main(){
    int k, end_index = 0, max_sum=-1;	// max_sum设置为-1非常关键,若设置为0，过不了测试点5(例：-2 0 0 0 -1)
    int max_subseq_sum[MAX_K];  // 存放到第i个位置为止的最大子序列和
    int start_idx[MAX_K];       // 用于存放到第i个位置为止的最大子序列的起始位置
    int data[MAX_K];            // 存放输入的数据
	memset(max_subseq_sum, 0, sizeof(max_subseq_sum));  // 数组初始置0

    cin >> k;

    for(int i = 0; i < k; i++){
        cin >> data[i];
        if(i == 0){
            max_subseq_sum[i] = data[i];
            start_idx[i] = i;
        }
        else{	// 到第i个位置为止的最大子序列和=max(data[i],max_subseq_sum[i-1]+data[i])
            if(data[i] > max_subseq_sum[i-1] + data[i]){	// data[i]较大时，i作为序列起始位置
                max_subseq_sum[i] = data[i];
                start_idx[i] = i;
            }
            else{	// max_subseq_sum[i-1]+data[i]较大时，起始位置与到i-1为止的最大子序列和的起始位置的相同
                max_subseq_sum[i] = max_subseq_sum[i-1] + data[i];
                start_idx[i] = start_idx[i-1];
            }
        }
        if(max_subseq_sum[i] > max_sum){    	// 找到最大值的结束位置
            max_sum = max_subseq_sum[i];
            end_index = i;
        }
    }

    if(max_subseq_sum[end_index] < 0)
        printf("0 %d %d", data[0], data[k-1]);
    else
        printf("%d %d %d", max_sum, data[start_idx[end_index]], data[end_index]);

	return 0;
}