PAT (Advanced Level) Practise 1007 Maximum Subsequence Sum

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最新推荐文章于 2020-06-06 16:33:06 发布

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本文探讨了最大子序列和问题的解决方案，该问题是寻找给定整数序列中具有最大和的连续子序列。通过一个具体示例展示了如何确定最大子序列及其总和，并提供了完整的C++代码实现。

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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

题意：给出一串数字，问连续的一串数字最大是多少，若最大值为负值，则输出0，并输出这串数字头尾的数字

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>

using namespace std;

int n,l,r,a[100009];

int main()
{
    scanf("%d",&n);
    int flag=0,ans=-1;
    for(int i = 0; i < n; i++)
    {
        scanf("%d",&a[i]);
        if(a[i]>=0) flag=1;
    }
    if(!flag) printf("0 %d %d\n",a[0],a[n-1]);
    else
    {
        for(int i=0;i<n;i++)
        {
            int sum=0,j;
            for(j=i;j<n;j++)
            {
                if((sum+=a[j])<0) break;
                if(sum>ans)
                {
                    ans=sum;
                    l=a[i];
                    r=a[j];
                }
            }
            i=j;
        }
        printf("%d %d %d\n",ans,l,r);
    }
    return 0;
}