PTA-数据结构-01-复杂度2 Maximum Subsequence Sum

最新推荐文章于 2024-05-19 15:25:35 发布

sharemon

最新推荐文章于 2024-05-19 15:25:35 发布

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本文链接：https://blog.youkuaiyun.com/sharemon/article/details/99489545

本文详细介绍了一种寻找整数序列中最大子序列和的算法，通过实例解释了如何确定子序列的起始和结束位置，同时处理了所有元素为负数的特殊情况。输入包括整数序列的长度和具体数值，输出则是最大子序列的和及其首尾元素。

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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

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int n;
    scanf("%d", &n);

    int * a = (int *)malloc(sizeof(int) * n);
    int i;
    for (i=0;i<n;i++)
    {
        scanf("%d", a+i);
    }

    int maxSum = -1;
    int thisSum = 0;
    int tryStart = 0;
    int start = 0;
    int end = 0;
    for (i=0;i<n;i++)
    {
        thisSum += a[i];
        if (thisSum > maxSum)
        {
            maxSum = thisSum;
            start = tryStart;
            end = i;
        }
        if (thisSum < 0)
        {
            thisSum = 0;
            tryStart = i+1;
        }
    }

    if (maxSum == -1)
    {
        start = 0;
        end = n-1;
        maxSum = 0;
    }

    printf("%d %d %d", maxSum, a[start], a[end]);

    return 0;
}