C++求解最大子序列和、位置问题

最新推荐文章于 2022-08-22 11:22:14 发布

转载最新推荐文章于 2022-08-22 11:22:14 发布 · 242 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Logic09/p/4146218.html

文章标签：

#c/c++

本文详细解析了最大连续子序列和问题的算法实现，通过一个具体例子说明如何找到具有最大元素和的连续子序列，并展示了一种解决此问题的C++代码实现。输入包含一系列整数，算法将输出最大子序列的和及其起始和结束位置。

Maximum Subsequence Sum

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

？？？？？？为什么测试是部分正确

// MaxSubSet.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include "stdio.h"
#include "stdlib.h"
#include <iostream>
//using namespace std;
void MaxSubsequenceSum(int a[],int n);
int _tmain(int argc, _TCHAR* argv[])
{
int n = 0;
scanf("%d",&n);
int *a = new int[n];
for(int i = 0;i<n;i++)
scanf("%d",&a[i]);
MaxSubsequenceSum(a,n);
system("pause");
return 0;

}
void MaxSubsequenceSum(int a[],int N)
{
int Thissum,Maxsum,i,j;

Maxsum =a[0];
int l = 0;
int n = 0;
int m= 0;
for(i = 0;i<N;i++)
{
if(a[i]<0)
l++;
Thissum =0;
for(j = i;j<N;j++)
{
Thissum += a[j];
if(Thissum > Maxsum)
{
Maxsum = Thissum;
n = i;
m = j;
}
}
}
if(l == N)
std::cout<<0<<std::endl;
else
std::cout<<Maxsum<<" "<<a[n]<<" "<<a[m]<<std::endl;
}