Maximum Subsequence Sum

最新推荐文章于 2024-03-17 14:54:33 发布

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最新推荐文章于 2024-03-17 14:54:33 发布

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本文探讨了最大子序列和问题的解决方法，提供了一个高效的算法实现，并通过具体例子进行了说明。该算法能够处理大规模数据集，适用于不同规模的整数序列。

最大子序列问题

给定K个整数组成的序列{ N1, N2 , ..., NK}，“连续子列”被定义为{ Ni , Ni+1 , ..., Nj }，其中 1≤i≤j≤K。“最大子列和”则被定义为所有连续子列元素的和中最大者。例如给定序列{ -2, 11, -4, 13, -5, -2 }，其连续子列{ 11, -4, 13 }有最大的和20。现要求你编写程序，计算给定整数序列的最大子列和。

本题旨在测试各种不同的算法在各种数据情况下的表现。各组测试数据特点如下：

数据1：与样例等价，测试基本正确性；
数据2：102个随机整数；
数据3：103个随机整数；
数据4：104个随机整数；

数据5：105个随机整数；

输入格式：

输入第1行给出正整数K (≤100000)；第2行给出K个整数，其间以空格分隔。

输出格式：

在一行中输出最大子列和。如果序列中所有整数皆为负数，则输出0。

输入样例：

-2 11 -4 13 -5 -2

输出样例：

代码如下：

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

int MaxSubseqSum4(int A[], int N);

int main(){
	int N;
	scanf("%d", &N);
	int *A = (int*)malloc(sizeof(int)*N);
	
	for(int i = 0; i < N; i++) {
		scanf("%d",A+i);
	} 
	printf("%d", MaxSubseqSum4( A, N ));
	
	free(A);
	return 0;
}

int MaxSubseqSum4( int A[], int N )  
{   int ThisSum, MaxSum;
    int i;
    ThisSum = MaxSum = 0;
    for( i = 0; i < N; i++ ) {
          ThisSum += A[i]; /* 向右累加 */
          if( ThisSum > MaxSum )
                  MaxSum = ThisSum; /* 发现更大和则更新当前结果 */
          else if( ThisSum < 0 ) /* 如果当前子列和为负 */
                  ThisSum = 0; /* 则不可能使后面的部分和增大，抛弃之 */
    }
    return MaxSum;  
}

Maximum Subsequence Sum

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

代码：

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

int main(){
	int N;
	scanf("%d", &N);
	int *A = (int*)malloc(sizeof(int)*N);
	int isAllNegative = 1; 
	
	int min = 0, max = 0, zancun = 0;
	
	int ThisSum, MaxSum;
    int i;
    ThisSum = MaxSum = 0;
	
	
	for(int i = 0; i < N; i++) {
		scanf("%d",A+i);
		if(*(A+i) >= 0)
			isAllNegative = 0;
	}
	
	if(isAllNegative == 1){
		printf("%s %d %d", "0", *(A), *(A+(N-1)));
		return 0;
	}
	
	for( i = 0; i < N; i++ ) {
			
          ThisSum += A[i]; /* 向右累加 */
          if( ThisSum > MaxSum ){
          	MaxSum = ThisSum; /* 发现更大和则更新当前结果 */
          	max = i;
          	min = zancun;
          }
          else if(ThisSum == MaxSum && ThisSum == 0){
          	min = zancun;
          	max = i;
          }
                  
          else if( ThisSum < 0 ){/* 如果当前子列和为负 */
          	ThisSum = 0;/* 则不可能使后面的部分和增大，抛弃之 */
          	zancun = i + 1;
		  } 
                   
    }
	printf("%d %d %d", MaxSum, *(A+min), *(A+max));
	
	free(A);
	return 0;
}