【杭电OJ1003】Max Sum

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最新推荐文章于 2025-01-22 21:38:14 发布

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本文探讨了最大子序列和问题，这是一种常见的编程竞赛题目，旨在寻找数组中具有最大和的连续子序列。文章提供了两种解决方案：一种是时间复杂度为O(n^3)的暴力穷举法，另一种是时间复杂度为O(n)的动态规划法，后者更为高效。通过实例输入和输出，详细解释了如何实现这些算法。

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 331336 Accepted Submission(s): 78887

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input

2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5

Sample Output

Case 1: 14 1 4 Case 2: 7 1 6

#include<stdio.h>
#define maxn 100000
/*
	暴力穷举 （时间复杂度n的3次方）
    效率低下，运行超时
*/
int main(){
	int n,num,i,j,k,count=1;
	int a[maxn];
	scanf("%d",&n);
	while(n--){
		
		scanf("%d",&num);
		
		for(i=0;i<num;i++)
			scanf("%d",&a[i]);
			
		int maxsum = -9999999;
		int start = 0, end = 0;
	
		for(i=0;i<num;i++)	//Time complexity: O(n^3)
			{
				for(j=i;j<num;j++)	// pay attention this: j starts from i not i+1
				{	
					int sum = 0;
					for(k=i;k<=j;k++)
					{
						sum += a[k];
					}
					if(sum>maxsum)
					{
						maxsum = sum;
						start = i;
						end = j;
					}
				}
	 
			}
		printf("Case %d:\n",count++);
		printf("%d %d %d\n",maxsum,start+1,end+1);
	}
}

#include<stdio.h>
/*
	动态规划 
*/
int main(){
	int n,num,i,j,k,count=1;
	int a;
	scanf("%d",&n);
	while(n--){
		
		scanf("%d", &num);
		int maxsum = -99999,start = 0,end = 0,sum=0,temp=0;
		for(i = 0;i < num; i++)
		{
			scanf("%d", &a);
			sum += a;
			if(sum > maxsum){
				maxsum = sum;
				start = temp;
				end = i;
			}
			if(sum<0){ //小于零时特殊处理
				sum = 0;
				temp =	i+1;
			}
		} 
		printf("Case %d:\n",count++);
		printf("%d %d %d\n",maxsum,start+1,end+1);
		if(n){
			printf("\n");
		}
	}
}