HDU-1003 Max Sum

最大子序列和问题解析

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本文介绍了一个经典的编程挑战——最大子序列和问题，并提供了一种有效的算法解决方案。该问题要求从给定的一系列整数中找出具有最大和的连续子序列及其起始和结束位置。文章通过一个具体示例解释了算法的工作原理，并附带了完整的源代码。

A - Max Sum
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit

Status

Practice

HDU 1003
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>
int main()
{
	int a[100100],t,n,sum,f,ans,k,z,w=0,c;
	scanf("%d",&t);
	c=t;
	while(t--){
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
		sum=0;k=f=z=1;ans=a[1];
		for(int i=1;i<=n;i++){
			if(sum>=0)
				sum+=a[i];
			else{
				sum=a[i];
				k=i;
			}
			if(ans<sum){
				ans=sum;
				z=i;
				f=k;
			}
		}
		printf("Case %d:\n%d %d %d\n",++w,ans,f,z);
		if(w<c) printf("\n");
	}
}