在线处理

 

  • 定义： 在线 的意思是指每输入一个数据就进行 即是处理 ，在任何一个地方终止输入，算法都能能正确给出当前解

  • 特点：数据只需一次扫描，便可给出结果，无需在主机内存进行存储；并且在任何时刻，此算法都能根据输入给出正确结果。
  • 此算法非常高效，时间复杂度仅为0(N)

 

 

例题： HDU 1003 Max Sum 

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 299181    Accepted Submission(s): 71008

 

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>
 
int main(){
	int t, n, maxLeft, maxRight, maxSum, temp;
	int thisLeft, thisSum;
	scanf("%d", &t);
	for(int id = 1; id <= t; ++id){
		scanf("%d", &n);
		scanf("%d", &maxSum);
		thisLeft = maxLeft = maxRight =  0;
		thisSum = maxSum;
		if(thisSum < 0){ thisSum = 0; thisLeft = 1; }
		for(int i = 1; i < n; ++i){
			scanf("%d", &temp);
			thisSum += temp;
			if(thisSum > maxSum){
				maxSum = thisSum;
				maxLeft = thisLeft;
				maxRight = i;
			}
			if(thisSum < 0){
				thisLeft = i + 1;
				thisSum = 0;
			}
		}
		printf("Case %d:\n%d %d %d\n", id, maxSum, maxLeft + 1, maxRight + 1);
		if(id != t) printf("\n");
	}
	return 0;
}