菜鸟上路，杭电OJ1003 最大子串问题

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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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

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

我做这道题时试了三种做法：

前两种分别是：

把输入存进数组，再遍历，不用说，超时！

第二种，把输入存入数组，从头开始，对于sum大于0者，逐个往后加上去，对于出现sum小于0者，则放弃当前子串，sum置为0，从当前子串的下一个位置开始算sum。

问题是还是超时，因为初始化和存入数组用了太多时间了！

第三种直接逐个处理输入的数字，同步更新起始位置，终止位置以及sum和max，就AC了。可见有时候在考虑一下又能节省出许多时间！！

代码：

#include <iostream>using namespace std;int main(){int j;int n,m;int s=0;int temp=0;int start,end;int max=-1001;int sum=0;int q=1;cin>>n;while(n--){sum=0;temp=0;max=-1001;s=0;cin>>m;while(m--){cin>>j;sum+=j;if(sum>max){max=sum;start=s;end=s+temp;}temp++;if(sum<0){sum=0;s=s+temp;temp=0;}}if(q!=1)cout<<endl;cout<<"Case "<<q++<<":"<<endl;cout<<max<<' '<<start+1<<' '<<end+1<<endl;;}}

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