hdu1003 Max Sum

水题一枚，开始写出了一个复杂度n平方的代码，提交 果然tle，

然后观察数据 再优化了一些具体情况，还是tle，

这样的优化还是治标不治本啊，，

后来按捺不住我又看了别人的代码，，原来可以把复杂度弄成o（n）；

只需要遍历1次即可。

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158459    Accepted Submission(s): 37061

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

上代码

#include <iostream>#include <cstring>#include <cstdio>#define maxn 100010using namespace std;int num[maxn];int main(){    int t;    int cnt=0;    cin>>t;    while(t--)    {        memset(num,0,sizeof(num));        int n;        cin>>n;        for(int i=1;i<=n;i++)            cin>>num[i];        __int64 sum=0; int start=1;__int64 max=(-1)*maxn*1000;//因为可能都是负数，所以设置初始max尽可能的小

int st=1;int end=1;//初始化        for(int i=1;i<=n;i++)        {            if(sum>=0)                sum+=num[i];//如果之前的和大于等于0，继续加            else            {                sum=num[i];//否则，舍弃前面所有的数据，从这一项开始                st=i;//记录起始点            }            if(sum>max)            {                start=st;                max=sum;                end=i;            }        }        if(cnt>0)            cout<<endl;        cout<<"Case "<<++cnt<<':'<<endl;        cout<<max<<' '<<start<<' '<<end<<endl;            }    return 0;}