Max Sum（杭电OJ1003）解题报告

Max Sum

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

解题思路：刚开始用最笨的方法试了一下，暴力搜索，全部走一边，时间花太多

了，超时了，对此改进了搜索的方式，只要两个循环就搞定了，时间加快了，但是相
对于来说还不是最快的，还是会超时的，一个个向后面走，只要和大于Max的，就记
录的，代码如下：
#include<iostream>
usingnamespacestd;
intmain()
{
intn;
cin>>n;
intpoint1,point2;
intstep=1;
while(n)
{
intmax=-99999;
intm,data[100000];
cin>>m;
for(inti=1;i<=m;i++)
cin>>data[i];
for(i=1;i<=m;i++)
{
intsum=0;
for(intj=i;j<m;j++)
{
sum+=data[j];
if(sum>max)
{
max=sum;
point1=i;
point2=j;
}
}
}
if(step!=1)
cout<<endl;
cout<<"Case"<<step<<":"<<endl;
cout<<max<<""<<point1<<""<<point2<<endl;
step++;
n--;
}
return0;
}
交上去还是超时的，最近用到动态规划，对算法进行改进了，起始点还是从头开始的，

一直到后面搜索，一直和为小于零，起始点就从开始小于零的后一位开始并把sum改为
零，再搜索的过程中，一遇到大的数据就记录下来，把其计为起始点和终点的，这里
面主要考虑到，当你搜索到一个位置的，它的和不小于零的，那对于后面来说，加上
去还是会变大的，不会给变小的，所以要再搜索下去的，走一边就KO了。代码如下：
#include<iostream>
usingnamespacestd;
#defineMin-999999
intmain()
{
intdata[100000],start,end;
intm;
intstep=1;
cin>>m;
while(m--)
{
intn;
cin>>n;
for(inti=1;i<=n;i++)
cin>>data[i];
intmax=Min;
intk=1;
intsum=0;
for(i=1;i<=n;i++)
{
sum=sum+data[i];
if(sum>max)
{
max=sum;
start=k;
end=i;
}
if(sum<0)
{
sum=0;
k=i+1;
}
}
if(step!=1)
cout<<endl;
cout<<"Case"<<step<<":"<<endl;
cout<<max<<""<<start<<""<<end<<endl;
step++;
}
return0;
}