[ACM] hdu 1003 Max Sum（最大子段和模型）
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 135262 Accepted Submission(s): 31311
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
Author
Ignatius.L
解题思路：
当全部数都为负数时，最大子段和为0.
int MaxSum(int num[],int n)
{
int sum=0,b=0;
int i;
for (i=1;i<=n;i++)
{
if(b>0)
b+=num[i];
else
b=num[i];
if(b>sum)
sum=b;
}
return sum;
}
仅仅求最大和，没有保存位置。
保存位置：start为起 end为末
int start=1,end=1,s=1,e=1;
int sum=0,max=num[1];//不能让max=0
for(int i=1;i<=n;i++)
{
e=i;
sum=sum+num[i];
if(max<sum)
{
max=sum;
start=s;
end=e;
}
if(sum<0)
{
s=i+1;
sum=0;
}
}
本题须要保存起始位置。以下代码假设全是负数，输出最小的那个位置
代码：
#include <iostream>
using namespace std;
const int maxn=100002;
int num[maxn];
int n;
int main()
{
int t;cin>>t;int c=1;
while(t--)
{
cin>>n;
for(int i=1;i<=n;i++)
cin>>num[i];
int start=1,end=1,s=1,e=1;//这里start end一定要赋值为1
int sum=0,max=num[1];//不能让max=0
for(int i=1;i<=n;i++)
{
e=i;
sum=sum+num[i];
if(max<sum)
{
max=sum;
start=s;
end=e;
}
if(sum<0)
{
s=i+1;
sum=0;
}
}
cout<<"Case "<<c++<<":"<<endl;
cout<<max<<" "<<start<<" "<<end<<endl;
if(t)
cout<<endl;
}
return 0;
}