Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 159198 Accepted Submission(s): 37255
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.
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：
这道题，看起来很水，也就是个dp求最大连续子序列的和。。那么只要定义好状态就可以了，dp[i] 表示的是以a[i]为结尾的最大子序列的和。
那么状态转移方程为：dp[i] = max( dp[i-1]+a[i],a[i] )
不说了，直接上代码：
# include<cstdio>
# include<iostream>
using namespace std;
# define MAX 100000
int mat[MAX];
struct node
{
int l;
int r;
}seg[MAX];
int main(void)
{
int n;
int icase = 0;
int t;cin>>t;
while ( t-- )
{
icase++;
cin>>n;
for ( int i = 1;i <= n;i++ )
{
cin>>mat[i];
}
seg[1].l = 1;
seg[1].r = 1;
for ( int i = 2;i <= n;i++ )
{
if ( mat[i] <= mat[i-1]+mat[i] )
{
mat[i] = mat[i-1]+mat[i];
seg[i].l = seg[i-1].l;
seg[i].r = i;
}
else
{
seg[i].l = seg[i].r = i;
}
}
int maxID = 1;
int _max = mat[1];
for ( int i = 2;i <= n;i++ )
{
if ( mat[i] > _max )
{
_max = mat[i];
maxID = i;
}
}
printf("Case %d:\n",icase);
printf("%d %d %d\n",_max,seg[maxID].l,seg[maxID].r);
if ( t != 0 )
cout<<endl;
}
return 0;
}