HDU 1003 Max Sum
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 211310 Accepted Submission(s): 49611
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
解题心得：
这个题是一个简单地动态规划题，题意：输入n个数，求它的最大的子序列和。 首先写出状态转移方程： sum[i]=max{sum[i-1]+a[i],a[i]}。
s数组是记录开始位置，每当sum加上一个a[i]，值小于0的时候，s取用新的值。 ans是记录结束位置，每当sum加上一个a[i]，值大于等于0的时候，更新一下。
我又在格式问题上出错了，以后要更加注意格式问题，最后一行不需要换行！
也可以参考这个人的思路，http://blog.csdn.net/code_pang/article/details/7772200，通过枚举发现的规律。
最后是代码：
#include <iostream> #include <cstdio> using namespace std; int main() { int t; int n; int a[100005];//存储序列 int sum[100005];//存储以每个数为结尾的子序列和 int s[100005];//存储开始位置 int ans;//结束位置 scanf("%d",&t); for(int i=1;i<=t;i++){ scanf("%d",&n); for(int i1=0;i1<n;i1++){ scanf("%d",&a[i1]); } ans=0; sum[0]=a[0]; s[0]=0; for(int j=1;j<n;j++){ if(sum[j-1]>=0){ sum[j]=sum[j-1]+a[j]; s[j]=s[j-1]; }else{ sum[j]=a[j]; s[j]=j; } if(sum[ans]<sum[j]) ans=j; } if(i<t){ printf("Case %d:\n%d %d %d\n",i,sum[ans],s[ans]+1,ans+1); printf("\n"); }else{ printf("Case %d:\n%d %d %d\n",i,sum[ans],s[ans]+1,ans+1); } } return 0; }
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