HDU-1003 Max Sum（经典DP）

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

定义一个数组sum，
sum[i]含义：所有以a[i]（i=3，表示以3结尾）为结尾的序列的序列和构成一个集合，此集合的最大值就是sum[i]。

如1,2,3.所有以a[3]=3为结尾的序列的序列和集合是{6,5,3},因而sum[3]=6.
sum的状态转移方程：sum[i] = max{sum[i-1]+a[i], a[i]}
ans必定是sum[0···(k-1)]之一。由于要记录起始位置和结束位置，引入s数组记录获得sum的序列的起始元素的位置，而由sum的定义，sum[i]的结束位置是i不用另外记录。

#include <iostream>#include <cstdlib>#include <cstdio>#include <cmath>#include <iomanip>#include <string>#include <set>#include <map>#include <queue> #include <stack>#include <vector>#include <algorithm>#include <sstream> #include <cstring>using namespace std;int main(){    int t,t1;    cin>>t;    t1=0;    while(t1<t)    {        int n,ans=1;        cin>>n;        int a[n+1],sum[n+1],s[n+1]; //s保存以i结尾的最大序列的首元素位置         for(int i=1;i<=n;i++){            cin>>a[i];        }        sum[1]=a[1];        s[1]=1;     //初始化         for(int i=2;i<=n;i++){            if(sum[i-1]>=0){                sum[i]=sum[i-1]+a[i];                s[i]=s[i-1];            }            else{                sum[i]=a[i];                s[i]=i;            }            if(sum[ans]<sum[i]){                ans=i;            }        }        cout<<"Case "<<t1+1<<":\n"<<sum[ans]<<" "<<s[ans]<<" "<<ans<<"\n";        t1++;        if(t1!=t)            cout<<endl;     }    return 0;}