Problem A

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.<br>

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).<br>

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.<br>

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

題意：求子序列最大和

思路：基礎動態規劃

感想：最近總是搞系統，這個也沒時間做

AC程式碼：

#include<iostream>
#include<string.h>
#include<set>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<numeric>
#include<math.h>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<iomanip>
#include<cstdio>
using namespace std;
int main()
{
    //freopen("r.txt","r",stdin);
    int n,m,sum,maxx,j,k;
    int a[100005];
    cin>>n;
    int ans=1,step;
    while(n--)
    {
        cin>>m;
        for(int i=0;i<m;i++)
        {
            scanf("%d",&a[i]);
        }
        sum=0;
        maxx=-1001;
        step=1;
        for(int i=0;i<m;i++)
        {
            sum+=a[i];
                if(sum>maxx)
                    {
                        j=step;
                        k=i+1;
                        maxx=sum;
                    }
            if(sum<0)
            {
                sum=0;
                step=i+2;
            }

        }
        cout<<"Case "<<ans++<<":"<<endl;
        cout<<maxx<<" "<<j<<" "<<k<<endl;
        if(n>0) cout<<endl;
    }
    return 0;
}