G - Max Sum

HDU - 1003

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

題意：找到和最大的子序列，輸出和以及首位元素的位置。

解法：對于一個數有兩種選擇，加入前邊的序列，或者令起一個序列，即如果前邊序列和加上這個數之后仍比這個數小，那么從這個數開始令起一個序列。所以題目可以轉化成求各個局部的最大值，再在這些最大值里找最大值。

代碼：

#include<iostream>
using namespace std;
int a[100005];
int main()
{
    int num;
    cin>>num;
    for(int k=1;k<=num;k++){
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)
            cin>>a[i];
        int sum=a[1];
        int ans=a[1];
        int x=1,y=1,z=1;
        for(int i=2;i<=n;i++){
            if(a[i]>sum+a[i]){
                sum=a[i];
                z=i;
            }
            else
                sum+=a[i];
            if(sum>ans){
                ans=sum;
                x=z;
                y=i;
            }
        }
        cout<<"Case "<<k<<":"<<endl;
        cout<<ans<<" "<<x<<" "<<y<<endl;
        if(k!=num)
            cout<<endl;
    }
    return 0;
}