HDOJ_1003_Max Sum

题目：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

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这题的核心就是求最大子串和，是经典动态规划算法，具体的算法核心可以参考该链接http://blog.nlogn.cn/programming-pearls-the-maximum-sum-of-substring/，该博主用图示的方法将算法过程讲的非常清晰透彻。我在这里用几句话简述就是：

对于第i个元素是否要添加入子串中，看前k个元素的和sum[k],与第i个元素的和是否小于0.

    若sum[k]+a[i]<0;则不选择第i个元素，并重新定位子串开始位置。
    若sum[k]+a[i]>0;则选择第i个元素加入子串中，若sum[k]+a[i]>maxsum,则更新maxsum的值，并更新子串end位置。

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代码如下：

#include <iostream>
#include <vector>       

using namespace std; 

int main(){
    int t,n,k;
    int casenum = 0;
    cin>>t;
    while(t--){
        cin>>n;

        int copy_n = n;
        casenum++;
        int start = 0;
        int end = 0;
        int maxsum = -1000;
        int sum = 0;//连续字串的和（不是最大和） 
        int temp =0;
        int s=0;
        vector<int> nums;
        while(n--){
            cin>>k;
            nums.push_back(k);
        }
        int i ;
        maxsum = nums[0];
        for(i = 0;i<copy_n;i++){//找最大字串和 
            sum = sum+nums[i]; 

            if(sum>=maxsum){

                maxsum = sum;
                start = s; 
                end = start + temp;
            }
            temp++;
            if(sum<0){
                sum = 0;
                s = s+temp;
                temp = 0;
            }


        }
        cout<<"Case "<<casenum<<":"<<endl;
        cout<<maxsum<<" "<<start+1<<" "<<end+1<<endl;
        if(t!=0) cout<<endl;
    }
}