Max Sum（杭电OJ1003）解题报告

 

Max Sum

 

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

解题思路： 刚开始用最笨的方法试了一下，暴力搜索，全部走一边，时间花太多

了，超时了，对此改进了搜索的方式，只要两个循环就搞定了，时间加快了，但是相

对于来说还不是最快的，还是会超时的，一个个向后面走，只要和大于Max的，就记

录的，代码如下：

#include
using namespace std;
int main()
{
    int n;
    cin>>n;
    int point1,point2;
    int step=1;
    while(n)
    {
        int max=-99999;
        int m,data[100000];
        cin>>m;
        for(int i=1;i<=m;i++)
            cin>>data[i];
        for(i=1;i<=m;i++)
        {
            int sum=0;
            for(int j=i;j
            {
                sum+=data[j];
                if(sum>max)
                {
                    max=sum;
                    point1=i;
                    point2=j;
                }
            }
        }
        if(step!=1)
           cout<
        cout<<"Case "<":"<
        cout<" "<" "<
        step++;
        n--;
    }
    return 0;
}

交上去还是超时的，最近用到动态规划，对算法进行改进了，起始点还是从头开始的，

一直到后面搜索，一直和为小于零，起始点就从开始小于零的后一位开始并把sum改为

零，再搜索的过程中，一遇到大的数据就记录下来，把其计为起始点和终点的，这里

面主要考虑到，当你搜索到一个位置的，它的和不小于零的，那对于后面来说，加上

去还是会变大的，不会给变小的，所以要再搜索下去的，走一边就KO了。代码如下：

#include
using namespace std;
#define Min -999999
int main()
{
    int data[100000],start,end;
    int m;
    int step=1;
    cin>>m;
    while(m--)
    {
        int n;
        cin>>n;
        for (int i=1; i<=n;i++)
            cin>>data[i];
        int max = Min;
        int k=1;
        int sum = 0;
        for (i=1; i<=n; i++)
        {
            sum = sum + data[i];
            if (sum > max)
            {
                max = sum;
                start=k;
                end=i;
            }
            if(sum<0)
            {
                sum=0;
                k=i+1;
            }
        }
        if(step!=1)
            cout<
        cout<<"Case "<":"<
        cout<" "<" "<
        step++;
    }
    return 0;
}