C++学习（1）：最大子段和（多种解法）

问题：给定由n个数（可能为负数）组成的序列a1,a2,a3,...,an,求该序列子段和的最大值。

第一种解法：（最容易考虑的方法，将所有的子段一一相加，然后比较）

#include<iostream>
using namespace std;
int Maxsum(int n,int *a,int &besti,int &bestj)
{
    int sum = 0;
    for(int i=1;i<=n;i++)
    {
        int thissum = 0;
        for(int j=i;j<=n;j++)
        {
            thissum +=a[j];
            if(thissum>sum)
            {
                sum = thissum;
                besti = i;
                bestj = j;
            }
        }
    }
    return sum;
}

int main()
{
    int Case = 1;
    int T;//共有多少组数
    cin>>T;
    while(T>0)
    {
        T--;
        int sum;
        int N = 9;//一组有多少个数据
        cin>>N;
        int* array = new int[N+1];
        for(int i=1;i<=N;i++)
        {
            cin>>array[i];
        }
        int besti,bestj;
        sum = Maxsum(N,array,besti,bestj);
        cout<<"Case "<<Case<<":"<<endl;
        cout<<sum<<" "<<besti<<" "<<bestj<<endl;

        if(T>0)
        {
            cout<<endl;
        }
        Case++;
    }
    return 0;
}

 第二种解法：（利用分治算法，将序列从中间分为两个部分a[1:n/2]和a[n2+1:n]，分别求最大子段和，然后比较。）

#include<iostream>
using namespace std;
int MaxSubSum(int *a,int left,int right,int &besti,int &bestj)
{
    int sum = 0;
    if(left==right)
    {
        sum = a[left];
        besti  = bestj = left;
    }
    else
    {
        int center = (left+right)/2;
        int leftsum = MaxSubSum(a,left,center,besti,bestj);
        int rightsum = MaxSubSum(a,center+1,right,besti,bestj);

        int s1 = -10001;
        int lefts = 0;
        for(int i = center;i>=left;i--)
        {
            lefts +=a[i];
            if(lefts>=s1)
            {
                s1 = lefts;
                besti = i;
                
            }
        }

        int s2 = -10001;
        int rights = 0;
        for(int j = center+1;j<=right;j++)
        {
            rights +=a[j];
            if(rights>s2)
            {
                s2 = rights;
                bestj = j;
            }
        }

        sum = s1+s2;

        if(sum<leftsum)
        {
            sum = MaxSubSum(a,left,center,besti,bestj);
        }
        if(sum<rightsum)
        { 
            sum = MaxSubSum(a,center+1,right,besti,bestj);
        }
    }
    return sum;
}

int MaxSum(int n,int *a,int &besti,int &bestj)
{
    return MaxSubSum(a,1,n,besti,bestj);
}

int main()
{
    int Case = 1;
    int T;//共有多少组数
    cin>>T;
    while(T>0)
    {
        bool flag = true;
        T--;
        int sum;
        int N;//一组有多少个数据
        cin>>N;
        int* array = new int[N+1];
        for(int i=1;i<=N;i++)
        {
            cin>>array[i];
        }
        int besti,bestj;
        //判断是否全部为负数
        for(int j = 1;j<=N;j++)
        {
            if(array[j]>0) flag = false;
        }
        if(!flag)
        {
            sum = MaxSum(N,array,besti,bestj);
            cout<<"Case "<<Case<<":"<<endl;
            cout<<sum<<" "<<besti<<" "<<bestj<<endl;
        }else
        {
            sum = array[1];
            besti = bestj = 1;
            cout<<"Case "<<Case<<":"<<endl;
            cout<<sum<<" "<<besti<<" "<<bestj<<endl;
        }
        
        if(T>0)
        {
            cout<<endl;
        }
        Case++;
    }
    return 0;
}