排序应用举例

问题描述：

在给定结合S中，是否存在之和与给定数x相等的两数，有则输出。

#include<iostream>
using namespace std;
const int maxn = 5000;
int main()
{
    void merge_sort(int a[], int beg, int end);
    void put(int a[], int n);
    int binary_sort(int A[], int y, int beg, int end);
    int a[maxn], x, n;
    cout << "Enter two numbers: ";
    cin >> x >> n;
    cout << "Enter some numbers:" << endl;
    for(int i = 0; i < n ; ++i)
        cin >> a[i];
    merge_sort(a, 0, n-1);
    int index = 0;
    for(int i = 0; i < n; ++i)
    {
        index = binary_sort(a, x-a[i], 0, n-1);
        if(index != -1 && index != i)
            cout << "YES!!!     "<< a[i] << " + " <<a[index] << " == " << x << endl;
    }
    if(index == -1)
        cout << "NO!!!" << endl;
    put(a, n);
    return 0;
}

void merge(int a[], int beg, int mid, int end)
{
    //合并
    int n1 = mid - beg + 1, n2 = end - mid;
    int L[n1+1], R[n2+1];
    for(int i = 0; i < n1; ++i)
        L[i] = a[beg+i];
    L[n1] = 50000;   //哨兵元素
    mid = mid + 1;
    for(int i = 0; i < n2; ++i)
        R[i] = a[mid+i];
    R[n2] = 50000;   //哨兵元素
    int i=0, j =0, k = beg;
    //------------------块一----------------------------------
    for(int k = beg; k <= end; ++k)
    {
        //当同时遇到哨兵意味着所有牌已经合并
        if(L[i] <= R[j])
        {
            a[k] = L[i];
            ++i;
        }
        else
        {
            a[k] = R[j];
            ++j;
        }
    }
    //------------------块一----------------------------------
}

void merge_sort(int a[], int beg, int end)
{
    //将n个元素分成各个汗n/2个元素的子问题
    int mid = (beg+end)/2;
    if(beg < end)
    {
        merge_sort(a, beg, mid);
        merge_sort(a, mid+1, end);
        merge(a, beg, mid, end);
    }
}

void put(int a[], int n)
{
    for(int i = 0; i < n; ++i)
        cout << a[i] << "  ";
    cout << endl;
}

int binary_sort(int a[], int y, int beg, int end)
{//二分查找
    int mid;
    while(beg < end)
    {
        mid = (beg+end)/2;
        if(y == a[mid])
            return mid;
        if(y > a[mid])
            beg = mid + 1;
        else
            end = mid -1;
    }
    return -1;
}

//时间复杂度T(n) = Θ(nlgn);
/*

8   8
1 2 6 4 7 9 3 5

*/