题目描述

blablabla

样例

#include<iostream>

using namespace std;

const int N = 1000010;
int n;
int q[N], tmp[N];
void merge_sort(int q[], int l , int r) {

     if(l >= r) return;

     int mid = r + l >> 1;
     merge_sort(q, l, mid);
     merge_sort(q, mid + 1, r);//md，可算弄懂了。递归，呵呵‘
     int k = 0, i = l , j = mid + 1;
     while(i <= mid && j <= r) {
         if(q[i] <= q[j]) tmp[k++] = q[i++];
         else tmp[k++] = q[j++];
     }
     while(i <= mid) tmp[k++] = q[i++];
     while(j <= r) tmp[k++] = q[j++];
     for( i = l, j = 0; i <= r; i++, j++)
     q[i] = tmp[j];
}

int main() {

    cin >> n;
    for(int i = 0; i < n; i++) cin >> q[i];
    merge_sort(q, 0, n - 1);
    for(int i = 0; i < n; i++) printf("%d ",q[i]);
    return 0;
}

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla