阅读程序1

#include <bits/stdc++.h>
using namespace std;
const int mod = 1024;
int fastmi(int a, int b) {
int res = 1;
while (b) {
if (b%2==1) res = (res*a)%mod;
  a = (a*a)%mod;
  b = b / 2;
}
 return res % mod;
}
int main() {
 int a, b;
 cin >> a >> b;
 cout << fastmi(a, b);
 return 0;
}

这个程序是快速幂，这个不应该错

阅读程序2

#include <bits/stdc++.h>
using namespace std;
const int maxn = 20;
int a[maxn];
bool vis[maxn];
int n;
void dfs(int cur) {
 if (cur == n) {  
  for (int i = 0; i < n; ++i)  cout << a[i] << " ";
  cout << endl;
  return; 
 } 
 for (int i = 1; i <= n; i++) {  
  if (!vis[i]) { 
   a[cur] = i; 
   vis[i] = true;  
   dfs(cur+1); 
   vis[i] = false; 
  }
 }
}
int main() {
 cin >> n;
 dfs(0);
 return 0;
}

这个代码是在用DFS枚举全排列，在枚举子集

阅读程序3

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n;
int arr[N], reg[N];
void sort(int start, int end) {
    if (start >= end)
        return;
    int len = end - start, mid = (len >> 1) + start;
   int start1 = start, end1 = mid;
   int start2 = mid + 1, end2 = end;
   sort(start1, end1);
   sort(start2, end2);
   int k = start;
   while (start1 <= end1 && start2 <= end2)
       reg[k++] = arr[start1] < arr[start2] ? arr[start1++] : arr[start2++];
   while (start1 <= end1)
       reg[k++] = arr[start1++];
   while (start2 <= end2)
       reg[k++] = arr[start2++];
   for (k = start; k <= end; k++)
       arr[k] = reg[k];
}
int main() {
 cin >> n;
 for (int i=0; i<n; ++i) cin >> arr[i];
 sort(0, n-1);
 for (int i=0; i<n; ++i) cout << arr[i] << " ";
 return 0;
}

这题是归并排序

完善程序1

#include <bits/stdc++.h>
using namespace std;
const int N = 110;
char s[N];
int n;
void solve(int l, int r)
{
    if (l > n || r > n || r > l) return;   
    if (l == n && r == n)
    {
        cout << s << endl;
        return;
    }
    s[l + r] = '(';   
    solve(l+1, r);
    s[l+r] = ')';
    solve(l, r + 1);
}
int main()
{
     scanf("%d", &n);
     solve(0, 0);
 return 0;
}

输入n，输出所有由n对小括号组成有效括号字符串。用的是递归和回溯算法

完善程序2

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
vector<int> g[N];    
int q[N], hh, tt;    
int degree[N];     
int del[N];           
int main() {
    int n, a, b;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d%d", &a, &b);
        g[a].push_back(b), g[b].push_back(a);    
        degree[a]++, degree[b]++;    
    }
    for (int i = 1; i <= n; ++i) 
        if (degree[i] < 2) q[tt++] = i, del[i] = true;  
    while (hh < tt) {    
        int u = q[hh++];    
        for (int v: g[u])     
            if (!del[v] && --degree[v] < 2) 
                q[tt++] = v, del[v] = true;
    }    
    for (int i = 1; i <= n; ++i) 
        if (!del[i]) printf("%d ", i);   
    return 0;
}

给定一个无向图，要找出环上顶点。用了拓扑排序算法，用于找出无向图中的所有环上的节点（即那些无法通过拓扑排序删除的节点）