一.递归

1.数的计算

评测系统

#include <iostream>
int countCombinations(int n) { //计算当然组合种数if (n == 1) {return 1;}int count = 1;//数字本身就是一个有效组合for (int i = 1; i <= n / 2; i++) {count += countCombinations(i);//自身+当前数所产生的组合种数}return count;
}
using namespace std;
int main()
{int n;cin >> n;cout<<countCombinations(n);
}

2.计算函数值

#include<iostream>
using namespace std;
int s(int x) {if (x == 0)return 1;else if (x % 2 == 0) {return s(x / 2);}else {return s(x - 1) + 1;}
}
int main() {int x;cin >> x;cout << s(x);
}

3.约瑟夫环

评测系统

#include <iostream>
using namespace std;
int f(int n,int k){if(n==1)return 1;elsereturn (f(n-1,k)+k-1)%n+1;
}
int main()
{int n,k;cin>>n>>k;cout<<f(n,k);return 0;
}

4.金额查错


评测系统

解析：假设错误的总金额是 100 元，而明细账目清单上的金额总和是 120 元，那么可能遗漏的金额组合应该总和为 20 元，因为 120 - 100 = 20。题目要求找出所有可能的组合，使得这些组合的金额总和为 20 元。

第一次：count作为金额求和的结果，寻找可行的子串

#include <iostream>
#include <vector>
using namespace std;
void f(int i, int sum, int count, vector<int>& a, vector<int>& subset, vector<vector<int>>& result) {if (sum == count) {result.push_back(subset);return;}if (i == a.size() || sum > count) {return;}subset.push_back(a[i]);f(i + 1, sum + a[i], count, a, subset, result);//包含当前元素subset.pop_back();//不包含当前元素f(i + 1, sum, count, a, subset, result);
}
int main()
{int total,n;cin >> total>> n;vector<int> a(n);int sum = 0;for (int i = 0; i < n; i++) {cin >> a[i];sum += a[i];}int count=sum - total;vector<vector<int>> result;//存放最终结果vector<int> subset;//寻找满足条件的子集f(0, 0, count, a, subset, result);for (const auto& x : result) {for (int x2 : x) {cout << x2 << " ";}cout << endl;}
}

再考虑次序和去重问题，得到最终代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
void f(int i, int sum, int count, vector<int>& a, vector<int>& subset, vector<vector<int>>& result) {if (sum == count) {result.push_back(subset);return;}if (i == a.size() || sum > count) {return;}for (int j = i; j < a.size(); ++j) {if (j > i && a[j] == a[j - 1])continue;subset.push_back(a[j]);f(j + 1, sum + a[j], count, a, subset, result);subset.pop_back();}
}
int main()
{int total,n;cin >> total>> n;vector<int> a(n);int sum = 0;for (int i = 0; i < n; i++) {cin >> a[i];sum += a[i];}sort(a.begin(), a.end());//排序int count=sum - total;vector<vector<int>> result;//存放最终结果vector<int> subset;//寻找满足条件的子集f(0, 0, count, a, subset, result);for (const auto& x : result) {for (int x2 : x) {cout << x2 << " ";}cout << endl;}
}

二.进制转换

1.任意进制转十进制：x=xk+a

如十六转十

评测系统

#include <iostream>
using namespace std;
int main()
{string s = "2021ABCD";int a[100];//存放十六进制的每个数for (int i = 0; i < s.length(); i++) { //调整十六进制数if (s[i] >= '0' && s[i] <= '9') {a[i] = s[i]-'0';}else {a[i] = 10+s[i] - 'A';}}int x=0;//输出的十进制数for (int i = 0; i < s.length(); i++) { //【转换代码】x = x * 16 + a[i];}cout << x;
}

2.十进制转任意进制：通过数组a输出

while(x){a[cnt++]=x%k;x=x/k;
}
reverse(a,a+cnt);

评测系统

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int main()
{int T;cin >> T;while (T--) {int N, M;cin >> N >> M;string s;cin >> s;long long int a[100];//【先转成十进制】for (int i = 0; i < s.length(); i++) {if (s[i] >= '0' && s[i] <= '9') {a[i] = s[i] - '0';}else {a[i] = 10 + s[i] - 'A';}}long long int x = 0;//输出的十进制数for (int i = 0; i < s.length(); i++) {x = x * N + a[i];}if (M == 10) {cout << x << endl;}else { //【十进制转M进制】string b[100];int cnt = 0;while (x) {if (x % M >= 10) {b[cnt++] = x % M - 10 + 'A';}elseb[cnt++] = to_string(x % M);x = x / M;}reverse(b, b + cnt);//翻转//输出for (int i = 0; i < cnt; i++) {cout << b[i];}cout << endl;}}
}