T1. 八进制回文数

#include <iostream>
typedef long long LL;
using namespace std;
int a[30];
//转八进制
LL f(LL m) {LL sum = 0, n = 0;while(m) {a[++ n] = m % 8;m /= 8;}for(int i = n; i >= 1; i --) sum = sum * 10 + a[i];return sum;
}
//数字反转
LL r(LL m)
{LL sum = 0;while(m) { sum = sum * 10 + m % 10;m /= 10;}return sum;
}
int main()
{int n;cin >> n;//枚举平方根for(int i = 1; i <= n / i; i ++){LL m = f(i * i);if(m == r(m)) cout << i * i << " ";}
}

T2. 主要成分

#include <iostream>
#include <cstring>
#include <unordered_map>
using namespace std;
const int N = 1e6 + 10;
int a[N];
int main()
{int n;scanf("%d", &n);unordered_map<int, int> h;for(int i = 0; i < n; i ++){scanf("%d", &a[i]);if(h.count(a[i])) h[a[i]] ++;else h[a[i]] = 1;}for(int i = 0; i < n; i ++) {if(h[a[i]] >= n / 2) {cout << a[i];return 0;}}cout << "No";return 0;
}

T3. 简单算数题

#include <iostream>
#include <unordered_map>
#include <stack>
using namespace std;
stack<int> stk, op;
void eval() {int b = stk.top(); stk.pop();int a = stk.top(); stk.pop();char p = op.top(); op.pop();if(p == '+') stk.push(a + b);else if(p == '-') stk.push(a - b);else if(p == '*') stk.push(a * b);else if(p == '/') stk.push(a / b);
}
int main()
{string s;getline(cin, s);//运算符优先级unordered_map<char, int> pr {{'+', 1}, {'-', 1}, {'*', 2}, {'/', 2}};for(int i = 0; i < s.size(); i ++) {if(s[i] == ' ') continue;else if(isdigit(s[i])){int j = i, x = 0;while(isdigit(s[j])) x = x * 10 + s[j ++] - '0';stk.push(x);i = j - 1;}else {//计算栈中优先级大于等于当前运算符while(!op.empty() && pr[op.top()] >= pr[s[i]]) eval();op.push(s[i]);   }}while(!op.empty()) eval();cout << stk.top();return 0;
}

T4. 数独填数

#include <iostream>
using namespace std;
char g[100][100];
//sum[x]表示x字符的出现总数，row[i][x]表示第i行中x字符有没有出现过
//col[i][x]表示第i列中x字符有没有出现过
//f[i][x]表示第i个大格子中x字符有没有出现过
int sum[150], row[10][150], col[10][150], f[10][150]; 
int check(int i, int j, char x) {if(sum[x] > 8) return 0;if(row[i][x]) return 0;if(col[j][x]) return 0;// i / 3 * 3 + j / 3; 大格子if(f[i / 3 * 3 + j / 3][x]) return 0;return 1;
}
void dfs(int n) {if(n > 80) {for(int i = 0; i < 9; i ++){for(int j = 0; j < 9; j ++)cout << g[i][j];cout << endl;}exit(0);}int i = n / 9, j = n % 9; //i行j列//已经填过数了if(g[i][j] != '.') {dfs(n + 1);return ;}//枚举该位置可以填的数for(int x = '1'; x <= '9'; x ++){if(!check(i, j, x)) continue;g[i][j] = x;sum[x] ++;row[i][x] = 1;col[j][x] = 1;f[i / 3 * 3 + j / 3][x] = 1;dfs(n + 1);g[i][j] = '.';sum[x] --;row[i][x] = 0;col[j][x] = 0;f[i / 3 * 3 + j / 3][x] = 0;}}
int main()
{char x;for(int i = 0; i < 9; i ++){for(int j = 0; j < 9; j ++){cin >> x;g[i][j] = x;sum[x] ++;row[i][x] = 1;col[j][x] = 1;f[i / 3 * 3 + j / 3][x] = 1;}}dfs(0);return 0;
}