#include using namespace std; //十进制转十六进制 int main() { long long n; cin >> n; string s; s = ""; if (n == 0) cout << n; else { while (n) { string temp = "0"; int k = n % 16; if (k < 10) { temp[0] = ('0' + k); s = temp + s; } else { temp = (k - 10 + 'A'); s = temp + s; } n /= 16; } cout << s; } return 0; }

#include #include using namespace std; //十进制转十六进制 int main() { long long n; cin >> n; cout << setiosflags(ios::uppercase) << hex << n << endl; //std::cout << std::hex << n << std::endl; return 0; }

#include using namespace std; //回文数判断 int huiwen(int x) { int rex = 0, temp; //rex为逆序数 int orx = x; //x为原始数据 while (x != 0) { temp = x % 10; //temp为余数 rex = rex * 10 + temp; x /= 10; } if (rex == orx) return 1; else return 0; } //计算所有位数之和 int weishu_sum(int x) { int sum = 0; while (x != 0) { sum += x % 10; x /= 10; } return sum; } int main() { int n; cin >> n; for (int i = 10000; i <= 999999; i++) { if (huiwen(i)&&(weishu_sum(i)==n)) { cout << i << endl; } } return 0; }

#include using namespace std; //回文数判断 int huiwen(int x) { int rex = 0, temp; //rex为逆序数 int orx = x; //x为原始数据 while (x != 0) { temp = x % 10; //temp为余数 rex = rex * 10 + temp; x /= 10; } if (rex == orx) return 1; else return 0; } int main() { for (int i = 1000; i <= 9999; i++) { if (huiwen(i)) cout << i << endl; } return 0; }

#include using namespace std; //水仙花数判断 int shuixianhua(int x) { int a, b, c; a = x / 100; b = x / 10 % 10; c = x % 10; return (x == a * a * a + b * b * b + c * c * c); } int main() { for (int i = 100; i <= 999; i++) { if (shuixianhua(i)) cout << i << endl; } return 0; }

#include using namespace std; #define N 40 int main() { int a[N][N]; int n; cin >> n; for (int i = 1; i <= n; i++) a[i][1] = a[i][i] = 1; for (int i = 3; i <= n; i++) { for (int j = 2; j <= i-1; j++) { a[i][j] = a[i - 1][j - 1] + a[i - 1][j]; } } for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) { cout << a[i][j] << " "; } cout << endl; } return 0; }

#include using namespace std; //杨辉三角 int yanghui(int i,int j) { if ((j == 1) || (i == j)) return 1; else { return yanghui(i - 1, j - 1) + yanghui(i - 1, j); } } int main() { int n; cin >> n; for (int i = 1; i <= n; i++) { for (int j = 1; j <=i ; j++) { cout << yanghui(i, j) << " "; } cout << endl; } return 0; }

#include using namespace std; #define N 1000 //查找数列元素位置 int find(int a[], int item,int length) { for (int i = 0; i < length; i++) { if (item == a[i]) return i+1; } return - 1; } int main() { int a[N]; int n; cin >> n; for (int i = 0; i < n; i++) { cin >> a[i]; } int item; cin >> item; cout << find(a, item, n); return 0; }

#include #include using namespace std; int main() { for (int i = 0; i <= 31; i++) { cout << bitset<5>(i) << endl; } return 0; }

1. 年份是4的倍数而不是100的倍数；
   
2. 年份是400的倍数。
   

#include using namespace std; int IsRunnian(int x) { if (x % 400 == 0) return 1; else { if ((x % 4 == 0) && (x % 100 != 0)) return 1; else return 0; } } int main() { int x; cin >> x; if (IsRunnian(x)) cout << "yes"; else cout << "no"; return 0; }

#include #include using namespace std; #define M 10000int main() { int a[M]; //定义数组 int n; //输入的阶乘数 int s; //记录每个元素乘以阶乘后的值 int c = 0; //记录需要的进位 cin >> n; memset(a, 0, sizeof(a)); //数组初始化为0，memset函数需要string.h头文件 a[0] = 1; //第一位为1for (int i = 2; i <= n; i++) {//循环每个元素乘以阶乘 for (int j = 0; j < M; j++) { s = a[j] * i + c; //计算元素乘以阶乘的值 a[j] = s % 10; //计算每一位的元素 c = s / 10; //计算进位 } } for (int i = M-1; i >= 0; i--) { if (a[i] != 0) { for (int j = i; j >= 0; j--) { cout << a[j]; } //注意这里一次执行结束之后就要跳出循环 //因为一个for循环之后就会输出不为零开始的整个计算结果 //要是不break,那么会循环重复输出剩下的首位不为零的计算结果 break; } } cout << endl; return 0; }

#include #include using namespace std; #define M 101 int getN(char c) { if (c == 0) return 0; else return c - '0'; } int main() { char a[M] = { 0 }, b[M] = {0}; //定义数组 int c[M]; memset(c, 0, sizeof(c)); cin >> a; cin >> b; int len1 = strlen(a); int len2 = strlen(b); int s; int k = 0; for (int i = 0; i < len1 / 2; i++) { char temp = a[i]; a[i] = a[len1 - 1 - i]; a[len1 - 1 - i] = temp; } for (int i = 0; i < len2 / 2; i++) { char temp = b[i]; b[i] = b[len2 - 1 - i]; b[len2 - 1 - i] = temp; } for (int i = 0; i < M-1; i++) { s = getN(a[i]) + getN(b[i]) + k; c[i] = s % 10; k = s / 10; } for (int i = M - 1; i >= 0; i--) { if (c[i] != 0) { for (int j = i; j >= 0; j--) cout << c[j]; break; } } return 0; }

#include using namespace std; #define M 100 int findmin(int a[], int n) { int min=0; for (int i = 0; i < n; i++) { if (a[i] < a[min]) min = i; } return min; } int main() { int n; cin >> n; int a[M]; for (int i = 0; i < n; i++) { cin >> a[i]; } int num = n; int sum = 0; while (num != 1) { int x; //最小值的下标 x = findmin(a, num); //寻找最小值，并将它换到最后一个 int temp = a[x]; a[x] = a[num - 1]; a[num - 1] = temp; int y; y = findmin(a, num - 1); //寻找倒数第二小的值，并将它换到倒数第二的位置 temp = a[y]; a[y] = a[num - 2]; a[num - 2] = temp; sum = sum + a[num - 1] + a[num - 2]; //花费统计 a[num - 2] = a[num - 1] + a[num - 2]; //加入两者之和 num--; //数量减一 } cout << sum << endl; return 0; }

#include using namespace std; int main() { int a, b, n; int yushu; cin >> a >> b >> n; yushu = a % b; //初始化yushu，yushu是求每次相除的余数 for (int i = 1; i <= n; i++) { yushu = yushu%b*10; if (yushu % b == a % b) {//如果下一次循环的余数等于初始余数，说明接下来重复之前的计算 n %= i; //n缩小(降低时间复杂度的关键) i = 0; //重新遍历 } } for (int i = 0; i < 3; i++){ cout << yushu / b; yushu = yushu % b * 10; } return 0; }

1. 各组的核桃数量必须相同
   
2. 各组内必须能平分核桃（当然是不能打碎的）
   
3. 尽量提供满足1,2条件的最小数量（节约闹革命嘛）
   

#include using namespace std; int gcd(int a, int b) { //欧几里得算法（辗转相除法） while (a!=b) { if (a > b)a = a - b; else b = b - a; } return a; } int gongbeishu(int a, int b) { return a / gcd(a, b) * b; } int main() { int x, y,z; cin >> x >> y>>z; cout << gongbeishu(gongbeishu(x,y),z) << endl; //最小公倍数=两数乘积/最大公约数 return 0; }

//回形取数 #include using namespace std; #define M 200 int Array[M][M]; int vis[M][M] = { 0 }; int main() { int m, n; cin >> m >> n; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { cin >> Array[i][j]; vis[i][j] = 1; } } int tol = 0, i = 0, j = 0; while (m * n > tol) { //1.上往下 while (i < m && vis[i][j] == 1) { cout << Array[i][j] << " "; vis[i][j] = 0; i++; tol++; } i--; j++; //恢复 //2.左往右 while (j < n && vis[i][j] == 1) { cout << Array[i][j] << " "; vis[i][j] = 0; j++; tol++; } j--; i--; //3.下往上 while (i >= 0 && vis[i][j] == 1) { cout << Array[i][j] << " "; vis[i][j] = 0; i--; tol++; } i++; j--; //右往左 while (j >=0 && vis[i][j] == 1) { cout << Array[i][j] << " "; vis[i][j] = 0; j--; tol++; } j++; i++; } return 0; }

//2n皇后问题 #include using namespace std; int mp[10][10]; //棋盘数组 int num; //放置方法个数 int n; //棋盘维数//检测是否符合放置皇后的条件 int check(int x, int y, int s) { for (int i = 0; i < x; i++) {//检测该列是否放置过相同颜色的皇后 if (mp[i][y] == s)return 0; } for (int i = x - 1, j = y - 1; i >= 0 && j >= 0; i--, j--)//检测左对角线是否放置过相同颜色的皇后 { if (mp[i][j] == s)return 0; } for (int i = x-1, j = y+1; i >=0 && j < n; i--, j++) {//检测右对角线是否放置过相同颜色的皇后 if (mp[i][j] == s)return 0; } return 1; }void dfs(int x, int s) { if (x == n) {//表示放置次数达到n if (s == 2)dfs(0, 3); else num++; return; } for (int j = 0; j < n; j++) { if (mp[x][j] != 1)continue; //如果该位置不能放置皇后，结束循环 if (check(x, j, s) == 0)continue; //不符合情况，结束循环 else mp[x][j] = s; //可以放置皇后 dfs(x + 1, s); //下一层 mp[x][j] = 1; //回溯 }} int main() { cin >> n; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { cin >> mp[i][j]; } } dfs(0, 2); cout << num; return 0; }

//FJ的字符串 #include using namespace std; string FJ(int count) { if (count == 1) return "A"; else return FJ(count - 1) + (char)('A'+count-1) +FJ(count - 1); } int main() { int n; cin >> n; cout << FJ(n) << endl; return 0; }

#include using namespace std; int main() { int n; cin >> n; string res = "A"; for (int i = 1; i < n; i++) { res = res + char('A' + i) + res; } cout << res << endl; return 0; }

//Sine之舞 #include #include #include using namespace std; string def(int n) {//不含右括号 if (n == 1) return "sin(1"; else { string res; int k = -1; for (int i = 1; i < n; i++) { char ch; if (k == -1) ch = '-'; else ch = '+'; stringstream ss; ss << n; string s; ss >> s; res = def(n - 1) + ch + "sin(" + s; } return res; } } string An(int n) { string res = def(n); for (int i = 0; i < n; i++) { res += ")"; } return res; } int k=0; string sef(int n) { if (n == 1) { k++; stringstream ss; ss << k; string s; ss >> s; return An(1) +"+"+ s; } else { k++; stringstream ss; ss << k; string s; ss >> s; return sef(n - 1) + ")" + An(n) +"+"+ s; } } string Sn(int n) { string res = sef(n); for (int i = 0; i < n-1; i++) { res = "(" + res; } return res; }int main() { int n; cin >> n; //cout << An(n) << endl; cout << Sn(n) << endl; }

//报时助手 #include using namespace std; string print(int n) { switch (n) { case 0: return "zero"; break; case 1: return "one"; break; case 2: return "two"; break; case 3: return "three"; break; case 4: return "four"; break; case 5: return "five"; break; case 6: return "six"; break; case 7: return "seven"; break; case 8: return "eight"; break; case 9: return "nine"; break; case 10: return "ten"; break; case 11: return "eleven"; break; case 12: return "twelve"; break; case 13: return "thirteen"; break; case 14: return "fourteen"; break; case 15: return "fifteen"; break; case 16: return "sixteen"; break; case 17: return "seventeen"; break; case 18: return "eighteen"; break; case 19: return "nineteen"; break; case 20: return "twenty"; break; case 30: return "thirty"; break; case 40: return "forty"; break; case 50: return "fifty"; break; default: return print(n / 10 * 10) + " " + print(n % 10); break; } } void baoshi(int h, int m) { if (m == 0) cout << print(h) << " o'clock"; else cout << print(h) + " " + print(m); } int main() { int h, m; cin >> h>>m; baoshi(h, m); return 0; }

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