Volume 1. Maths - Misc
import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { // TODO Auto-generated method stub BigInteger num1, num2; int n; int kc = 1000000000; int l, r, mid, t; Scanner in = new Scanner(System.in); while (in.hasNext()) { n = in.nextInt(); num1 = in.nextBigInteger(); l = 0; r = kc; while (true)//二分 { mid = (l + r) >> 1; num2 = BigInteger.valueOf(mid); num2 = num2.pow(n); t = num2.compareTo(num1); if (t == 0) { System.out.println(mid); break; } else if (t > 0) r = mid - 1; else l = mid + 1; } } } }
#include <iostream> #include <fstream> #include <cstring> #include <string> #include <algorithm> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cmath> using namespace std; int main() { long long int n, nt, mid, x, y; while (cin>>n, n) { x = y = nt = ceil(sqrt(n)); mid = (nt * nt + (nt - 1) * (nt - 1) + 1) >> 1;/**< long long 防止mid超了 */ //cout<<mid<<endl; if (n != mid) { if (n > mid) { if (nt & 1)/**< n为奇数， 从左到右，再下 在左边*/ { x = x - n + mid; } else/**< 从下到上再到左 在下边*/ { y = y - n + mid; } } else { if (nt & 1)/**< n为奇数， 从左到右，再下 在下边*/ { y = y + n - mid; } else/**< 从下到上再到左 在左边*/ { x = x + n - mid; } } } cout<<x<<' '<<y<<endl; } return 0; }
#include <iostream> #include <fstream> #include <cstring> #include <string> #include <algorithm> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cmath> using namespace std; char s[8], ts[8]; int a[][6] =/**< 枚举，总共24种情况， 在枚举中可以发现(1, 6)(2, 5)(3, 4)两两一组， 当其中一组换位置时，另外两组中必须只有一组需要换位置， 或者但两组互换时，其中一组也要换位置 */ { 1, 6, 2, 5, 3, 4, 1, 6, 5, 2, 4, 3, 1, 6, 3, 4, 5, 2, 1, 6, 4, 3, 2, 5, 6, 1, 5, 2, 3, 4, 6, 1, 2, 5, 4, 3, 6, 1, 4, 3, 5, 2, 6, 1, 3, 4, 2, 5, 2, 5, 6, 1, 3, 4, 2, 5, 1, 6, 4, 3, 2, 5, 3, 4, 1, 6, 2, 5, 4, 3, 6, 1, 5, 2, 1, 6, 3, 4, 5, 2, 6, 1, 4, 3, 5, 2, 4, 3, 1, 6, 5, 2, 3, 4, 6, 1, 3, 4, 2, 5, 6, 1, 3, 4, 5, 2, 1, 6, 3, 4, 1, 6, 2, 5, 3, 4, 6, 1, 5, 2, 4, 3, 2, 5, 1, 6, 4, 3, 5, 2, 6, 1, 4, 3, 6, 1, 2, 5, 4, 3, 1, 6, 5, 2 }; bool fun() { int i, j; for (i = 1; i < 24; i++) { for (j = 0; j < 6; j++) if (s[a[0][j]] != ts[a[i][j]]) break; if (j == 6) return true; } return false; } int main() { int i = 1; while (cin>>s[i++]) { for (; i < 7; i++) cin>>s[i]; s[i] = '\0'; for (i = 1; i < 7; i++) cin>>ts[i]; ts[i] = '\0'; cout<<(fun() ? "TRUE" : "FALSE")<<endl; if (cin.get() == EOF) break; i = 1; } return 0; }
10025 - The ? 1 ? 2 ? ... ? n = k problem
1 /**< 思路： 2 先记f(n) = (n*(n+1))/2; 3 现将k转换为正数，求k对应于f(n)的最小上界n， 4 从n开始， 5 第一步:如果f(n) - k这个数是偶数,则输出n，否则 6 n++，继续第一步。 7 8 i = f(n) - k>=0肯定成立， 9 如果i为要使得f(n) = ?1?2?...?n = k; 10 就要在f(n)中将i的一半为正，一半为负即可。所以i必定为偶数。 11 */ 12 #include <iostream> 13 #include <cmath> 14 using namespace std; 15 16 int main() 17 { 18 int n, k, i; 19 cin>>n; 20 while (n--) 21 { 22 cin>>k; 23 if (k == 0)//为0输出3 24 { 25 i = 3; 26 } 27 else 28 { 29 if (k < 0) 30 { 31 k = -k; 32 } 33 i = sqrt(k * 2); 34 if (((i * (i + 1)) / 2 - k) < 0) 35 i++; 36 while (((i * (i + 1)) / 2 - k) & 1) 37 { 38 i++; 39 } 40 } 41 42 cout<<i<<endl; 43 if (n) 44 { 45 cout<<endl; 46 } 47 } 48 return 0; 49 }
#include <iostream> #include <cmath> using namespace std; int arr[55]; int main() { int n, avg, s, t = 1; while (cin>>n, n) { s = avg = 0; for (int i = 0; i < n; i++) { cin>>arr[i]; avg += arr[i]; } avg /= n; for (int i = 0; i < n; i++) { s += abs(arr[i] - avg); } cout<<"Set #"<<t++<<endl; cout<<"The minimum number of moves is "<<s / 2<<"."<<endl<<endl; } return 0; }
//这样就过了 #include <iostream> #include <string> using namespace std; string s; int main() { int n, len; cin>>n; char r; while(n--) { cin>>s; len = s.size() - 1; if (len <= 1) { r = '+'; } else if (s[len] == '5' && s[len - 1] == '3') { r = '-'; } else if (s[len] == '4' && s[0] == '9') { r = '*'; } else { r = '?'; } cout<<r<<endl; } return 0; } //这样始终过不了，实在是想不明白 #include <iostream> #include <string> using namespace std; string s; int main() { int n, len; cin>>n; char r; while(n--) { cin>>s; len = s.size() - 1; if (len <= 1) { r = '+'; } else if (s[len] == '5') { r = '-'; } else if (s[len] == '4') { r = '*'; } else { r = '?'; } cout<<r<<endl; } return 0; }
#include <iostream> #include <cmath> using namespace std; int main() { int t, x, y, ans; cin>>t; while(t--) { cin>>x>>y; ans = 0; x = abs(x - y); if (x <= 3) { ans = x; } else { y = sqrt(x); while (x - (1 + y) * y / 2 < (y - 1) * y / 2) { y--; } while (x - (1 + y) * y / 2 - (y - 1) * y / 2 > 1) { ans++; x -= y; } if (x - (1 + y) * y / 2 - (y - 1) * y / 2 == 1) { ans++; } ans += 2 * y - 1; } cout<<ans<<endl; } return 0; }
#include <iostream> using namespace std; int main() { double h, u, d, f; int days; double dis, dre; while (cin>>h>>u>>d>>f, h) { days = 1; dis = 0; dre = u * f / 100; while (true) { dis += u; if (dis > h) { cout<<"success on day "<<days<<endl; break; } dis -= d; if (dis < 0) { cout<<"failure on day "<<days<<endl; break; } u -= dre; if (u < 0) { u = 0; } days++; } } return 0; }
#include <iostream> using namespace std; int main() { long long int n; while (cin>>n, n >= 0) { if (n <= 1) cout<<"0%"<<endl; else cout<<25 * n<<"%"<<endl; } return 0; }
10790 - How Many Points of Intersection?
/* f(a, b)=f(a-1,b)+¡¾(a-1)*(b-1) + 0¡¿* b/2 f(a-1, b)=f(a-2,b)+¡¾(a-2)*(b-1) + 0¡¿* b/2 f(a-2, b)=f(a-3,b)+¡¾(a-3)*(b-1) + 0¡¿* b/2 f(2, b)=0+¡¾(b-1)¡¿* b/2 F(a,b) =(b-1)*b/2 *¡¾1+ ...+ (a-3) + (a-2)+(a-1)¡¿ =(b-1)*(b)/2 * a(a-1)/2 */ #include <iostream> using namespace std; int main() { long long a, b, t = 1, bb; while (cin>>a>>b, a | b) { if (b & 1) { bb = (b - 1) / 2; } else { bb = b / 2; b = b - 1; } cout<<"Case "<<t++<<": "<<a * (a - 1) / 2 * b * bb<<endl; } return 0; }
