[POJ 1289] The Cat in the Hat · 数学

大概就是给你两个数A和B，有且仅有一组n，k满足：

然后输出      和    

暴力枚举k即可。

#include <stdio.h>#include <algorithm>#include <iostream>#include <cmath>using namespace std;#define ll long longll A,B,n,k,sq,x,y,ans1,ans2;ll sqr( ll t , ll k ) {ll p = 1 ;while ( k -- ) p *= t ;return p ;}int main() {cin >> A >> B ;while ( 1 ) {if ( A + B == 0 ) break ;if ( B == 1 ){cout << "0 " << A << endl ;cin >> A >> B ;continue ;}for (ll i = 1 ; i <= 31 ; i++ ) {double h1,h2 ;h1 = pow( A , (double) 1 / (double) i ) - 1 ;   h2 = pow( B , (double) 1 / (double) i ) ;   if ( fabs( h1 - h2) < 1e-14 ) {   n = (ll) ( h1 + 0.5 ) ; k = i ;   break ;    }}ans1 = sqr( n , k ) / ( n - 1 ) ;ans2 = 0 ;for (ll i = 0 ; i <= k ; i ++ ) ans2 += sqr( n + 1 , k - i ) * sqr( n , i ) ;cout << ans1 << " " << ans2 << endl ;cin >> A >> B ;}return 0;}