/*
* 解题思路:
* 题意: 输入最开始的猫的高度和最后干活猫的数量 , 求有多少只猫不用干活和所有猫的高度和是多少
* 把题意理清就好( 设 H , M为输入的两个值 , K为产生了多少代 , N为每代产生多少只猫 )
* N ^ k = M , H * ( 1/( N+1 ) ) ^K = 1;
* K = log( M ) / log( N ) = log( H ) / log( N+1 );
* N从1开始遍历,直到上面两个等式相减绝对值小于1e-10 则找到N值
* 然后即可求出K值,最后做些相加运算即可
*/
#include <math.h> #include <stdio.h> int main( ) { int i,m,n,sum1,sum2,num,tmp,x; while( scanf("%d%d",&m,&n) && m && n ) { num = 1; while( fabs( log(num)/log( num+1 ) - log( n )/log( m ) ) > 1e-10 ) num++; x = (int)(log( m )/log( num+1 )+0.5); sum1 = sum2 = 0; for( i=0;i<x;i++ ) { tmp = (int)(pow( num , i)); sum1 += tmp; sum2 += m*tmp; m /= (num+1); } sum2 += (int)(pow( num , i )); printf("%d %d\n",sum1,sum2); } return 0; }
原文:http://blog.csdn.net/u011886588/article/details/19117595