看了半天，没思路，摘了一位dn的，学习了！
View Code 先看对数的性质,loga(b ^ c) = c * loga(b),loga(b * c) = loga(b) + loga(c);假设给出一个数10234432,那么log10( 10234432 ) = log10( 1.0234432 * 10 ^ 7 ) = log10( 1.0234432 ) + 7 ;log10( 1.0234432 )就是log10( 10234432 )的小数部分.log10( 1.0234432 ) = 0.010063744 10 ^ 0.010063744 = 1.023443198 那么要取几位就很明显了吧 ~ 先取对数(对10取),然后得到结果的小数部分bit,pow( 10.0 ,bit)以后如果答案还是 < 1000那么就一直乘10。注意偶先处理了0 ~ 20项是为了方便处理 ~ 这题要利用到数列的公式:an = ( 1 / √ 5 ) * [(( 1 + √ 5 ) / 2 ) ^ n - (( 1 - √ 5 ) / 2 ) ^ n](n = 1 , 2 , 3 .....)取完对数log10(an) =- 0.5 * log10( 5.0 ) + (( double )n) * log(f) / log( 10.0 ) + log10( 1 - (( 1 - √ 5 ) / ( 1 + √ 5 )) ^ n)其中f = (sqrt( 5.0 ) + 1.0 ) / 2.0 ;log10( 1 - (( 1 - √ 5 ) / ( 1 + √ 5 )) ^ n) -> 0 所以可以写成log10(an) =- 0.5 * log10( 5.0 ) + (( double )n) * log(f) / log( 10.0 );最后取其小数部分。#include < iostream > #include < cmath > using namespace std; int fac[ 21 ] = { 0 , 1 , 1 }; const double f = (sqrt( 5.0 ) + 1.0 ) / 2.0 ; int main(){ double bit; int n,i; for (i = 3 ;i <= 20 ;i ++ )fac[i] = fac[i - 1 ] + fac[i - 2 ]; // 求前20项 while (cin >> n) { if (n <= 20 ) { cout << fac[n] << endl; continue ; } bit =- 0.5 * log( 5.0 ) / log( 10.0 ) + (( double )n) * log(f) / log( 10.0 ); // 忽略最后一项无穷小 bit = bit - floor(bit); bit = pow( 10.0 ,bit); while (bit < 1000 )bit = bit * 10.0 ; printf( " %d\n " ,( int )bit); } return 0 ;}