c++ strchr函数_strchr函数返回_模拟实现strchr函数(43)

其中f=(sqrt(5.0)1.0)/2.0;

因为log10(1-((1-√5)/(1√5))^n)趋近于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>

#include<cstdio>

using namespace std;

const double s = (sqrt(5.0)1.0)/2;

int main()

{

int n,i;

doublebit;

int fac[21] = { 0 , 1 };

for(i = 2; i < 21; i)

fac[i] = fac[i-1] fac [i-2];

while(cin >> n)

{

if(n <= 20) {

cout << fac[n] << endl;

continue;

}

else{

bit = -0.5*log(5.0)/log(10.0)((double)n)*log(s)/log(10.0);//调用公式

bit = bit - floor(bit); //取小数部分└(^o^)┘

bit = pow(10.0,bit);

while(bit < 1000) //要求四位，所以要将小数点右边的数移到左边直到符合要求

bit = 10.0 * bit;

cout << (int)bit << endl;

}

}

return 0;

}

{

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,37, 41, 43, 47, 53, 59, 61, 67, 71, \

73, 79, 83, 89, 97, 101, 103, 107, 109,113, 127, 131, 137, 139, 149, 151, \

157, 163, 167, 173, 179, 181, 191, 193,197, 199, 211, 223, 227, 229, 233, \

239, 241, 251, 257, 263, 269, 271, 277,281, 283, 293, 307, 311, 313, 317, \

331, 337, 347, 349, 353, 359, 367, 373,379, 383, 389, 397, 401, 409, 419, \

421, 431, 433, 439, 443, 449, 457, 461,463, 467, 479, 487, 491, 499, 503, \

509, 521, 523, 541, 547, 557, 563, 569,571, 577, 587, 593, 599, 601, 607, \

613, 617, 619, 631, 641, 643, 647, 653,659, 661, 673, 677, 683, 691, 701, \

709, 719, 727, 733, 739, 743, 751, 757,761, 769, 773, 787, 797, 809, 811, \

821, 823, 827, 829, 839, 853, 857, 859,863, 877, 881, 883, 887, 907, 911, \

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