Multiplying by Rotation
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Multiplying by Rotation
大意:
有一些数字的乘法计算很奇特,如179487*4=717948;
即把最后一位提到最前面;
现在给出n1,n2,n3,n1代表计算是的进制,n2代表第一个乘数的最后以为,n3代表第三个乘数,求出n2的长度;
要点:
以179487*4=717948为例,由于.....7*4=7.....;
4×7 = 28,得出右数第1个数为8,即.......87×4 = 7.....8;
4×8 = 32,得出第2个数为(28/10)+(32%10)= 4,即.....487×4 = 7.......48;
4*4 = 16,得出第3个数为(32/10)+(16%10)= 9,即....9487×4 = 7.......948;
4*9 = 36,得出第4个数为(16/10)+(36%10) = 7,即...79487×4 = 7....7948;
4*7 = 28,得出第5个数为((36/10)+(28%10)) % 10 = 1,即..179487*4 = 7...17948;
4×1 = 4, 得出第6个数为(28/10 + ((36/10)+(28%10)) / 10)+(4%10)= 7,由于4+3=7没有再进位,计算结束;
代码:
#include <iostream>using namespace std;int main(){int base, last, mul;while (cin >> base >> last >> mul){int cur = last * mul;int m1, m2, count = 1;while (cur != last){m1 = cur / base;m2 = cur % base;cur = m2 * mul + m1;count ++;}cout << count << endl;}}
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