Description
On
the planet Zoop, numbers are represented in base 62, using the digits
0, 1,
. . . , 9, A, B, . . . , Z, a, b, . . . , z
where
A (base 62) = 10 (base
10)
B (base 62) = 11 (base 10)
...
z (base 62) = 61 (base 10).
Given the digit representation of a number x in base 62, your goal is to
determine if x is divisible by 61.
Input
The
input test file will contain multiple cases. Each test case will be given by a
single string containing only the digits ‘0’ through ‘9’, the uppercase letters
‘A’ through ‘Z’, and the lowercase letters ’a’ through ’z’. All strings will
have a length of between 1 and 10000 characters, inclusive. The end-of-input is
denoted by a single line containing the word “end”, which should not be
processed. For example:
1v3
2P6
IsThisDivisible
end
Output
For
each test case, print “yes” if the number is divisible by 61, and “no”
otherwise. For example:
yes
no
no
In the first example, 1v3 = 1
× 622 + 57 × 62 + 3 = 7381, which is divisible by 61.
In the
second example, 2P6 = 2 × 622 + 25 × 62 + 6 = 9244, which is not
divisible by 61.
Sample Input
1v3
2P6
IsThisDivisible
end
Sample Output
yes
no
no
Source
62进制的数是否能被61整除。
输入的数达10000位，但是可以一边运算一边mod。
#include <stdio.h> #include <string.h> int convertToInt(char a){ if(‘0‘<=a && a<=‘9‘)return a-‘0‘; if(‘A‘<=a && a<=‘Z‘)return a-‘A‘+10; if(‘a‘<=a && a<=‘z‘)return a-‘a‘+36; } int main() { char ch[10001]; while( scanf("%s",ch), strcmp("end",ch)!=0 ){ int mo=0; int temp=1; for(int i=0; ch[i]!=‘\0‘; i++,temp*=62){ mo+=convertToInt(ch[i])*temp; mo%=61; temp%=61; } if(mo==0){ puts("yes"); }else{ puts("no"); } } return 0; }
原文：http://www.cnblogs.com/chenjianxiang/p/3554763.html