The last three digits of the n-th-square of M

The n-th of positive integer m may be a very large number, and we only ask for the last three digits of that number

Example 1:

For example, input 5 and 3, 5 3 is 125, the output is 125

Example 2:

For example, the input 2 and 10 2 10 is 1024, the output is 24

Example 3:

For example, the input 111 and 5 111 5 is 116850581551, the output is 551

Method One:

Because M is a relatively large number, and the n-th is likely to go out of scope, we can multiply the tree by multiplying the number as a string, multiplying by the phase. But this method is more cumbersome. Because we only take the last three bits, we multiply it by modulo 1000, leaving only the last three digits.

int last_three_number (unsigned int m,unsigned int n) {if (0==n) return 1; if (0==m) return 0; unsigned int i=0,result=1; fo R (; i<n;i++) {   result*=m;   if (1000<=result)   {   result%=1000;   }} return result;

Method Two:

On the basis of method one, multiplication is no longer m*m. Instead, the N/2 power of M times the power of the N/2 of M.

The code is as follows:

int last_three_number1 (unsigned int m,unsigned int n) {if (0==n) return 1; if (0==m) return 0; unsigned int i=0,result=1; w Hile (n!=0) {if (n%2==1) result= (result*m)%1000; m= (m*m)%1000; n/=2;} return result; }

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The last three digits of the n-th-square of M