标签：other turn not orm ati erp idea color blog
Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array. Example1: a = 2 b = [3] Result: 8 Example2: a = 2 b = [1,0] Result: 1024
ab % k = (a%k)(b%k)%k
Since the power here is an array, we‘d better handle it digit by digit.
One observation:
a^1234567 % k = (a^1234560 % k) * (a^7 % k) % k = (a^123456 % k)^10 % k * (a^7 % k) % k
put it other way:
Suppose f(a, b) calculates a^b % k; Then translate above formula to using f :
f(a,1234567) = f(a, 1234560) * f(a, 7) % k = f(f(a, 123456),10) * f(a,7)%k;
Implementation of this idea:
1 public class Solution { 2 public int superPow(int a, int[] b) { 3 return superPow(a, b, b.length, 1337); 4 } 5 6 public int superPow(int a, int[] b, int len, int k) { 7 if (len == 1) return powMud(a, b[0], k); 8 return powMud(superPow(a, b, len-1, k), 10, k) * powMud(a, b[len-1], k) % k; 9 } 10 11 public int powMud(int x, int y, int k) { 12 int res = 1; 13 for (int i=y; i>0; i--) { 14 x = x % k; 15 res = (res * x) % k; 16 } 17 return res; 18 } 19 }
Another clear solution:
1 public class Solution { 2 public int superPow(int a, int[] b) { 3 int res = 1; 4 int j = 0; 5 a = a % 1337; 6 while(j < b.length){ 7 int rem = pow(a,b[j]); 8 // left shift b; 9 res = (rem * pow(res,10)) % 1337; 10 j++; 11 } 12 return res; 13 } 14 // write a pow function to avoid overflow 15 public int pow(int a, int b){ 16 int res = 1; 17 while(b > 0){ 18 // module 1337 every time to avoid overflow 19 res = (res * a) % 1337; 20 b--; 21 } 22 return res; 23 } 24 }
标签：other turn not orm ati erp idea color blog
原文地址：http://www.cnblogs.com/EdwardLiu/p/6108098.html
