Lintcode:fast Power Problem Solving report

Fast Power

Original title Link: http://lintcode.com/en/problem/fast-power/#

Calculate the An % b where a, B and n is all 32bit integers.

Example

For 231% 3 = 2

For 1001000% 1000 = 0

Challenge

O (LOGN)

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Solution 1:

In fact, the problem should be suppose n > 0.

We use the multiplication rule of modulo operation: http://baike.baidu.com/view/4887065.htm

(A * b)% P = (a% p * b% p)% p (3)

A^n% B is decomposed into (a^ (N/2) * a^ (N/2) * (a))%b = ((a^ (N/2) * a^ (N/2))%b * (a)%b)%b = ((a^ (N/2)%b * a^ (N/2)%b)%b * (a)%b)%b

The implementation is as follows:

Note that 2 base case:n = 0 N = 1 are handled in particular. Because n = 1 o'clock, a POW (a, B, 1) will be decomposed, and this will be repeatedly called.

1 classSolution {2     /*3 * @param A, B, n:32bit integers4 * @return: An integer5      */6     /*7 * @param A, B, n:32bit integers8 * @return: An integer9      */Ten      Public Static intFastpower (intAintBintN) { One         //Write your code here A         LongRET =Pow (A, b, n); -          -         return(int) ret; the     } -      -     //Suppose n > 0 -      Public Static LongPowintAintBintN) { +         if(A = = 0) { -             return0; +         } A          at         //The base case. -         if(n = = 0) { -             return1%b; -         } -          -         if(n = = 1) { in             returnAb; -         } to          +         LongRET = 0; -          the         //(A * b)% P = (a% p * b% p)% p (3) *ret = POW (A, B, N/2); $RET *=ret;Panax Notoginseng          -         //This step is to prevent overflow theRET%=b; +          A         if(n% 2 = = 1) { theRET *= POW (A, B, 1); +         } -          $         //perform the take-out Operation $RET = ret%b; -          -         returnret; the     } -};

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Solution 2:

Or you can write the Pow (A, B, 1) directly as a% B. The following solution takes base case:n = 1 off.

1 //Solution 2:2     //Suppose n > 03      Public Static LongPowintAintBintN) {4         if(A = = 0) {5             return0;6         }7         8         //The base case.9         if(n = = 0) {Ten             return1%b; One         } A          -         LongRET = 0; -          the         //(A * b)% P = (a% p * b% p)% p (3) -ret = POW (A, B, N/2); -RET *=ret; -          +         //This step is to prevent overflow -RET%=b; +          A         if(n% 2 = = 1) { atRET *= (a%b); -         } -          -         //perform the take-out Operation -RET = ret%b; -          in         returnret; -}

View Code

GITHUB:

Https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/lintcode/math/FastPower.java

Lintcode:fast Power Problem Solving report