## Thursday, July 7, 2016

### LeetCode 372 - Super Pow

http://bookshadow.com/weblog/2016/07/07/leetcode-super-pow/
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


## 题目描述：

https://all4win78.wordpress.com/2016/07/07/leetcode-372-super-pow/

    public int superPow(int a, int[] b) {
        int magicNum = 1337;
        a = a % magicNum;
        int prod = 1;
        for (int i = b.length - 1; i >= 0; i--) {
            int temp = 1;
            for (int j = 0; j <= 9; j++) {
                if (b[i] == j) {
                    prod = (prod * temp) % magicNum;
                }
                temp = (temp * a) % magicNum;
            }
            a = temp;
        }
        return prod;
    }

http://www.voidcn.com/blog/niuooniuoo/article/p-6093171.html

(a*b)%c
= { (mc+a%c)*(nc+b%c ) }%c
= { mcnc + (nc)*(a%c) + (mc)*(b%c) + (a%c)*(b%c) } % c   其中 mcnc + (nc)*(a%c) + (mc)*(b%c)可以整除c
= {(a%c)*(b%c)}%c。
 public int superPow(int a, int[] b) {
int res = 1;
for (int i = 0; i < b.length; i++) {
res = pow(res, 10) * pow(a, b[i]) % 1337;
}
return res;
}

public int pow(int a, int b) {
if (b == 0) return 1;
if (b == 1) return a % 1337;
return pow(a % 1337, b / 2) * pow(a % 1337, b - b / 2) % 1337;
}
https://discuss.leetcode.com/topic/51328/7ms-java-solution-using-fast-power-algorithm
private static final int NUM = 1337;
public int superPow(int a, int[] b) {
int ans = 1;
//not expecting to be a part of input
if(b==null||b.length==0)
return 0;
a = a%NUM;
int len = b.length;
for (int i = 0; i < len; i++) {
ans = ((pow(ans,10)*pow(a,b[i]))%NUM);
}
return ans;
}

private int pow(int a, int b){
if(b==1)
return a;
if(b==0)
return 1;
int x = pow(a,b/2)%NUM;
x = (x*x)%NUM;
if((b&1)==1)
x = (x*a)%NUM;
return x;
}
http://xiadong.info/2016/07/leetcode-372-super-pow/
a ^ (n + m) = (a ^ n) * (a ^ m)

http://www.cnblogs.com/grandyang/p/5651982.html

1. a^b % 1337 = (a%1337)^b % 1337
2. xy % 1337 = ((x%1337) * (y%1337)) % 1337, 其中xy是一个数字如:45, 98等等

X^678 = ((X^670 % 1337) * (X^8 % 1337)) % 1337 = (pow((X^670 % 1337), 10) * (X^8 % 1337)) % 1337

https://www.hrwhisper.me/leetcode-super-pow/

c mod m = (a ⋅ b) mod m  = [(a mod m) ⋅ (b mod m)] mod m

private int mod = 1337;
public int superPow(int a, int[] b) {
int n = b.length;
int ans = 1;
for (int i = n - 1; i >= 0; i--) {
ans = ans * quick_pow(a, b[i]) % mod;
a = quick_pow(a, 10);
}
return ans;
}
int quick_pow(int a, int b) {
int ans = 1;
a %= mod;
while (b > 0) {
if ((b & 1) !=0) ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans;
}

def superPow(self, a, b): """ :type a: int :type b: List[int] :rtype: int """ ans, pow = 1, a for x in range(len(b) - 1, -1, -1): ans = (ans * (pow ** b[x]) % 1337) % 1337 pow = (pow ** 10) % 1337 return ans
https://discuss.leetcode.com/topic/50543/8ms-java-solution-using-fast-power
http://www.luozhipeng.com/?p=527
private static final int M = 1337;

public int normalPow(int a, int b) {
int result = 1;
while (b != 0) {
if (b % 2 != 0)
result = result * a % M;
a = a * a % M;
b /= 2;
}
return result;
}
public int superPow(int a, int[] b) {
a %= M;
int result = 1;
for (int i = b.length - 1; i >= 0; i--) {
result = result * normalPow(a, b[i]) % M;
a = normalPow(a, 10);
}
return result;
}
X. Pigeonhole principle
https://discuss.leetcode.com/topic/51176/7ms-java-solution
if a^k % m = a^t % m, then a^(k+1) % m must be equal to a^(t+1).
Let's say a^14 %m = a^27 %m, then a^28 % m is equal to a^15%m,
a^k %m = a^(k-13)%m.
let's have an array to keep the power of 'a' and the index is mod value, example:
a=2, b=0, modToPower[1] = 0;
a=2, b =1, modToPower[2] = 1;
a=2, b =2, modToPower[4] = 2;
...
the length of modToPower is 1337(it can be extended to any other positive integer).
and once there is b=x, and modToPower[a^b%1337] is existing, then we can exit the loop.
regarding array 'b', like above if a^k % m = a^(k-13)%m, we can easily map the value represented by array 'b' to mod by 13.
like a^[1,2,3] = a^123 = a^110 =...= a^6;
public int superPow(int a, int[] b) {
if (a <= 0 || b == null || b.length == 0) {
return 0;
}
int modNum = 1337;
a %= modNum;
int[] log = new int[modNum];
for (int i = 0; i < log.length; i++) {
log[i] = -1;
}
int startNum = 1;
Map<Integer, Integer> powerToModNum = new HashMap<Integer, Integer>();
int powerMod;
for (int startIndex = 0; ; startIndex++) {
startNum %= modNum;
if (log[startNum] == -1) {
log[startNum] = startIndex;
powerToModNum.put(startIndex, startNum);
startNum = (startNum*a)%modNum;

} else {
powerMod = startIndex - log[startNum%modNum];
break;
}
}

int j = 0;
for (int power : b) {
j = (j*10 + power%powerMod)%powerMod;
}
return powerToModNum.get(j);
}
https://discuss.leetcode.com/topic/51460/java-solution-pigeonhole-principle
The idea is to find a cycle in the power. As the number is modded by 1337, there must be a duplication among the power of from 1 to 1337. This tells which position in cycle b corresponds to.

public
int superPow(int a, int[] b)
{ int []pows = new int[1337]; // max cycle is 1337 Set<Integer> set = new HashSet<Integer>(); // pigeon hole principle dictates that must be a duplicate among the power from 1 to 1337 if moded by 1337 int cycle = 0; int val = 1; for (int i = 0; i < 1337; i++) { val = (int)(((long)val * a) % 1337); // cycle found if (set.contains(val)) break; set.add(val); pows[cycle++] = val; } // b: String -> BigInteger StringBuilder str = new StringBuilder(); for(int v: b) str.append(v); BigInteger bVal = new BigInteger(str.toString()); bVal = bVal.subtract(new BigInteger("1")).mod(new BigInteger("" + cycle)); return pows[bVal.intValue()]; }

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