LeetCode: Super Pow
2016-07-27 23:15
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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:
Example2:
问题描述:求解(a)^b%1337的值,b是一个非常大的整数,这里用一个数组表示.
分析:当时以为1337这个数字比较特殊,找了一会没发现什么规律,后来用了暴力解法,但是超时是肯定的.
最后发现用一个数学公式 (a * c) % b = (a % b) * (c % b)
所以题目就可以 表示为a^(10j + k) = (a^j)^10 * a^k
从b的高位到低位计算,代码如下:
int power(int x, int n) {
if (n == 0) return 1;
if (n == 1) return x % 1337;
return power(x % 1337, n / 2) * power(x % 1337, n - n / 2) % 1337;
}
int superPow(int a, int* b, int bSize) {
if (a == 1) {
return a;
}
long long res = 1;
for (int i = 0; i < bSize; ++i) {
res = (power(res, 10) * power(a, b[i])) % 1337;
}
return res;
}
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
问题描述:求解(a)^b%1337的值,b是一个非常大的整数,这里用一个数组表示.
分析:当时以为1337这个数字比较特殊,找了一会没发现什么规律,后来用了暴力解法,但是超时是肯定的.
最后发现用一个数学公式 (a * c) % b = (a % b) * (c % b)
所以题目就可以 表示为a^(10j + k) = (a^j)^10 * a^k
从b的高位到低位计算,代码如下:
int power(int x, int n) {
if (n == 0) return 1;
if (n == 1) return x % 1337;
return power(x % 1337, n / 2) * power(x % 1337, n - n / 2) % 1337;
}
int superPow(int a, int* b, int bSize) {
if (a == 1) {
return a;
}
long long res = 1;
for (int i = 0; i < bSize; ++i) {
res = (power(res, 10) * power(a, b[i])) % 1337;
}
return res;
}
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