KY16 求root(N, k)|模拟暴力解法

模拟暴力解法失败，这个方法正确是正确的，但是提交的时候会运行超时

#include <stdio.h>
#include <iostream>
#pragma warning(disable:4996);
using namespace std;
const int maxn = 100010;
struct BigInteger {int digit[maxn];int length;BigInteger();BigInteger(int x);BigInteger(const BigInteger& b);BigInteger operator*(const BigInteger& b);BigInteger operator/(const BigInteger& b);BigInteger operator%(const BigInteger& b);BigInteger operator-(const BigInteger& b);bool operator<(const BigInteger& b);BigInteger convertToK(const BigInteger& k);int convertToint(const BigInteger& b);int sum();void printBigInteger();
};BigInteger::BigInteger()
{fill(digit, digit + maxn, 0);length = 0;
}BigInteger::BigInteger(const BigInteger& b)
{length = b.length;for (int i = 0; i < length; i++){digit[i] = b.digit[i];}
}BigInteger::BigInteger(int x)
{fill(digit, digit + maxn, 0);length = 0;while (x != 0){digit[length++] = x % 10;x /= 10;}
}bool BigInteger::operator<(const BigInteger& b)
{if (length < b.length)return true;if (length > b.length)return false;if (length == b.length){for (int i = length - 1; i >= 0; i--){if (digit[i] > b.digit[i])return false;}return true;}
}BigInteger BigInteger::operator-(const BigInteger& b)
{int carray = 0;BigInteger answer;answer.length = length;for (int i = 0; i < length; i++){int now = digit[i] - b.digit[i] - carray;carray = 0;if (now < 0){now = now + 10;carray = 1;}answer.digit[i] = now;}while (answer.digit[answer.length - 1] == 0 && answer.length > 1) {answer.length--;}return answer;
}BigInteger BigInteger::operator*(const BigInteger& b)
{BigInteger ans;ans.length = length + b.length;for (int i = 0; i < length; i++){for (int j = 0; j < b.length; j++){ans.digit[i + j] += digit[i] * b.digit[j];}}int carray = 0;for (int i = 0; i < ans.length; i++){int now = carray + ans.digit[i];ans.digit[i] = now % 10;carray = now / 10;}//如果还有进位if (carray > 0)ans.digit[ans.length++] = carray;//然后开始除掉多余的0while (ans.digit[ans.length-1] == 0 && ans.length > 1){ans.length--;}return ans;
}BigInteger BigInteger::operator/(const BigInteger& b)
{//从高位开始相除BigInteger answer;BigInteger remainder;answer.length = length;BigInteger tempB(b);for (int i = length - 1; i >= 0; i--){//看看余数的大小if (!(remainder.length == 1 && remainder.digit[0] == 0)){//如果remainder不是0的话for (int j = remainder.length - 1; j >= 0; j--){remainder.digit[j + 1] = remainder.digit[j];}remainder.length++;}remainder.digit[0] = digit[i];while (tempB < remainder){remainder = remainder - tempB;answer.digit[i]++;}}while (answer.digit[answer.length - 1] == 0 && answer.length > 1){answer.length--;}return answer;
}BigInteger BigInteger::operator%(const BigInteger& b)
{//从高位开始相除BigInteger answer;BigInteger remainder;answer.length = length;BigInteger tempB(b);for (int i = length - 1; i >= 0; i--){//看看余数的大小if (!(remainder.length == 1 && remainder.digit[0] == 0)){//如果remainder不是0的话for (int j = remainder.length - 1; j >= 0; j--){remainder.digit[j + 1] = remainder.digit[j];}remainder.length++;}remainder.digit[0] = digit[i];while (tempB < remainder){remainder = remainder - tempB;answer.digit[i]++;}}return remainder;
}int BigInteger::convertToint(const BigInteger& b)
{int ans = 0;int exp = 1;for (int i = 0; i < b.length; i++){ans += exp * b.digit[i];exp *= 10;}return ans;
}BigInteger BigInteger::convertToK(const BigInteger& k)
{BigInteger answer;BigInteger tempB(*this);while (!(tempB.length == 1 && tempB.digit[0] == 0)){BigInteger kkk = tempB % k;answer.digit[answer.length++] = convertToint(BigInteger(tempB % k));tempB = tempB / k;}return answer;
}
void BigInteger::printBigInteger()
{for (int i = length - 1; i >= 0; i--){printf("%d", digit[i]);}
}int BigInteger::sum()
{int answer = 0;for (int i = 0; i < length; i++){answer += digit[i];}return answer;
}BigInteger faseExp(int a, int b)
{BigInteger tempA(a);BigInteger Expans(1);while (b != 0){if (b % 2 == 1){Expans = Expans * tempA;}tempA = tempA * tempA;b = b / 2;}return Expans;
}void Root(BigInteger n, BigInteger k)
{if (n < k){n.printBigInteger();return;}else {BigInteger ansK = n.convertToK(BigInteger(k));int newN = ansK.sum();BigInteger N(newN);Root(newN, k);}
}int main()
{int a, b, k;scanf("%d %d %d", &a, &b, &k);BigInteger ans = faseExp(a, b);BigInteger BigK(k);Root(ans, k);}

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KY16 求root(N, k)|模拟暴力解法

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