高精度下二进制转为十进制的办法？

#include<stdio.h>
#include<stdlib.h>
class highAcuInt {
public:
	unsigned int* value;
	highAcuInt() {
		value = new unsigned int[100]{ 0 };
	}
	highAcuInt(highAcuInt* old) {
		this->value = new unsigned int[100]{ 0 };
		for (int i = 0; i < 100; i++) {
			this->value[i] = old->value[i];
		}
	}
	void setvalue(int i, unsigned int a) {
		this->value[i] = a;
	}
	int getvalue(int i) {
		return this->value[i];
	}
	static highAcuInt bitwiseAND(highAcuInt& h1, highAcuInt& h2) {//按位与
		highAcuInt result = new highAcuInt();
		for (int i = 0; i < 100; i++) {
			result.value[i] = h1.value[i] & h2.value[i];//数组每个元素按位与
		}
		return result;
	}
	highAcuInt highAccuLShift(unsigned int a) {//左移运算
		unsigned int n = a / 32;//每个左移32位应视为整体
		unsigned int s = a % 32;//其余部分单独处理
		highAcuInt result = new highAcuInt(this);
		if (n != 0) {
			for (int i = n; i < 100; i++) {
				result.value[i - n] = result.value[i];
			}//左移n*32位
			for (int i = 1; i <= n; i++) {
				result.value[100 - i] = 0;
			}//左移时右侧填充0
		}
		for (int i = 0; i < 99 - n; i++) {
			int hw = 0;
			result.value[i] = (result.value[i] << s);
			result.value[i] += (result.value[i + 1] >> 32 - s);
		}//处理其余部分。将i+1单元左移被舍弃的部分存入i位置的低位
		result.value[99] = (result.value[99] << s);//最低位单独处理
		return result;
	}
	highAcuInt highAccuRShift(unsigned int a) {//右移运算，本题没什么用，建议忽视
		unsigned int n = a / 32;//每个右移32位应视为整体
		unsigned int s = a % 32;//其余部分单独处理
		highAcuInt result = new highAcuInt(this);
		if (n != 0) {
			for (unsigned int i = 99 - n; i < 100; i--) {
				result.value[i + n] = result.value[i];
			}//右移n*32位
			for (int i = 0; i < n; i++) {
				result.value[i] = 0;
			}//右移时左侧填充0（由于该类是无符号数，因此是逻辑右移）
		}
		for (int i = 99; i >= n; i++) {
			result.value[i] = (result.value[i] >> s);
			//result.value[i] += (result.value[i + 1] << 32 - s);
		}//处理其余部分。将i+1单元右移被舍弃的部分存入i位置的高位
		return result;
	}
	static highAcuInt highAccuPlus(highAcuInt& h1, highAcuInt& h2) {//加法运算
		highAcuInt result = new highAcuInt();
		int cf = 0;//存放进位标志
		for (unsigned int i = 99; i < 100; i--) {//注：使用i--遍历数组的特殊写法，为了避免unsigned恒大于0
			unsigned int temp = h1.value[i] + h2.value[i] + cf;//类似全加器的原理
			if (
				(temp < h1.value[i] || temp < h2.value[i]) ||
				(h1.value[i] == 0xFFFFFFFF && h2.value[i] == 0xFFFFFFFF)//判断无符号数的加法溢出
				)
				cf = 1;//若溢出则记录进位
			else cf = 0;
			result.value[i] = temp;
		}
		return result;
	}
	static highAcuInt highAccuMultiply(highAcuInt& h1, highAcuInt& h2) {//无符号乘法运算
		int n0 = 0;
		highAcuInt result = new highAcuInt();
		highAcuInt mask = new highAcuInt();
		mask.value[0] = 0x80000000;//mask用于判断h2最高位是否为0
		highAcuInt temp = bitwiseAND(h2, mask);
		for (int i = 0; i < 3200; i++) {
			result = result.highAccuLShift(1);
			if (temp.value[0] != 0) {
				result = highAccuPlus(result, h1);
			}
			h2 = h2.highAccuLShift(1);//乘法电路原理，使用加法和移位实现无符号数乘法
			temp = bitwiseAND(h2, mask);
		}
		return result;
	}
	static void highAccuFactorial(int a) {
		highAcuInt result = new highAcuInt();
		result.value[99] = 1;
		highAcuInt h = new highAcuInt();
		h.value[99] = 1;
		for (int i = 1; i <= a; i++) {
			h.value[99] = i;
			result = highAccuMultiply(result, h);
		}
		printf("0x%.8x%.8x%.8x%.8x%.8x", result.value[90], result.value[91], result.value[92], result.value[93], result.value[94]);
		printf("%.8x%.8x%.8x%.8x%.8x\n", result.value[95], result.value[96], result.value[97], result.value[98], result.value[99]);
	}
};
int main() {
	int a = 50;
	highAcuInt::highAccuFactorial(a);
}