超级福音！！！高进度模板

两大传送门

1. 原址传送门（有详细讲解）
2. 发现地址传送门

废话不多说，直接上代码！

#include <bits/stdc++.h>
struct BigInt {
	static const int maxlength = 1005;
	int num[maxlength], len;
	void clean() {
		memset(num, 0, sizeof(num));
		len = 1;
	}
	BigInt() {
		clean();
	}
	void read() {
		char str[maxlength];
		scanf("%s", str);
		len = strlen(str);
		for (int i = 1; i <= len; i++) {
			num[i] = str[len - i] - '0';
		}
	}
	void write() {
		for (int i = len; i; i--) {
			printf("%d", num[i]);
		}
		putchar('\n');
	}
	void itoBig(int x) {
		clean();
		while (x != 0) {
			num[len++] = x % 10;
			x /= 10;
		}
		if (len != 1) {
			len--;
		}
	}
	bool operator <(const BigInt &cmp) const {
		if (len != cmp.len) {
			return len < cmp.len;
		}
		for (int i = len; i; i--) {
			if (num[i] != cmp.num[i]) {
				return num[i] < cmp.num[i];
			}
		}
		return 0;
	}
	bool operator >(const BigInt &cmp) const {
		return cmp < *this;
	}
	bool operator <=(const BigInt &cmp) const {
		return !(cmp < *this);
	}
	bool operator !=(const BigInt &cmp) const {
		return cmp < *this || *this < cmp;
	}
	bool operator ==(const BigInt &cmp) const {
		return !(cmp < *this || *this < cmp);
	}
	BigInt operator +(const BigInt &A) const {
		BigInt S;
		S.len = max(len, A.len);
		for (int i = 1; i <= S.len; i++) {
			S.num[i] += num[i] + A.num[i];
			if(S.num[i] >= 10) {
				S.num[i] -= 10;
				S.num[i + 1]++;
			}
		}
		while(S.num[S.len + 1]) S.len++;
		return S;
	}
	BigInt operator -(const BigInt &A) const {
		BigInt S;
		S.len = max(len, A.len);
		for (int i = 1; i <= S.len; i++) {
			S.num[i] += num[i] - A.num[i];
			if(S.num[i] < 0) {
				S.num[i] += 10;
				S.num[i + 1]--;
			}
		}
		while (!S.num[S.len] && S.len > 1) {
			S.len--;
		}
		return S;
	}
	BigInt operator *(const BigInt &A) const {
		BigInt S;
		if ((A.len == 1 && A.num[1] == 0) || (len == 1 && num[1] == 0)) {
			return S;
		}
		S.len = A.len + len - 1;
		for (int i = 1; i <= len; i++) {
			for (int j = 1; j <= A.len; j++) {
				S.num[i + j - 1] += num[i] * A.num[j];
				S.num[i + j] += S.num[i + j - 1] / 10;
				S.num[i + j - 1] %= 10;
			}
		}
		while (S.num[S.len + 1]) {
			S.len++;
		}
		return S;
	}
	BigInt operator /(const BigInt &A) const {
		BigInt S;
		if ((A.len == 1 && A.num[1] == 0) || (len == 1 && num[1] == 0)) {
			return S;
		}
		BigInt R, N;
		S.len = 0;
		for (int i = len; i; i--) {
			N.itoBig(10);
			R = R * N;
			N.itoBig(num[i]);
			R = R + N;
			int flag = -1;
			for (int j = 1; j <= 10; j++) {
				N.itoBig(j);
				if(N * A > R) {
					flag = j - 1;
					break;
				}
			}
			S.num[++S.len] = flag;
			N.itoBig(flag);
			R = R - N * A;
		}
		for (int i = 1; i <= (S.len >> 1); i++) {
			swap(S.num[i], S.num[len - i + 1]);
		}
		while (!S.num[S.len] && S.len > 1) {
			S.len--;
		}
		return S;
	}
	BigInt operator %(const BigInt &A) const {
		BigInt S;
		BigInt P = *this / A;
		S = *this - P * A;
		return S;
	}
};
int main(){
	
	return 0;
}