将M进制的数X转换为N进制的数输出。

link:https://www.nowcoder.com/practice/ae4b3c4a968745618d65b866002bbd32?tpId=40&tqId=30990&tPage=2&rp=2&ru=/ta/kaoyan&qru=/ta/kaoyan/question-ranking

输入涉及大数运算，在知道进制转换规则的基础上，进行大数运算，

#include <iostream>
#include <string.h>
#include <cstdio>
#include <vector>
using namespace std;
const int maxn = 1010;
struct bign
{
	int d[maxn], len;
	bign() {
		memset(d, 0, sizeof(d));
		len = 0;
	}
};
bign mutify(bign n, int x) {
	bign c;
	c.len = 0;
	int carry = 0;
	for (int i = 0; i < n.len; i++) {
		int temp = n.d[i] * x + carry;
		c.d[c.len++] = temp % 10;
		carry = temp / 10;
	}
	while (carry != 0) {
		c.d[c.len++] = carry % 10;
		carry /= 10;
	}
	return c;
}
bign add(bign n1, bign n2) {
	bign c;
	c.len = 0;
	int carry = 0;
	for (int i = 0; i < n1.len || i < n2.len; i++) {
		int temp = n1.d[i] + n2.d[i] + carry;
		c.d[c.len++] = temp % 10;
		carry = temp / 10;
	}
	if (carry != 0) {
		c.d[c.len++] = carry;
	}
	return c;
}
bign m2ten(string x, int m) {//m进制变10进制
	bign ans;
	ans.len = 1;
	bign product;
	product.d[0] = 1;
	product.len = 1;
	for (int i = x.size()-1; i >=0 ; i--)
	{
		int temp;
		if (x[i] >= '0' && x[i] <= '9') {
			temp = x[i] - '0';
		}
		else if (x[i] >= 'A' && x[i] <= 'Z') {
			temp = x[i] - 'A' + 10;
		}
		ans = add(ans, mutify(product, temp));
		product = mutify(product, m);
	}
	return ans;
}
bign divide(bign num, int x, int& r) {
	bign c;
	c.len = num.len;
	//从高位开始除
	for (int i = num.len - 1; i >= 0; i--) {
		r = num.d[i] + r * 10; //被除数的当前位
		if (r < x) {//被除数当前位不够除
			c.d[i] = 0;
		}
		else {
			c.d[i] = r / x;
			r = r % x;
		}
	}
	while (c.len - 1 >= 1 && c.d[c.len - 1] == 0)
		c.len--;
	return c;
}
void show(bign num) {
	for (int i = num.len - 1; i >= 0; i--)
	{
		if (num.d[i] > 9) {
			printf("%c", (char)(num.d[i] - 10 + 'A'));
		}
		else {
			printf("%d", num.d[i]);
		}
	}
	cout << endl;
}
bign ten2n(bign num, int n) {//10进制变n进制
	bign ans;
	if (n == 10) {
		return num;
	}
	else {
		do
		{
			int r = 0;
			num = divide(num, n, r);
			ans.d[ans.len++] = r;
		} while (!(num.d[0] == 0 && num.len == 1));
	}
	return ans;
}
int main()
{
	int m, n;
	string x;
	cin >> m >> n>>x;
	bign t = m2ten(x, m);
	bign res = ten2n(t, n);
	show(res);
}

标签：10,转换,进制,temp,int,len,bign,carry
来源： https://www.cnblogs.com/custoyth/p/12726277.html