进制转换

将n 进制数x，转换为m 进制数y输出.(n,m<=20)

输入格式

x<n>m

输出格式

x<n>=y<m>

样例输入

48<10>8

样例输出

48<10>=60<8>

没啥好讲的，先转化成10进制，再转化成其他进制。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
#define rep(i, a, n) for(int i = a; i <= n; ++i)
#define per(i, n, a) for(int i = n; i >= a; --i)
typedef long long ll;
const int maxn = 1e2 + 5;
char ax[20] ={'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'};
char a[maxn], num[maxn], nnum[maxn];
ll pre, now, chanum;
void retu()        //先转换成10进制 
{
  ll x = 1;
  per(i, strlen(num) - 1, 0)
    {
      int y = num[i] >= 'A' ? num[i] - 'A' + 10 : num[i] - '0';
      chanum += y * x;
      x *= pre;
    }
}
void change(ll num)        //在转换成n进制 
{
  int pos = 0;
  while(num > 0)
    {
      int x = num % now;
      nnum[++pos] = x > 9 ? ax[x - 10] : x + '0';
      num /= now;
    }
}
int main()
{
  scanf("%s", a);
  int i = 0;
  while(a[i] != '<') {num[i] = a[i]; i++;}
  ++i;
  while(a[i] != '>') {pre = pre * 10 + a[i] - '0'; i++;}
  ++i;
  while(i < strlen(a)) {now = now * 10 + a[i] - '0'; i++;}
  retu();
  change(chanum);
  printf("%s<%lld>=", num, pre);
  per(i, strlen(nnum + 1), 1) printf("%c", nnum[i]);
  printf("<%lld>
", now);
  return 0;
}