icpc 银川 II. Base62 任意进制的转换 短除法
作者:互联网
Problem Description
As we already know, base64 is a common binary-to-text encoding scheme. Here we define a special series of positional systems that represent numbers using a base (a.k.a. radix) of 2 to 62. The symbols ‘0’ – ‘9’ represent zero to nine, and ‘A’ – ‘Z’ represent ten to thirty-five,and ‘a’ – ‘z’ represent thirty-six to sixty-one. Now you need to convert some integer z in base x into base y.
Input
The input contains three integers x; y (2 ≤ x; y ≤ 62) and z (0 ≤ z < x120), where the integer z is given in base x.
Output
Output the integer z in base x.
Sample Input
16 2 FB
Sample Output
11111011
Analysis of ideas
利用短除法
Accepted code
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 1100;
int getnum(char ch) //字符转数字
{
if(ch <= '9') return ch-'0';
else if(ch <= 'Z') return 10+ch-'A';
else return 36+ch-'a';
}
char getch(int num) //数字转字符
{
if(num <= 9) return num+'0';
else if(num <= 35) return num-10+'A';
else return num-36+'a';
}
int n,m;
char str1[maxn],str2[maxn];
int t[maxn],ans[maxn];
void solve()
{
int len = strlen(str1);
for(int i = 0; i < len; i++) //先把字符串变成数组,高位->低位
t[i] = getnum(str1[i]);
int j = 0,k = 0;
while(j < len)
{
for(int i = j; i < len-1; i++) //除以m,把余数加到下一位
{
t[i+1] += t[i] % m * n;
t[i] /= m;
}
ans[k++] = t[len-1]%m; //个位数余m
t[len-1] /= m;
while(j < len && !t[j]) j++; //最高位是0,j++
}
for(int i = 0; i < k; i++) //逆序变成字符串
{
str2[i] = getch(ans[k-i-1]);
}
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
int t;
cin(t);
while(t--)
{
mem(ans,0),mem(str2,0); //wa
cin>>n>>m;
cin>>str1;
solve();
cout<<n<<' '<<str1<<endl;
cout<<m<<' '<<str2<<endl<<endl;
}
}
标签:int,Base62,len,icpc,II,++,base,include,define 来源: https://www.cnblogs.com/hezongdnf/p/11967074.html