AcWing 1346. 回文平方
原题链接
简单
作者:
眼角下的痣
,
2024-01-28 18:23:47
,
所有人可见
,
阅读 33
1.题解–扩展10进制转b进制 , a进制转10进制
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
char get(int x){
if(x <= 9) return x + '0';
return x - 10 + 'A';
}
//将10进制转化为b进制
string base(int n,int b){
string num;
while(n) num += get(n % b), n/=b;
reverse(num.begin() , num.end());
return num;
}
int uget(char c){
if(c <= '9') return c - '0';
return c - 'A' + 10;
}
//b进制转化为10进制
int base10(string num ,int b){
int res = 0;
for(auto c : num)
res = res * b + uget(c);
return res;
}
bool check(string s){//判断一个字符串是否为回文字符串
for(int i = 0, j = s.size() - 1 ; i < j; i++ ,j--)
if(s[i] != s[j])
return false;
return true;
}
int main(){
ios :: sync_with_stdio(0);cin.tie(0);
int b;
cin >> b;
for(int i = 1;i <= 300 ;i++){
auto num = base(i * i , b );
if(check(num)){
cout << base(i,b) << ' ' << num << endl;
}
}
/*
for(int i = 1;i <= 300 ;i++){
auto num = base(i , b);
cout << i << ' ' << base10(num, b) << endl;
}
*/
return 0;
}
2.进阶 – a进制转b进制
(1)a进制转10进制,10进制转b进制
(2)a进制直接转b进制
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
int main(){
ios :: sync_with_stdio(0);cin.tie(0);
int T;
cin >> T;
while(T--){
int a ,b;
string a_line ,b_line;
cin >> a>> b >> a_line;
vector<int> number;
for(auto c : a_line){
if(c >= '0' && c <= '9') number.push_back(c - '0');
if(c >= 'A' && c <= 'Z') number.push_back(c - 'A' + 10);
if(c >= 'a' && c <= 'z') number.push_back(c - 'a' + 36);
}
reverse(number.begin() , number.end());
vector<int> res;
while(number.size()){
int r = 0;
for(int i = number.size() - 1; i>= 0;i --){
number[i] += r * a;
r = number[i] % b;
number[i] /= b;
}
res.push_back(r);
while(number.size() && number.back() == 0) number.pop_back();
}
reverse(res.begin() , res.end());
for(auto x : res){
if(x <= 9) b_line += char(x + '0');
if(x >= 10 && x <= 35) b_line += char(x + 'A'- 10);
if(x >= 36 ) b_line += char(x + 'a'- 36 );
}
cout << a << ' ' << a_line << endl;
cout << b << ' ' << b_line << endl;
cout << endl;
}
return 0;
}