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- 题意是给定一个数的进制枚举另一个数可能与他相等的进制
- 注意这个进制有下界没有上界 即如果最大数字是9 那么他至少是十进制，最大是
zzz...zzz（10个z）能表示的最大数的进制 应该是36 ^ 10-1~=3 ^15longlong 可以存下20位左右- 这样顺序枚举进制一定会超时，所以我们使用二分的方法
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
int get(char c){ //字符串转对应整数
if(c<=‘9‘)return c-‘0‘;
else return c-‘a‘+10;
}
LL calc(string s, LL radix){ //二分时radix可能会超出int范围
LL res=0;
for(int i=0;i<s.size();i++){
if(((double)res*radix+get(s[i]))>1e16)return 1e18; //如果res大于1e16则一定无解，返回较大的数
res=res*radix+get(s[i]);
}
return res;
}
int main(){
string s1,s2;
int tag,rad;
cin>>s1>>s2>>tag>>rad;
if(tag==2)swap(s1,s2);
LL target=calc(s1,rad);
LL l=0,r=max(target,36LL); //target较小时，直接取36作为二分上界
for(auto it:s2)l=max(l,get(it)+1LL); //找出下界
while(l<r){
LL mid=l+r>>1;
if(calc(s2,mid)>=target)r=mid;
else l=mid+1;
}
if(calc(s2,r)!=target) cout<<"Impossible";
else cout<<r;
}
- 需要特判一下1不是质数
- 注意可以用到
cin>>n,n>0逗号表达式简化代码- 10进制转其他进制可能会存不下，用longlong 存数，转最多的二进制大概在15位左右
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long LL;
vector<int> ten_radix(int n,int radix){
vector<int> res;
while(n){
res.push_back(n%radix);
n/=radix;
}
if(n)res.push_back(n);
return res;
}
int radix_ten(vector<int> n,int rad){
int res=0;
for(int i=0;i<n.size();i++)res=res*rad+n[i];
return res;
}
bool check(int n){
if(n==1)return false;
for(int i=2;i<=n/i;i++){
if(n%i==0)return false;
}
return true;
}
int main(){
int n1,rad;
while(cin>>n1>>rad,n1>=1){
vector<int> n2= ten_radix(n1,rad);
int target= radix_ten(n2,rad);
if(check(n1)&&check(target))cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
}
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
char get(int x){
if(x<=9)return x+‘0‘;
else return x-10+‘A‘;
}
string change(int n){
vector<char> c;
while(n){
c.push_back(get(n%13));
n=n/13;
}
if(n)c.push_back(get(n));
while(c.size()<2)c.push_back(0+‘0‘); //少于两位要补零
string res;
for(int i=c.size()-1;i>=0;i--)res+=c[i];
return res;
}
int main(){
int n1,n2,n3;
cin>>n1>>n2>>n3;
cout<<"#";
cout<<change(n1)<<change(n2)<<change(n3);
}
//y 总果然写的还是简单多了qwq
#include <iostream>
using namespace std;
char get(int x)
{
if (x <= 9) return ‘0‘ + x;
return ‘A‘ + x - 10;
}
int main()
{
int a[3];
for (int i = 0; i < 3; i ++ ) scanf("%d", &a[i]);
cout << ‘#‘;
for (int i = 0; i < 3; i ++ ) cout << get(a[i] / 13) << get(a[i] % 13);
return 0;
}
作者：yxc
链接：https://www.acwing.com/activity/content/code/content/269939/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
char num[][5]={ "tret","jan","feb","mar","apr","may","jun","jly","aug","sep","oct","nov","dec",
"tam","hel","maa","huh","tou","kes","hei","elo","syy","lok","mer","jou"};
int get(string s){
for(int i=0;i<26;i++){
if(s==num[i]){
if(i<13)return i;
else return (i-12)*13;
}
}
return 0;
}
int main(){
int n;
cin>>n;
getchar(); //读掉末尾的换行
while(n--){
string s;
getline(cin,s);
if(s[0]<=‘9‘){
int x=stoi(s);
int high=x/13,low=x%13; //因为某位是0的时候不用输出，所以这里麻烦点
if(high){
if(low)cout<<num[x/13+12]<<" "<<num[x%13]<<endl;
else cout<<num[x/13+12]<<endl;
}
else cout<<num[x%13]<<endl;
}
else {
int k=s.find(‘ ‘);
int res=get(s.substr(0,k));
if(k!=-1)res+=get(s.substr(k+1));
cout<<res<<endl;
}
}
}
//sstream操作下的解题方法
#include <iostream>
#include <sstream>
using namespace std;
char names[][5] = {
"tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec",
"tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou",
};
int get(string word)
{
for (int i = 0; i < 25; i ++ )
if (names[i] == word)
{
if (i < 13) return i;
return (i - 12) * 13;
}
return -1; // 一定不会执行
}
int main()
{
int n;
cin >> n;
getchar();
while (n -- )
{
string line;
getline(cin, line);
stringstream ssin(line);
if (line[0] <= ‘9‘)
{
int v;
ssin >> v;
if (v < 13) cout << names[v] << endl;
else
{
cout << names[12 + v / 13];
if (v % 13 == 0) cout << endl;
else cout << ‘ ‘ << names[v % 13] << endl;
}
}
else
{
int res = 0;
string word;
while (ssin >> word) res += get(word);
cout << res << endl;
}
}
return 0;
}
作者：yxc
链接：https://www.acwing.com/activity/content/code/content/269965/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
- 注意特判0这种特殊情况
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
//转化成b进制
vector<int> change_b(int n,int b){
if(!n)return {0}; //如果是0，直接返回{0}即可
vector<int> res;
while(n){
res.push_back(n%b);
n=n/b;
}
return res;
}
//检查是否为回文数
bool check(vector<int> c){
for(int i=0, j=c.size()-1;i<j;i++,j--)
if(c[i]!=c[j])return false;
return true;
}
int main(){
int n,b;
cin>>n>>b;
vector<int> res=change_b(n,b);
if(check(res))cout<<"Yes"<<endl;
else cout<<"No"<<endl;
cout<<res[res.size()-1];
for(int i=res.size()-2;i>=0;i--)cout<<" "<<res[i];
}
原文：https://www.cnblogs.com/zzcxxoo/p/13252454.html