《算法笔记》3.5小节——入门模拟-＞进制转换

1022 D进制的A+B

#include <iostream>
using namespace std;
int maxn=32;

int main() {
    int z[maxn],num=0,a,b,d;
    cin>>a>>b>>d;
    int res=a+b;
    do{
        z[num++]=res%d;
        res=res/d;
    }while(res);
    for (int i = num-1; i >=0 ; i--) {
        cout<<z[i];
    }
    return 0;
}

问题 A: 又一版 A+B

#include <iostream>
using namespace std;

int main() {
    long long int  a,b,res;
    int m,z[100],num;
    while(cin>>m){
        if(m==0) break;
        cin>>a>>b;
        num=0;
        res=a+b;
        do{
            z[num++]=res%m;
            res=res/m;
        } while (res);
        for (int i = num-1; i >= 0; i--) {
            cout<<z[i];
        }
        cout<<endl;
    }
    return 0;
}

进制转换

• long long int大概是10的19次方左右，但是题目中给的是10的三十次方，超过了long long int的范围，所以要手动模拟除法。

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int divideByTwo(string &num){
    int carry=0;//余数
    for (int i = 0; i < num.length(); ++i) {
        int digit=num[i]-'0';
        int val=carry*10+digit;
        num[i]=(val/2)+'0';
        carry=val%2;
    }
    return carry;
}

bool isZero(const string &num){
    for (char c:num) {
        if(c!='0') return false;
    }
    return true;
}

int main(){
    string num;
    while(cin>>num){
        if(num=="0"){
            cout<<'0'<<endl;
            continue;
        }
        string binary;
        while(!isZero(num)){
            int reminder= divideByTwo(num);
            binary.push_back(reminder+'0');

            int pos=num.find_first_not_of('0');
            if(pos!= string::npos){
                num=num.substr(pos);
            }else num="0";
        }
        reverse(binary.begin(),binary.end());
        cout<<binary<<endl;
    }
    return 0;
}

数制转换

• 注意输入可能有字母，进制大于十的情况，要用数组存储数据

#include <iostream>
#include<algorithm>
using namespace std;
long long int other_to_ten(string s,int p){
    if(p==10) return stoll(s);
    long long res=0;
    for(char c:s){
        int digit;
        if(isdigit(c)){
            digit=c-'0';
        }else{
            digit= toupper(c)-'A'+10;
        }
        res=res*p+digit;
    }
    return res;
}

string ten_to_other(long long n,int p){
    if(p==10) return to_string(n);
    if(n==0) return "0";
    string res;
    while(n>0){
        int rem=n%p;
        char c=(rem<10)?('0'+rem):('A'+rem-10);
        res.push_back(c);
        n=n/p;
    }
    reverse(res.begin(),res.end());
    return res;
}

int main() {
    int a,b;
    string n;
    while(cin>>a>>n>>b){
        long long int res= other_to_ten(n,a);
        string ans= ten_to_other(res,b);
        cout<<ans<<endl;
    }
    return 0;
}

八进制

#include <iostream>
using namespace std;

int main() {
    long int n;
    while(cin>>n){
        int z[100],num=0;
        do{
            z[num++]=n%8;
            n=n/8;
        }while(n!=0);
        for (int i = num-1; i >= 0; --i) {
            cout<<z[i];
        }
        cout<<endl;
    }
    return 0;
}