進位制轉換

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 65581    Accepted Submission(s): 35574
 

Problem Description

輸入一個十進位制數N，將它轉換成R進位制數輸出。

Input

輸入資料包含多個測試例項，每個測試例項包含兩個整數N(32位整數)和R（2<=R<=16, R<>10）。

Output

為每個測試例項輸出轉換後的數，每個輸出佔一行。如果R大於10，則對應的數字規則參考16進位制（比如，10用A表示，等等）。

Sample Input

7 2 23 12 -4 3

Sample Output

111 1B -11

題意：將n化為m進位制。

Code:

1.stack

（進位制轉換輸出與資料結構中棧的後進先出原理一樣）

#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<list>
#define mod 998244353
#define Max 0x3f3f3f3f
#define Min 0xc0c0c0c0
#define mst(a) memset(a,0,sizeof(a))
#define f(i,a,b) for(int i=a;i<b;i++)
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 5;
int main(){
    //ios::sync_with_stdio(false);
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF){
        stack<char>ss;
        if(n == 0){
            printf("0\n");
            continue;
        }
        if(n < 0){
            n = -n;
            printf("-");
        }
        while(n){
            ss.push(n % m >= 10?(n % m - 10 + 'A'):(n % m + '0'));
            n /= m;
        }
        while(!ss.empty()){
            printf("%c", ss.top());
            ss.pop();
        }
        printf("\n");
    }
    return 0;
}
 

2.陣列逆序輸出實現

#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<list>
#define mod 998244353
#define Max 0x3f3f3f3f
#define Min 0xc0c0c0c0
#define mst(a) memset(a,0,sizeof(a))
#define f(i,a,b) for(int i=a;i<b;i++)
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 5;
int main(){
    //ios::sync_with_stdio(false);
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF){
        if(n == 0){
            printf("0\n");
            continue;
        }
        char str[1005];
        int len = 0;
        bool flag = false;
        if(n < 0){
            n = -n;
            flag = true;
        }
        while(n){
            str[len++] = n % m >= 10?(n % m - 10 + 'A'):(n % m + '0');
            n /= m;
        }
        //int index = 0;
        while(!str[len - 1] && len > 0)   len--;
        if(flag)  printf("-");
        for(int i = len - 1; i >= 0; i--){
            printf("%c", str[i]);
        }
        printf("\n");
    }
    return 0;
}