• poj 1220 NUMBER BASE CONVERSION


    NUMBER BASE CONVERSION
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 5976   Accepted: 2738

    Description

    Write a program to convert numbers in one base to numbers in a second base. There are 62 different digits: 
    { 0-9,A-Z,a-z } 
    HINT: If you make a sequence of base conversions using the output of one conversion as the input to the next, when you get back to the original base, you should get the original number. 

    Input

    The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines will have a (decimal) input base followed by a (decimal) output base followed by a number expressed in the input base. Both the input base and the output base will be in the range from 2 to 62. That is (in decimal) A = 10, B = 11, ..., Z = 35, a = 36, b = 37, ..., z = 61 (0-9 have their usual meanings). 

    Output

    The output of the program should consist of three lines of output for each base conversion performed. The first line should be the input base in decimal followed by a space then the input number (as given expressed in the input base). The second output line should be the output base followed by a space then the input number (as expressed in the output base). The third output line is blank. 

    Sample Input

    8
    62 2 abcdefghiz
    10 16 1234567890123456789012345678901234567890
    16 35 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
    35 23 333YMHOUE8JPLT7OX6K9FYCQ8A
    23 49 946B9AA02MI37E3D3MMJ4G7BL2F05
    49 61 1VbDkSIMJL3JjRgAdlUfcaWj
    61 5 dl9MDSWqwHjDnToKcsWE1S
    5 10 42104444441001414401221302402201233340311104212022133030
    

    Sample Output

    62 abcdefghiz
    2 11011100000100010111110010010110011111001001100011010010001
    
    10 1234567890123456789012345678901234567890
    16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
    
    16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
    35 333YMHOUE8JPLT7OX6K9FYCQ8A
    
    35 333YMHOUE8JPLT7OX6K9FYCQ8A
    23 946B9AA02MI37E3D3MMJ4G7BL2F05
    
    23 946B9AA02MI37E3D3MMJ4G7BL2F05
    49 1VbDkSIMJL3JjRgAdlUfcaWj
    
    49 1VbDkSIMJL3JjRgAdlUfcaWj
    61 dl9MDSWqwHjDnToKcsWE1S
    
    61 dl9MDSWqwHjDnToKcsWE1S
    5 42104444441001414401221302402201233340311104212022133030
    
    5 42104444441001414401221302402201233340311104212022133030
    10 1234567890123456789012345678901234567890
    

     

      1 #include<cstdio>
      2 #include<iostream>
      3 #include<cstring>
      4 #include<algorithm>
      5 #include<cmath>
      6 //#include<vector>
      7 //#include<queue>
      8 //#include<set>
      9 #define INF 0x3f3f3f3f
     10 #define N 100005
     11 #define re register
     12 #define Ii inline int
     13 #define Il inline long long
     14 #define Iv inline void
     15 #define Ib inline bool
     16 #define Id inline double
     17 #define ll long long
     18 #define Fill(a,b) memset(a,b,sizeof(a))
     19 #define R(a,b,c) for(register int a=b;a<=c;++a)
     20 #define nR(a,b,c) for(register int a=b;a>=c;--a)
     21 #define Min(a,b) ((a)<(b)?(a):(b))
     22 #define Max(a,b) ((a)>(b)?(a):(b))
     23 #define Cmin(a,b) ((a)=(a)<(b)?(a):(b))
     24 #define Cmax(a,b) ((a)=(a)>(b)?(a):(b))
     25 #define D_e(x) printf("
    &__ %d __&
    ",x)
     26 #define D_e_Line printf("-----------------
    ")
     27 #define D_e_Matrix for(re int i=1;i<=n;++i){for(re int j=1;j<=m;++j)printf("%d ",g[i][j]);putchar('
    ');}
     28 using namespace std;
     29 //The Code Below Is Bingoyes's Function Forest.
     30 Ii read(){
     31     int s=0,f=1;char c;
     32     for(c=getchar();c>'9'||c<'0';c=getchar())if(c=='-')f=-1;
     33     while(c>='0'&&c<='9')s=s*10+(c^'0'),c=getchar();
     34     return s*f;
     35 }
     36 Iv print(int x){
     37     if(x<0)putchar('-'),x=-x;
     38     if(x>9)print(x/10);
     39     putchar(x%10^'0');
     40 }
     41 /*
     42 Iv Floyd(){
     43     R(k,1,n)
     44         R(i,1,n)
     45             if(i!=k&&dis[i][k]!=INF)
     46                 R(j,1,n)
     47                     if(j!=k&&j!=i&&dis[k][j]!=INF)
     48                         Cmin(dis[i][j],dis[i][k]+dis[k][j]);
     49 }
     50 Iv Dijkstra(int st){
     51     priority_queue<int>q;
     52     R(i,1,n)dis[i]=INF;
     53     dis[st]=0,q.push((nod){st,0});
     54     while(!q.empty()){
     55         int u=q.top().x,w=q.top().w;q.pop();
     56         if(w!=dis[u])continue;
     57         for(re int i=head[u];i;i=e[i].nxt){
     58             int v=e[i].pre;
     59             if(dis[v]>dis[u]+e[i].w)
     60                 dis[v]=dis[u]+e[i].w,q.push((nod){v,dis[v]});
     61         }
     62     }
     63 }
     64 Iv Count_Sort(int arr[]){
     65     int k=0;
     66     R(i,1,n)
     67         ++tot[arr[i]],Cmax(mx,a[i]);
     68     R(j,0,mx)
     69         while(tot[j])
     70             arr[++k]=j,--tot[j];
     71 }
     72 Iv Merge_Sort(int arr[],int left,int right,int &sum){
     73     if(left>=right)return;
     74     int mid=left+right>>1;
     75     Merge_Sort(arr,left,mid,sum),Merge_Sort(arr,mid+1,right,sum);
     76     int i=left,j=mid+1,k=left;
     77     while(i<=mid&&j<=right)
     78         (arr[i]<=arr[j])?
     79             tmp[k++]=arr[i++]:
     80             (tmp[k++]=arr[j++],sum+=mid-i+1);//Sum Is Used To Count The Reverse Alignment
     81     while(i<=mid)tmp[k++]=arr[i++];
     82     while(j<=right)tmp[k++]=arr[j++];
     83     R(i,left,right)arr[i]=tmp[i];
     84 }
     85 Iv Bucket_Sort(int a[],int left,int right){
     86     int mx=0;
     87     R(i,left,right)
     88         Cmax(mx,a[i]),++tot[a[i]];
     89     ++mx;
     90     while(mx--)
     91         while(tot[mx]--)
     92             a[right--]=mx;
     93 }
     94 */
     95 char number[]="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz",str[N],str_new[N];
     96 int main(){
     97     int T=read();
     98     while(T--){
     99         int a=read(),b=read();
    100         scanf("%s",str),
    101         printf("%d %s
    ",a,str);
    102         int len=strlen(str),flag=1,digit=0;
    103         while(flag){
    104             flag=0;
    105             int res=0;
    106             R(i,0,len-1){
    107                 int num;
    108                 if(str[i]>='0'&&str[i]<='9')num=str[i]^'0';
    109                 if(str[i]>='A'&&str[i]<='Z')num=str[i]-'A'+10;
    110                 if(str[i]>='a'&&str[i]<='z')num=str[i]-'a'+36;
    111                 num+=res*a,res=num%b,str[i]=number[num/b];
    112                 if(str[i]!='0')
    113                     flag=1;
    114             }
    115             str_new[++digit]=number[res];
    116         }
    117         printf("%d ", b);
    118         nR(i,digit,1)
    119             printf("%c",str_new[i]);
    120         printf("
    
    ");
    121     }
    122     return 0;
    123 }
    124 /*
    125     Note:
    126         There is always a truth: High precision is the descendants of Satan.
    127 */
    View Code
  • 相关阅读:
    模拟赛T5 : domino ——深搜+剪枝+位运算优化
    校内模拟赛T5:连续的“包含”子串长度( nekameleoni?) —— 线段树单点修改,区间查询 + 尺取法合并
    C++[Tarjan求点双连通分量,割点][HNOI2012]矿场搭建
    C++二分图匹配基础:zoj1002 FireNet 火力网
    迭代加深搜索 C++解题报告 :[SCOI2005]骑士精神
    Linux 常用命令
    Microservices and exception handling in Java with Feign and reflection
    微服务摘要
    Javac编译与JIT编译
    GC调优思路
  • 原文地址:https://www.cnblogs.com/bingoyes/p/10186852.html
Copyright © 2020-2023  润新知