poj 1220 NUMBER BASE CONVERSION

Posted 2021-01-30 bingoyes

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