The problem instance _c language of writing C language program to transform

Topic

Topic Description:
Converts the number X of M into the number output of N.
Input:
The first line entered consists of two integers: M and N (2<=m,n<=36).
The following line enters a number of x,x that is M, and now requires that you convert the M-ary number x into N-ary output.
Output:
The number of n-ary representations of the output x.
Sample input:
16 10
F
Sample output:
15
Tips:
When entered, the letter portion is uppercase, the output is lowercase, and there is large data.

Ideas

Large integer multiplication into a 10-binary array
Large integer division into a specified number

AC Code

  #include <stdio.h> #include <stdlib.h> #include <string.h> #define LEN Cha 
  R Str[len], Another[len]; 
    
  int Ten[len]; 
  int Switchtoten (); 
    
  void Switchtoanother (int k, int n); 
    
    int main () {int m, n, K; 
        
      while (scanf ("%d%d", &m, &n)!= EOF) {scanf ("%s", str); 
    
      k = Switchtoten (m);  
    Switchtoanother (k, n); 
  return 0; 
    
    int Switchtoten (int m) {int I, J, Len, K, C; 
    Initialize len = strlen (str); 
    K = 1; 
    
    memset (ten, 0, sizeof (ten)); 
      Convert to 10 number for (i = 0; i < len i + +) {for (j = 0; J < K; J + +) {ten[j] *= m; 
      } if (Str[i] >= ' 0 ' && str[i] <= ' 9 ') {ten[0] + = str[i]-' 0 '; 
      }else if (Str[i] >= ' A ' && str[i] <= ' Z ') {ten[0] + = str[i]-' a ' + 10; 
     }else if (Str[i] >= ' A ' && str[i] <= ' z ') {   Ten[0] + = str[i]-' a ' + 10; 
      
        for (j = c = 0; J < K; J + +) {Ten[j] + = C; 
          if (Ten[j] >=) {c = ten[j]/10;   
        TEN[J]%= 10; 
        }else {c = 0; 
        } while (c) {ten[k + +] = c% 10; 
      C/= 10; 
    }//flip array int temp; 
      for (i = 0, j = k-1 I < J; i + +, J-) {temp = Ten[i]; 
      Ten[i] = Ten[j]; 
    TEN[J] = temp; 
  } return k; 
    
    } void Switchtoanother (int k, int n) {int sum, I, R, T, D; 
    sum = 1; 
    R = 0; 
    
    memset (another, 0, sizeof (another)); 
    
      while (sum) {sum = 0; 
        for (i = 0; i < K; i + +) {d = ten[i]/n; 
    
        sum = D; 
          if (i = = k-1) {t = ten[i]% n; 
          if (t >= 0 && T <= 9) {Another[r] = t + ' 0 '; }else {Another[r] = t-10 + 'A '; 
        } R + +; 
        }else {ten[i + 1] = = Ten[i]% n * 10; 
      } Ten[i] = D; 
    }//print is output for (i = r-1 i >= 0; I-) {printf ("%c", Another[i]); 
  printf ("\ n"); 
 }

   /**************************************************************
         problem:1080
        User:wangzhengyi
 & nbsp;      language:c
        result:accepted
        time:170 Ms
        memory:920 kb
    ****************************************************************/ 
    
The following is attached a binary, octal, decimal, hexadecimal conversion of the C language Program (a bit long, well, look at the function n in the same portion of XD) ~

#include <stdio.h> #include <string.h> #include <stdlib.h> void d_b (int);
void D_o (int);
void d_x (int);
void b_d (int);
void b_o (int);
void b_x (int);
void O_b (int);
void o_d (int);
void o_x (int);
void X_b (char r[],int k);
void X_d (char r[],int k);
void X_o (char r[],int k);
 void Main () {int i,j,k=0;
 int q;
 Char r[10];
 printf ("+=============================================================+\n");      printf ("|
 Welcome to use the |\n conversion program ");
 printf ("+=============================================================+\n"); printf ("Note: This version only makes positive integer conversion!!")
 ");
 do {q=0;
 Fflush (stdin);
 printf ("\ n Select the system you want to convert: \n0, exit \n1, binary \n2, octal \n3, decimal \n4, hexadecimal \ n");
 scanf ("%d", &i);
   switch (i) {case 1:printf ("\ n Select a converted system: \n0, Exit \n1, binary \n2, octal \n3, decimal \n4, hexadecimal \ n");
   scanf ("%d", &j); Switch (j) {case 1:printf ("\ n) no conversion between the same system!"
    \ n ");
    Q=1;
   Break
    Case 2:printf ("Please enter the number you want to convert:");
    scanf ("%d", &k);
    B_o (k);
    Q=1;
   Break Case 3:printf ("\ nplease Enter the number you want to convert: ");
    scanf ("%d", &k);
    B_d (k);
    Q=1;
   Break
    Case 4:printf ("Please enter the number you want to convert:");
    scanf ("%d", &k);
    B_x (k);
    Q=1;
   Break Case 0:printf ("Thank you for using!!")
    
   ");
  } break;
   Case 2:printf ("\ n Select the converted system: \n0, Exit \n1, binary \n2, octal \n3, decimal \n4, hexadecimal \ n");
   scanf ("%d", &j); Switch (j) {case 2:printf ("\ n) no conversion between the same system!"
    \ n ");
    Q=1;
   Break
    Case 1:printf ("Please enter the number you want to convert:");
    scanf ("%d", &k);
    O_b (k);
    Q=1;
   Break
    Case 3:printf ("Please enter the number you want to convert:");
    scanf ("%d", &k);
    O_d (k);
    Q=1;
   Break
    Case 4:printf ("Please enter the number you want to convert:");
    scanf ("%d", &k);
    O_x (k);
    Q=1;
   Break Case 0:printf ("Thank you for using!!")
   ");
  } break;
   Case 3:printf ("\ n Select the converted system: \n0, Exit \n1, binary \n2, octal \n3, decimal \n4, hexadecimal \ n");
   scanf ("%d", &j); Switch (j) {case 3:printf ("\ n) no conversion between the same system!"
    \ n ");
   
    Q=1;
   Break
    Case 1:printf ("Please enter the number you want to convert:");
    scanf ("%d", &k);
    D_b (k);
    Q=1;
   Break Case 2:printf("Please enter the number you want to convert:");
    scanf ("%d", &k);
    D_o (k);
    Q=1;
   Break
    Case 4:printf ("Please enter the number you want to convert:");
    scanf ("%d", &k);
    D_x (k);
    Q=1;
   Break Case 0:printf ("Thank you for using!!")
   ");
  } break;
   Case 4:printf ("\ n Select the converted system: \n0, Exit \n1, binary \n2, octal \n3, decimal \n4, hexadecimal \ n");
   scanf ("%d", &j); Switch (j) {case 4:printf ("\ n) no conversion between the same system!"
    \ n ");
    Q=1;
   Break
    Case 1:printf ("Please enter the number you want to convert:");
    Fflush (stdin);
    Gets (R); for (k=0;;
    k++) {if (r[k]== ') break;
    } x_b (R,k);
    Q=1;
   Break
    Case 2:printf ("Please enter the number you want to convert:");
    Fflush (stdin);
    Gets (R); for (k=0;;
    k++) {if (r[k]== ') break;
    } x_o (R,k);
    Q=1;
   Break
    Case 3:printf ("Please enter the number you want to convert:");
    Fflush (stdin);
    Gets (R); for (k=0;;
    k++) {if (r[k]== ') break;
    } x_d (R,k);
    Q=1;
   Break Case 0:printf ("Thank you for using!!")
    
   ");
  } break; Case 0:printf ("\ n Thank you for using it!")
  
 \ n ");
}}while (Q==1);
 }The following are: binary conversion to decimal, octal, hexadecimal.
 void B_d (int a) {int i,s=0;
 int result=0;
  for (i=1;a!=0;i*=2) {if (a%10>1) {S=1;
 Break
  else {result+= (a%10) *i;
 A=A/10; } if (s==1) printf ("Your input is wrong!")
 Please re-enter \ n ");
else printf ("\ n converted number:%d\n", result);
 } void B_o (int a) {int i,j,k,s=0;
 int p[30];
 int result=0;
  for (i=1;a!=0;i*=2) {if (a%10>1) {S=1;
 Break
  else {result+= (a%10) *i;
 A=A/10;
 } for (j=0;result!=0;j++) {p[j]=result%8;
 RESULT=RESULT/8; if (s==1) printf ("Your input is wrong!")
 Please re-enter \ n ");
 else {printf ("\ n converted number is:");
 for (k=j-1;k>=0;k--) {printf ("%d", p[k]);
 printf ("\ n");
 } void b_x (int a) {int i,j,k,s=0;
 Char p[30];
 int result=0;
  for (i=1;a!=0;i*=2) {if (a%10>1) {S=1;
 Break
  else {result+= (a%10) *i;
 A=A/10;
 } for (j=0;result!=0;j++) {p[j]=result%16;
 RESULT=RESULT/16;
   if (p[j]>10) {switch (P[j]) {case 10:p[j]= ' A ';
  Break
   Case 11:p[j]= ' B ';
  Break
   Case 12:p[j]= ' C '; Break;
   Case 13:p[j]= ' D ';
  Break
   Case 14:p[j]= ' E ';
  Break
   Case 15:p[j]= ' F ';
  Break
 } else p[j]+=48; if (s==1) printf ("Your input is wrong!")
 Please re-enter \ n ");
 else {printf ("\ n converted number is:");
 for (k=j-1;k>=0;k--) {printf ("%c", P[k]);
 printf ("\ n");
The following///////are: octal is converted to binary, decimal, hexadecimal.
 void O_b (int a) {int i,j,k,s=0;
 int result=0;
 int p[30];
  for (i=1;a!=0;i*=8) {if (a%10>7) {S=1;
 Break
  else {result+= (a%10) *i;
 A=A/10;
 } for (j=0;result!=0;j++) {p[j]=result%2;
 RESULT=RESULT/2; if (s==1) printf ("Your input is wrong!")
 Please re-enter \ n ");
 else {printf ("\ n converted number is:");
 for (k=j-1;k>=0;k--) {printf ("%d", p[k]);
 printf ("\ n");
 } void O_d (int a) {int i,s=0;
 int result=0;
  for (i=1;a!=0;i*=8) {if (a%10>7) {S=1;
 Break
  else {result+= (a%10) *i;
 A=A/10; } if (s==1) printf ("Your input is wrong!")
 Please re-enter \ n ");
 else {printf ("\ n converted number is:%d\n", result);
 } void o_x (int a) {int i,j,k,s=0;
 Char p[30];
 int result=0; for (i=1;a!=0;i*=8) {if (a%10>7) {S=1;
 Break
  else {result+= (a%10) *i;
 A=A/10;
 } for (j=0;result!=0;j++) {p[j]=result%16;
 RESULT=RESULT/16;
 if (p[j]<10) p[j]+=48;
   else {switch (P[j]) {case 10:p[j]= ' A ';
  Break
   Case 11:p[j]= ' B ';
  Break
   Case 12:p[j]= ' C ';
  Break
   Case 13:p[j]= ' D ';
  Break
   Case 14:p[j]= ' E ';
  Break
   Case 15:p[j]= ' F ';
  Break }} if (s==1) printf ("Your input is incorrect!")
 Please re-enter \ n ");
 else {printf ("\ n converted number is:");
 for (k=j-1;k>=0;k--) {printf ("%c", P[k]);
 printf ("\ n");
The following///////are: 16 Convert to decimal, binary, octal.
 void X_d (char a[],int k) {int i,j,s=0;
 int result=0;
 int B[50];
   
 for (i=0;i<k;i++) {if (a[i]<= ' 9 ' &&a[i]>= ' 1 ') {b[i]=a[i]-48;
   else {switch (A[i]) {case ' a ': b[i]=10;
  Break
   Case ' B ': b[i]=11;
  Break
   Case ' C ': b[i]=12;
  Break
   Case ' D ': b[i]=13;
  Break
   Case ' E ': b[i]=14;
  Break
   Case ' F ': b[i]=15;
  Break
   Case ' a ': b[i]=10;
  Break
   Case ' B ': b[i]=11; Break;
   Case ' C ': b[i]=12;
  Break
   Case ' d ': b[i]=13;
  Break
   Case ' E ': b[i]=14;
  Break
   Case ' F ': b[i]=15;
  Break
   
  Default:s=1;
 {Result+=b[j]*i for (i=1,j=k-1;j>=0;j--, i*=16)}}} if (s==1) printf ("Your input is wrong!")
 Please re-enter \ n ");
 else {printf ("\ n Convert number is:%d", result);
 } void X_b (char a[],int k) {int i,j,s=0;
 int result=0;
 int B[50];
 int p[30];
 for (i=0;i<k;i++) {if (a[i]<= ' 9 ' &&a[i]>= ' 1 ') b[i]=a[i]-48;
   else {switch (A[i]) {case ' a ': b[i]=10;
  Break
   Case ' B ': b[i]=11;
  Break
   Case ' C ': b[i]=12;
  Break
   Case ' D ': b[i]=13;
  Break
   Case ' E ': b[i]=14;
  Break
   Case ' F ': b[i]=15;
  Break
   Case ' a ': b[i]=10;
  Break
   Case ' B ': b[i]=11;
  Break
   Case ' C ': b[i]=12;
  Break
   Case ' d ': b[i]=13;
  Break
   Case ' E ': b[i]=14;
  Break
   Case ' F ': b[i]=15;
  Break
   
  Default:s=1;
 {Result+=b[j]*i for (j=k-1,i=1;j>=0;j--, i*=16)}}} for (j=0;result!=0;j++) {P[j]=result%2
 RESULT=RESULT/2; if (s==1) printf ("Your input is wrong!")
 Please re-enter \ n ");
 else {printf ("\ n converted number is:");
 for (k=j-1;k>=0;k--) {printf ("%d", p[k]);
 printf ("\ n");
 } void X_o (char a[],int k) {int i,j,s=0;
 int result=0;
 int b[30];
 int p[30];
 for (i=0;i<k;i++) {if (a[i]<= ' 9 ' &&a[i]>= ' 1 ') b[i]=a[i]-48;
   else {switch (A[i]) {case ' a ': b[i]=10;
  Break
   Case ' B ': b[i]=11;
  Break
   Case ' C ': b[i]=12;
  Break
   Case ' D ': b[i]=13;
  Break
   Case ' E ': b[i]=14;
  Break
   Case ' F ': b[i]=15;
  Break
   Case ' a ': b[i]=10;
  Break
   Case ' B ': b[i]=11;
  Break
   Case ' C ': b[i]=12;
  Break
   Case ' d ': b[i]=13;
  Break
   Case ' E ': b[i]=14;
  Break
   Case ' F ': b[i]=15;
  Break
  Default:s=1;
 {Result+=b[j]*i for (j=k-1,i=1;j>=0;j--, i*=16)}}}
 for (j=0;result!=0;j++) {p[j]=result%8;
 RESULT=RESULT/8; if (s==1) printf ("Your input is wrong!")
 Please re-enter \ n ");
 else {printf ("\ n converted number is:");
 for (k=j-1;k>=0;k--) {printf ("%d", p[k]);printf ("\ n");
The following///////are: decimal conversion to binary, octal, hexadecimal.
 void d_b (int a) {int j,k;
 int p[30];
 for (j=0;a!=0;j++) {p[j]=a%2;
 A=A/2;
 printf ("\ n converted number is:");
 for (k=j-1;k>=0;k--) {printf ("%d", p[k]);
printf ("\ n");
 } void D_o (int a) {int j,k;
 int p[30];
 for (j=0;a!=0;j++) {p[j]=a%8;
 A=A/8;
 printf ("\ n converted number is:");
 for (k=j-1;k>=0;k--) {printf ("%d", p[k]);
printf ("\ n");
 } void d_x (int a) {int j,k;
 int p[30];
 for (j=0;a!=0;j++) {p[j]=a%16;
 A=A/16;
 if (p[j]<10) p[j]+=48;
   else {switch (P[j]) {case 10:p[j]= ' A ';
  Break
   Case 11:p[j]= ' B ';
  Break
   Case 12:p[j]= ' C ';
  Break
   Case 13:p[j]= ' D ';
  Break
   Case 14:p[j]= ' E ';
  Break
   Case 15:p[j]= ' F ';
  Break
 } printf ("\ n converted number is:");
 for (k=j-1;k>=0;k--) {printf ("%c", P[k]);
printf ("\ n");
  

 }