The sequence does not contain 1, the number containing 1 is removed (the administrator does not like the ranking problem)

Problem description, the administrator is a single dog, in the order of a sequence of numbers, see 1 is uncomfortable, so every 1 on the Carry

Examples:

1 2 3 4 5 6 7 8 9 10 11 12 ... 100 101 102) 103 104

Administrator's record:

If it contains 0 cases

0 2 3 4 5 6 7 8 9 20 22 23 24 .... 99 200 202

Input: 1

Output: 2

Input: 11

Output: 23

#include <stdio.h>#include<string.h>Charstr[ +];//Input Stringintstart[ +],ans[ +],res[ +];//dividend, quotient, remainder//pre-and post-conversion input systemConst intOldbase =Ten;Const intNewbase =9;voidChange () {//Restore individual digits to digital form        intI,len =strlen (str); start[0] =Len;  for(i=1; i<= len;i++)        {            if(str[i-1] >='0'&& str[i-1] <='9') {Start[i]= str[i-1] -'0'; }        } }voidsolve () {memset (res,0,sizeof(res));//remainder is initialized to null        inty,i,j; //modulus n Take the remainder method, (the general rule is the first surplus is low, the remainder is high)         while(start[0] >=1)        {//as long as the dividend is still greater than or equal to 1, then continue to "modulo 2 take the remainder"y=0; I=1; ans[0]=start[0]; //             while(I <= start[0]) {y= y * oldbase +Start[i]; Ans[i+ +] = y/newbase; Y%=newbase; } res[++res[0]] = y;//The remainder of this round of operationsi =1; //to the starting point of the next round of business             while((i<=ans[0]) && (ans[i]==0)) i++; //clear this round of use of the dividendmemset (Start,0,sizeof(start)); //The quotient of this round becomes the next round of dividend             for(j = i;j <= ans[0];j++) start[++start[0]] =Ans[j]; memset (ans,0,sizeof(ans));//clear this round of quotient, prepare for the next round of operations                                                                } }voidoutput () {//reverse output from high to low        inti; intnum[9]={0,2,3,4,5,6,7,8,9};  for(i = res[0];i >=1;--i) {printf ("%d", Num[res[i]]); } printf ("\ n"); }intMain () {scanf ("%s", str);            Change ();            Solve ();            Output (); return 0;}

Summary of the problem, personal thinking, in fact, this is a number of the conversion problem, in the series only, 0,2,3,4,5,6,7,8,9 that is only 9 bits,

This is a 10 binary conversion to 9 binary problem, we set up an array num[9]; corresponding input 0 to 9, then the problem is reduced to a large number divided by the number of digits!

The problem is simple. A large number of conversions can be made.

The sequence does not contain 1, the number containing 1 is removed (the administrator does not like the ranking problem)