Nyoj 57 6174 Questions

6174 Problem time limit:MS | Memory limit:65535 KB Difficulty:2

Describe

Let's say you have a four-digit number that each has a different number, you get a after all the numbers from the big to the small, you get B from small to large, and then you replace the number with a-a, and you continue to do so. For example, starting from 1234, you can get 4321-1234=3087, 8730-378=8352, 8532-2358=6174, and then back to its own! Now you need to write a program to determine how many times a four-digit number of such operations can occur, and the number of times the operation is calculated

For example, input 1234 execution order is 1234->3087->8352->6174->6174, output is 4

Input

the first line is N, which represents the n set of test data.
The next n lines each write a four-digit number that each of you has a different number.

Output

How many times does the action described above appear to loop

Sample input

11234

Sample output

4

1#include <iostream>2#include <algorithm>3#include <string.h>4 using namespacestd;5 Chara[5],b[5];intIa,ib;6 7 intMain ()8 {9     intn,count,temp=0;TenCin>>N; One      while(n--) A     { -Count=0; -          for(intI=0;i<4; i++) theCin>>A[i]; - Loop: -ia=ib=0; -Sort (a,a+4); +          for(intI=0;i<4; i++) -         { +ia*=Ten; Aia+= (a[i]-'0'); at         } -          for(intI=3; i>=0; i--) -         { -ib*=Ten; -ib+= (a[i]-'0'); -         } inib=ib-ia; -count++; to         if(temp!=IB) +         { -temp=IB; the          for(intI=0;i<4; i++) *         { $A[i]= (ib%Ten+'0');Panax NotoginsengIb/=Ten; -         } the             GotoLoop; +         } A         Else the         { +cout<<count<<Endl; -         } $     } $  -}

1#include <iostream>2#include <algorithm>3#include <stdio.h>4 using namespacestd;5 intMain ()6 {7     intK;8Cin>>K;9      while(k--)Ten     { One         intn,a[4],n1,n2; Ascanf"%d",&n); -         ints=1; -          while(n!=6174) the         { -a[0]=n%Ten; -a[3]=n/ +; -a[1]=n/Ten%Ten; +a[2]=n/ -%Ten; -Sort (a,a+4); +n1= +*a[3]+ -*a[2]+Ten*a[1]+a[0]; AN2= +*a[0]+ -*a[1]+Ten*a[2]+a[3]; atN=n1-N2; -s++; -         } -printf"%d\n", s); -     } -}

Nyoj 57 6174 Questions