Help little Li solve the problem C + + consolidates the number of digits in each location

Now there is a number a= 12345;

Want to get this number without a number use Division + modulo method to get

Principle: Except (/) to obtain the remainder of the Shangmo (%)

In this way, one of the required numbers is divided by the number of bits (a Chichong) and then divided into units.

Formula: A/(number of digits to be obtained)%10

Such as:

Million bit: 12345/10000%10 = 1;

thousand: 12345/1000%10 = 12%10 = (1...2) = 2;

Hundred: 12345/100%10 = 123%10= (12.....3) = 3;

10 bits: 12345/10%10 = 1234%10= (123.....4) = 4;

Digit: 12345/1%10 = 12345%10 = (1234...5) = 5;

1#include <iostream>2 using namespacestd;3 intMain () {4     inti;5     inta,b,c,d;6     7      for(i= -; i<= +; i++){8A = i/1%Ten;9b = i/Ten%Ten;Tenc = i/ -%Ten; One  A         if((a+b+c) = =5){ -D=i; -cout<<"and for the number of 5 there are:"<<d<<Endl; the         } -     } -  - return 0; +}

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Each number itself modulo a number larger than its own when equal to its own in the binary number is the largest i%i = 0;

Help little Li solve the problem C + + consolidates the number of digits in each location