First, design ideas
When discussing in class, the teacher's idea was to use the previous results to calculate the "1" of the larger number. But I do not think so, I am the input of the positive integer number of each bit on the decomposition of the calculation. such as ABC, first count C, plus B, and finally add a.
Second, the source code
1 //One.cpp:Defines the entry point for the console application.2 //3 4#include"stdafx.h"5#include"iostream.h"6 7 intFindones (intnum)8 {9 intones,mul,num1,num2,num3;Tenones=0; OneMul=1;//Divisor, 10 of the N-th square Anum1=num2=num3=0; - while(Num/mul)//when the rounding is not zero, the loop is calculated - { thenum1=num-(Num/mul) *mul;//the number on the latter one -num2= (Num/mul)%Ten;//number on the current number of digits -num3=num/(mul*Ten);//the number on the previous one - Switch(num2)//If the positive integer is abc,b as the current number of digits + { - Case 0://The remainder is 0 o'clock, containing A*mul 1; +ones+=num3*Mul; A Break; at Case 1://The remainder is 1 o'clock and contains a*mul+c+1; -ones+=num3*mul+num1+1; - Break; - default://the rest of the >1 are (a+1) *mul -ones+= (num3+1)*Mul; - Break; in } -mul*=Ten;//change the divisor and continue to calculate higher to } + returnones; - } the intMainintargcChar*argv[]) * { $ intNumber,ones;//number, number of 1Panax Notoginsengcout<<"Please enter a positive integer:"; -Cin>>Number ; theones=findones (number); +cout<<number<<"The number of "1" in"<<ones<<Endl; A return 0; the}
Iii. Results
Iv. Experience
When I hit the code, I thought it would be very good to write, want to use the idea of recursion to achieve, but did not be able to summarize the situation on each person. So draw on the idea of the Internet, there is a program is a one-time look at three, the current bit, high one and low one. The code is simple and perfect.
I find that every time I write this algorithm-oriented program, I am stretched out, or have to practice more, improve their ability to optimize the code.
In-the-box assignment-there are several "1" (C + +)