The beauty of programming should not be intimidated by the factorial, but the beauty of 2.2.

The beauty of programming should not be intimidated by the factorial, but the beauty of 2.2.

When I first saw this question, I thought it was the idea to use a string to represent the integer number. Later I saw that because it is a factorial of a number, if it is large, therefore, it is unreasonable to use the string representation. After reading the explanation of this question, the book gives a formula and then understands the meaning of the question.

First, I will post the function declaration and question requirements:

/* 2.2 do not be intimidated by the factorial * // * 2.2.1 N! The number of zeros at the end of */int dutcountof0infactoria1( int);/* 2.2.2 N! The binary value of */int dutpositionof1infactoria1( int );

We think that there is a number of N, so how many c (c> = 1 & c <= 9) exist in 1-N? We can use this formula to calculate:

The formula for calculating the number of numbers of x in 1-N is N/x + N/x2 + N/x3 +..., and the formula must converge at last.

Therefore, you can write the following code:

/* 2.2.1 N! The number of zeros at the end of the question * // the idea of solving this question is: 2*5 = 10. Therefore, we only need to confirm that there are several 5 in 1-N (because 2 has many) */int dutcountof0infactoria1n (int n) {if (n <= 0) return 0; int count = 0;/* the formula for calculating the number of x in 1-N is: N/x + N/x2 + N/x3 + ..., and this formula must eventually converge */while (n) {n + = n/5; n/= 5 ;} /* return the number of 0 at the end */return count;}/* 2.2.2 N! Binary represents the position of the second digit 1 * // * indicates the second digit 1 in the binary. Therefore, you need to know how many zeros exist next to it. Therefore, we only need to determine how many 2 s there are. */int dutpositionof1infactoria1( int n) {if (n <= 0) return 0; int count = 0; /* the formula for calculating the number of x in 1-N is: N/x + N/x2 + N/x3 + ..., and this formula must eventually converge */while (n) {count + = n/2; n/= 2;} return count + 1 ;}