Getting started with algorithm competitions 5.42 basics of mathematics-factors and factorial, getting started with algorithm Competitions

Getting started with algorithm competitions 5.42 basics of mathematics-factors and factorial, getting started with algorithm Competitions

Enter a positive integer n (2 <=n <= 100) and multiply the factorial n! = 1*2*3 *... * N is decomposed into the form of prime factor multiplication, and each output from small to large is (2, 3, 4, 5 ...) Index. For example, 825 = 3*5*5*11 should be expressed as (, 1 ), 0, 1, 2, 0, 1, 2, 3, 5, 7, and 11 respectively. The program should ignore prime numbers greater than the maximum factor (otherwise there will be an infinite number of zeros at the end ).
Sample input:
5
53
Sample output:
5! = 3 1 1
53! = 49 23 12 8 4 4 3 2 2 1 1 1 1 1 1

1 # include <stdio. h> 2 # include <string. h> 3 // determine the prime number 4/* do NOT use this if x is very large or small 5 * n too hour n = 1 will be incorrectly judged as a prime number, I * I may overflow 6 */7 int is_prime (int n) 8 {9 int I; 10 for (I = 2; I * I <= n; I ++) // judge the integer i11 if (n % I = 0) return 0 that cannot exceed sqrt (x); // once a factor greater than 1 is found, returns 0 (false) 12 return 1 immediately; // returns 1 (true) 13} 14 15 // prime number table 16 int prime [100], count = 0; 17 int main () 18 {19 // variable I, n and the exponent of each Prime Number 20 int I, n, p [100]; 21 // construct a prime number Table 22 for (int I = 2; I <= 100; I ++) 23 if (is_prime (I) prime [count ++] = I; 24 while (scanf ("% d", & n) = 1) 25 {26 printf ("% d! = ", N); 27 memset (p, 0, sizeof (p); // string. h28 int maxp = 0; // defines the maximum prime factor subscript and assigns the initial value 29 for (I = 1; I <= n; I ++) 30 {31 // I must be copied to the variable m, instead of directly modifying it 32 int m = I; 33 for (int j = 0; j <count; j ++) 34 while (m % prime [j] = 0) // repeatedly divided by prime [j], and accumulate p [I] 35 {36 m/= prime [j]; 37 p [j] ++; 38 if (j> maxp) maxp = j; // update the maximum prime factor subscript 39} 40} 41 // only loop to the maximum subscript 42 for (I = 0; I <= maxp; I ++) 43 printf ("% d", p [I]); 44 printf ("\ n"); 45} 46 return 0; 47}View Code

Analysis:
Because am × an = a (m + n), you only need to accumulate the exponent corresponding to all prime factors. Note that n <= 100, these prime factors will not exceed 100, ignore the last 0 when outputting the data.