POJ 2992 divisors to find the number of combinatorial factors

Title Source: POJ 2992 divisors

Test instructions: ...

Idea: The uniqueness of prime decomposition a number can be decomposed into several prime numbers multiplied P1^X1*P2^X2*...*PN^XN

Number of factors according to multiplication principle (x1+1) * (x2+1) *...* (xn+1)

Can not directly find the combination of the number of overflow can not be multiplied by the number of decomposition factor so that will time out

C (n,m) =n!/(m!* (n-m)!)

Another dp[i][j] represents the number of j factors in the factorial of I (J is prime)

Then the number of I primes is dp[n][i]-dp[m][i]-dp[n-m][i]

Last for loop from 1 to N enumeration I statistics

#include <cstdio> #include <cstring> #include <cmath> using namespace std;
const int MAXN = 500;
int VIS[MAXN];
int PRIME[MAXN];
int DP[MAXN][MAXN];
	Sieve primes void sieve (int n) {int m = sqrt (n+0.5);
	memset (Vis, 0, sizeof (VIS));
	Vis[0] = vis[1] = 1;
for (int i = 2, I <= m; i++) if (!vis[i]) for (int j = i*i; J <= N; j + = i) vis[j] = 1;
	} int Get_primes (int n) {sieve (n);
	int c = 0;
	for (int i = 2; I <= n; i++) if (!vis[i]) prime[c++] = i;
return C;
	} int get (int n, int m) {int sum = 0;
		while (n) {sum + = n/m;
	n/= m;
} return sum;
	} void pre (int n) {for (int i = 2; I <= n; i++) {for (int j = 2; J <= I; j + +) Dp[i][j] = Get (i, j);
	}} int main () {int c = get_primes (444);
	Pre (444);
	int n, m;
		while (scanf ("%d%d", &n, &m)! = EOF) {Long long ans = 1;
			for (int i = 2; I <= n; i++) {if (vis[i]) continue;
			C (n,m) =n!/(m!* (n-m)!)
		Ans *= dp[n][i]-dp[m][i]-dp[n-m][i] + 1; } printf ("%lld\n", ans);
} return 0; }