Hdu1299&&spoj-kpequ (Find the number of factors to find the solution)

Topic Link: HDU 1299 portal

Test instructions

The number of solutions to the 1/x+1/y=1/n of equations 1/3+1/2 and 1/2+1/3 are regarded as a set of solutions.

Analysis:

1/x+1/y = 1/n Set y = n + k;

==>1/x + 1/(n+k) =1/n;

==>x = n^2/k + N;

Because X is an integer, K is the approximate number of n^2.

The code is as follows:

 #include <iostream> #include <cstring> #include <cstdio> #include < algorithm>using namespace Std;const int maxn = 1e7+10;typedef long long ll;int p[maxn/10],cnt;bool vis[maxn];int fac[10    00],tot;void init () {cnt=0;    memset (vis,0,sizeof (VIS));            for (int i=2;i<maxn;i++) {if (!vis[i]) {p[cnt++]=i;        for (int j=i+i;j<maxn;j+=i) vis[j]=1;    }}}ll get (LL x) {tot=0;    memset (fac,0,sizeof (FAC));            for (int i=0;i<cnt&&p[i]*p[i]<=x;i++) {if (x%p[i]==0) {while (x%p[i]==0) fac[tot]++,x/=p[i];        tot++;    }} if (x>1) fac[tot++]=1;    LL ans = 1;    for (int i=0;i<tot;i++) ans=ans* (2*fac[i]+1); return ans;}    int main () {init ();    int T,cas=1;    scanf ("%d", &t);        while (t--) {LL n;        scanf ("%i64d", &n);    printf ("Scenario #%d:\n%i64d\n\n", cas++, (Get (n) +1)/2); } return 0;} 

Topic Link: spoj-kpequ Portal

Test instructions

The same question is almost the same as n replaced by n!, so high precision and n! factor decomposition is required.

Import Java.math.biginteger;import Java.util.scanner;public class Main {static int cnt; static int[] pri =new int[10010];s Tatic boolean[] Vis = new boolean[10010];p ublic static void init () {cnt=0;for (int i=0;i<10010;i++) vis[i] = False;for (in T i=2;i<10010;i++) {if (!vis[i]) {pri[cnt++]=i;for (int j=i+i;j<10010;j+=i) Vis[j]=true;}}} public static int calc (int n,int p) {if (n<p) return 0;else return calc (n/p,p) +n/p;} public static void Main (string[] args) {//TODO auto-generated method Stubinit (); Scanner cin = new Scanner (system.in), while (Cin.hasnext ()) {int N=cin.nextint (), if (n==0) break; BigInteger ans = biginteger.one;for (int i=0;i<cnt&&pri[i]<=n;i++) {int tmp = CALC (n,pri[i]); tmp = tmp*2+1; Ans=ans.multiply (biginteger.valueof (TMP));} System.out.println (ANS);}}

Hdu1299&&spoj-kpequ (Find the number of factors to find the solution)