Lightoj 1038-race to 1 Again (expected +DP)

Title Link: http://lightoj.com/volume_showproblem.php?problem=1038

Test instructions is: give you an n (1≤n≤105) each time n is randomly selected a factor d, and then let n=n/d, ask N to become 1 times the expectation;

When N = 2 o'clock 2 There are two factors:

E[2] = E[1]/2 + E[2]/2 + 1; So you can find e[2];

When n = 8 o'clock 8 There are 4 factors 1 2 4 8;

E[8] = E[1]/4 + E[2]/4 + E[4]/4 + e[8]/4+ 1; So you can find e[8];

......

we use e[i] to indicate that I becomes 1 of the times expected; then e[i] = e[a[1]]/cnt + e[a[2]]/cnt + ... + e[a[cnt]]/cnt + 1; (plus 1 is because this time except once );

Wherein CNT is the number of factors of I, a array of factor set I, if the order from small to large a[1] = 1, a[cnt] = i;

So the a[cnt in the upper formula] is replaced by I; the collation can be e[i] = (e[a[1]]+e[a[2]]+ ... +e[a[cnt-1]]+cnt)/(CNT-1);

#include <cstring>#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<stack>#include<vector>#include<queue>using namespacestd;#defineN 100005#defineMet (A, b) memset (A, B, sizeof (a))#defineMOD 110119typedefLong LongLL;DoubleDp[n];voidInit () {dp[1] =0;  for(intI=2; i<n; i++)    {        Doublesum =0; intCNT =0;  for(intj=1; j*j<=i; J + +)        {            if(I%j = =0) {CNT++; Sum+=Dp[j]; if(J*j! =i) {cnt++; Sum+ = dp[i/j];///J is the factor of I, i/j is also the factor of I;} }} sum+=CNT; Dp[i]= sum/(cnt-1); }}intMain () {Init (); intT, T =1, N; scanf ("%d", &T);  while(t--) {scanf ("%d", &N); printf ("Case %d:%.6f\n", t++, Dp[n]); }    return 0;}

View Code

Lightoj 1038-race to 1 Again (expected +DP)