Codeforces E-anya and Cubes block violence search

Say is given n cubes, cubes from 1 to n, each cube has a number, you have K chance to make the K number their own factorial, (natural use can be less than k chance, each cube is used up to 1 times), then find you from the N cubes to re-select any of the cubes to make Their and for S

N<25

Can know the direct use of N to enumerate the natural unbearable, we divide n numbers into two parts to calculate, the front N/2 number we enumerate they use J can get the number, and then the number of N-N/2 back to use this method, directly in the DFS enumeration at the time to determine whether s-s in the previous clock has occurred, Add up when they appear. chunked processing

1#include <iostream>2#include <algorithm>3#include <string.h>4#include <map>5#include <cstdio>6 using namespacestd;7typedefLong LongLL;8Map<ll,int>mp[ -];9LL a[ -];TenLL dp[ -]; OneLL time[1000000]; A intn,k; - LL S,ans; - voidDfsintPintK, LL s) { the     if(p==n/2){ -mp[k][s]++; -}Else{ -DFS (p+1, k,s); +         if(s+a[p]<=s) DFS (p+1, k,s+a[p]); -         if(k +1<=k && a[p]<= -&& S+dp[a[p]] <=s) DFS (p+1, K +1, s+Dp[a[p]]); +     } A } at voidDFS1 (intPintK, LL s) { -     if(p==N) { -           for(inti =0; i+k<=k; i++) -             if(Mp[i].count (S-s) >0) ans+=mp[i][s-s]; -}Else{ -DFS1 (p+1, k,s); in          if(s+a[p]<=s) DFS1 (p+1, k,s+a[p]); -          if(k +1<=k && a[p]<= -&& s+dp[a[p]]<=s) DFS1 (p+1, K +1, s+Dp[a[p]]); to     } + } - intMain () the { *  $ Panax Notoginsengdp[0]=1; -     for(LL i=1; i<= -; i++) dp[i]=dp[i-1]*i; thecn1=0; +scanf"%d%d%i64d",&n,&k,&S); A     for(intI=0; i<n; i++) thescanf"%i64d",&a[i]); +Dfs0,0,0); -DFS1 (n/2,0,0); $printf"%i64d\n", ans); $     return 0; -}

Codeforces E-anya and Cubes block violence search