Title Description
Input Format
Output Format
Sample
Data range and tips
Ideas:
You must make a table before doing the math. . . Consistent style solution to a problem, do not come to play a positive solution. . .
Pre-knowledge: a collection of all of the sub-set of 2 ^ n number of each element appears, and then the selected subset scheme number 2 ^ (2 ^ n) -1 for each set if there is, first, because a Save selected from the set and not empty set is selected from a heavy, a count less.
70% of the algorithm: DP so his mother is consequently capable , we will not count as a collection of n, k is the intersection of the program, then set out ah. . f [n] [k] is the answer, f [n] [k] = C (n, k) * f [nk] [0], meaning that the first elected k th reselection intersection is 0, then F [ n] [0] ye operator, it may be selected from the set of all numbers 2 ^ (2 ^ n) from the n -1-f [] [k ] obtained, however, f [n] [k] from the previous f [nk] [0] get, nothing wrong aftereffect did not have to, his mother would hit the table with you. . (WD great God here borrow Code)
1 #include<cstdio> 2 #define int long long 3 const long long mod=1000000007; 4 long long fac[1000005],inv[1000005],invv[1000005],x,y,n,k,f0[1000005],f[1000005]; 5 long long pow(long long b,long long t,long long modd,long long ans=1){ 6 for(;t;t>>=1 , b = b * b% mode) f (t & 1 ) ans = ans * b% mode; 7 return ans; 8 } 9 Signed main () { 10 scanf ( " %% LLD LLD " , & n & k); 11 fac [ 0 ] = inv [ 0 ] = invv [ 0 ] = f0 [ 0 ] = invv [ 1 ] = inv [ 1 ] = fac [ 1 ] = 1 ; f0 [ 1 ] = 2 ; 12 for ( int i = 2 ; to <= n; ++ i) fac [i] = fac [to-1 ] * as% v, invv [i] = (- v / i * invv [v% i]) v%, INV [i] = inv [i- 1 ] * invv [i]% v; 13 for ( int i = 1 ; i <= n; ++ i) { 14 f0 [i] = (pow ( 2 , pow ( 2 , i, received 1 ), v) - 1 + v)% v; 15 for ( int j = 1 ; j <= i; j ++) F [j] = fac [i] * inv [j] *% v inv [ij]% f0 * v [ij]% v, f0 [ i] = (f 0 [i] -f [j] + v)% v; 16 f [ 0 ] = f0 [i]; 17 } 18 printf ( "%lld\n",(f[k]+mod)%mod); 19 }
100% algorithm: Boy playing positive solution, before the election of k, n- = k, and now we are asking for is a collection of n, the intersection of 0 for the program, this time we again find the definition of a collection (I thought this died), a represents a determined collection element comprising a sum of the program, then there will be n sets to bloom like a flower, a variety of intersection 
, for example, two figures overlap the intermediate part E represents the intersection does not contain exactly two programs. Then we will require the size of a part (that is, the requirements of disjoint program), inclusion and exclusion seem to smell a hint of flavor. . . . 0 = Total intersection is - (> = 1) + (> = 2) .......
years = Σ (-1) i * ^ C (n, i) * (2 ^ (2 ^ (ni)) - 1);
PS: the so-called power of the power of recursion can be used, of course, you hit two quick power I have no idea. .
#include<iostream> #include<cstdio> #include<cmath> using namespace std; const int mn=1e6+20,mod=1e9+7; int f[mn],fac[mn],inv[mn],n,k; int qpow(int x,int k) { int ans=1; for(;k;k>>=1,x=1ll*x*x%mod) if(k&1) ans=1ll*ans*x%mod; return ans; } long long C(int n,int m) {return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;} int main() { scanf("%d%d",&n,&k);fac[0]=1; for(int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mod; inv[n]=qpow(fac[n],mod-2); for(int i=n;i;i--) inv[i-1]=1ll*inv[i]*i%mod; long long ans=0,tmp=C(n,k),tp=1;n-=k; int op=(n&1)?-1:1; for(int i=n;~i;i--) { ans=(ans+op*C(n,i)*tp%mod)%mod; tp=(tp*tp%mod+2*tp%mod)%mod; op=-op; } ans=ans*tmp%mod; printf("%lld\n",(ans+mod)%mod); } //for sigma i=0->n C(n,i)(2^(2^(n−i))−1) /* g++ 1.cpp -o 1 ./1 4 1 */