table of Contents
@description@
In a given range [0, 2 ^ N) of the random number generator, a given parameter A [0 ... 2 ^ N- 1].
The generator has \ (\ frac {A_i} { \ sum A} \) probability of generation i, are each generated independently.
There is now a X, an initial zero. Generating a random number for each operation and X exclusive v or v. Take
For each i ∈ [0, 2 ^ N), seeking operation of the X number of times equal to the first desired i.
The original title title face .
@solution@
Difficult to think expectations dp. Definition of DP [i] denotes the i reaches the desired number, then:
\[ dp[0] = 0 \\ dp[i] = (\sum_{j=0}^{2^N - 1}dp[j]\times p[i\oplus j]) + 1 \]
Wherein \ (P [I] = \ FRAC A_i {{} \} A SUM \) .
Simple approach is to Gaussian elimination. Apparently had not.
For routine optimization of Gaussian elimination using FIG structure (such as the DAG, the tree or chain) transfer, but the problem of transfer graph is complete, can not.
How to do? Observe the transfer type of structure and found that it actually is XOR convolution. So we try to take the generating function that set.
If notation generating function, and can be referred to as \ (DP \ Oplus P + the I = DP + K \ Times T \) , where \ (I [i] = 1 , T [i] = [i = 0] \) , \ (K \) is unknown.
Note that when n = 0 convolution is not established, it is necessary to fill in at the end of a \ (k \ Times T \) .
Variable bit shaped to give \ (DP \ Oplus (T - P) = the I - K \ Times T \) , both sides simultaneously fwt obtain \ (dp '\ times (T - P)' = I '- k \ times T' \) .
Noting \ ((T - P) ' \) of 0 is always (by definition is understood fwt) is 0, so the \ (I' - k \ times T '\) of 0 is also 0, whereby It can be solved for k.
But this way we do not know \ (dp '[0] \ ) How much is once again set the unknowns q. When inverse transform into operation with the unknown substituting it.
Then \ (DP \) series can be expressed as a linear function of q containing, according to \ (dp [0] = 0 \) can be solved for q inverse, so \ (DP \) series out on the solution.
@accepted code@
#include <cstdio>
const int MOD = 998244353;
const int INV2 = (MOD + 1) >> 1;
int add(int x, int y) {return (x + y >= MOD ? x + y - MOD : x + y);}
int sub(int x, int y) {return (x - y < 0 ? x - y + MOD : x - y);}
int mul(int x, int y) {return 1LL*x*y%MOD;}
int pow_mod(int b, int p) {
int ret = 1;
for(int i=p;i;i>>=1,b=mul(b,b))
if( i & 1 ) ret = mul(ret,b);
return ret;
}
struct node{
int k, b;
node() : k(0), b(0) {}
node(int _k, int _b) : k(_k), b(_b) {}
int get(int x) {return add(mul(k, x), b);}
friend node operator + (node a, node b) {
return node(add(a.k, b.k), add(a.b, b.b));
}
friend node operator - (node a, node b) {
return node(sub(a.k, b.k), sub(a.b, b.b));
}
friend node operator * (node a, int k) {
return node(mul(a.k, k), mul(a.b, k));
}
friend node operator / (node a, int k) {
return a * pow_mod(k, MOD - 2);
}
};
void fwt(node *A, int m, int type) {
int n = (1 << m), f = (type == 1 ? 1 : INV2);
for(int i=1;i<=m;i++) {
int s = (1 << i), t = (s >> 1);
for(int j=0;j<n;j+=s)
for(int k=0;k<t;k++) {
node x = A[j+k], y = A[j+k+t];
A[j+k] = (x + y)*f, A[j+k+t] = (x - y)*f;
}
}
}
node A[1<<18], B[1<<18], C[1<<18], f[1<<18];
int main() {
int N, M, S = 0; scanf("%d", &N), M = (1 << N);
for(int i=0;i<M;i++) scanf("%d", &A[i].b), S = add(S, A[i].b);
S = pow_mod(S, MOD - 2);
for(int i=0;i<M;i++) A[i].b = sub(i == 0 ? 1 : 0, mul(A[i].b, S));
for(int i=0;i<M;i++) B[i].b = 1;
C[0].b = MOD - 1;
fwt(A, N, 1), fwt(B, N, 1), fwt(C, N, 1);
int tmp = mul(B[0].b, pow_mod(C[0].b, MOD-2));
for(int i=1;i<M;i++)
f[i] = (B[i] - C[i]*tmp) / A[i].b;
f[0].k = 1; fwt(f, N, -1);
int x = sub(0, mul(pow_mod(f[0].k, MOD-2), f[0].b));
for(int i=0;i<M;i++) printf("%d\n", f[i].get(x));
}@details@
I feel like a practice fucks. . . But I am not clear official positive solution is wisdom. . .