思路: FFT

如果你知道多项式乘法，继而知道FFT，那题纯粹就是板子题，可惜当时比赛的时候，无人AC。

这题来简单抽象一下

对于概率分布，引入一个多项式

$$f(x)=a_0+a_1\ast x^1+a_2\ast x^2+...+a_{12}\ast x^{12}$$

这边 $x^t$, t其实对标了坐标轴t点

那么执行n次，然后刚好落在k点，不就是

$$f(x{)}^n=b_0+b_1\ast x^1+b_2\ast x^2+...+b_k\ast x^k+...+b_{12n}\ast x^{12n}中，第x^k的系b_{k}k$$

这题就这么简单

套用一下FFT板子，即可

#include <bits/stdc++.h>
using namespace std;const int mod = 1e9+7;namespace fft
{struct num{double x,y;num() {x=y=0;}num(double x,double y):x(x),y(y){}};inline num operator+(num a,num b) {return num(a.x+b.x,a.y+b.y);}inline num operator-(num a,num b) {return num(a.x-b.x,a.y-b.y);}inline num operator*(num a,num b) {return num(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}inline num conj(num a) {return num(a.x,-a.y);}int base=1;vector<num> roots={{0,0},{1,0}};vector<int> rev={0,1};const double PI=acosl(-1.0);void ensure_base(int nbase){if(nbase<=base) return;rev.resize(1<<nbase);for(int i=0;i<(1<<nbase);i++)rev[i]=(rev[i>>1]>>1)+((i&1)<<(nbase-1));roots.resize(1<<nbase);while(base<nbase){double angle=2*PI/(1<<(base+1));for(int i=1<<(base-1);i<(1<<base);i++){roots[i<<1]=roots[i];double angle_i=angle*(2*i+1-(1<<base));roots[(i<<1)+1]=num(cos(angle_i),sin(angle_i));}base++;}}void fft(vector<num> &a,int n=-1){if(n==-1) n=a.size();assert((n&(n-1))==0);int zeros=__builtin_ctz(n);ensure_base(zeros);int shift=base-zeros;for(int i=0;i<n;i++)if(i<(rev[i]>>shift))swap(a[i],a[rev[i]>>shift]);for(int k=1;k<n;k<<=1){for(int i=0;i<n;i+=2*k){for(int j=0;j<k;j++){num z=a[i+j+k]*roots[j+k];a[i+j+k]=a[i+j]-z;a[i+j]=a[i+j]+z;}}}}vector<num> fa,fb;vector<int> multiply(vector<int> &a, vector<int> &b){int need=a.size()+b.size()-1;int nbase=0;while((1<<nbase)<need) nbase++;ensure_base(nbase);int sz=1<<nbase;if(sz>(int)fa.size()) fa.resize(sz);for(int i=0;i<sz;i++){int x=(i<(int)a.size()?a[i]:0);int y=(i<(int)b.size()?b[i]:0);fa[i]=num(x,y);}fft(fa,sz);num r(0,-0.25/sz);for(int i=0;i<=(sz>>1);i++){int j=(sz-i)&(sz-1);num z=(fa[j]*fa[j]-conj(fa[i]*fa[i]))*r;if(i!=j) fa[j]=(fa[i]*fa[i]-conj(fa[j]*fa[j]))*r;fa[i]=z;}fft(fa,sz);vector<int> res(need);for(int i=0;i<need;i++) res[i]=fa[i].x+0.5;return res;}vector<int> multiply_mod(vector<int> &a,vector<int> &b,int m,int eq=0){int need=a.size()+b.size()-1;int nbase=0;while((1<<nbase)<need) nbase++;ensure_base(nbase);int sz=1<<nbase;if(sz>(int)fa.size()) fa.resize(sz);for(int i=0;i<(int)a.size();i++){int x=(a[i]%m+m)%m;fa[i]=num(x&((1<<15)-1),x>>15);}fill(fa.begin()+a.size(),fa.begin()+sz,num{0,0});fft(fa,sz);if(sz>(int)fb.size()) fb.resize(sz);if(eq) copy(fa.begin(),fa.begin()+sz,fb.begin());else{for(int i=0;i<(int)b.size();i++){int x=(b[i]%m+m)%m;fb[i]=num(x&((1<<15)-1),x>>15);}fill(fb.begin()+b.size(),fb.begin()+sz,num{0,0});fft(fb,sz);}double ratio=0.25/sz;num r2(0,-1),r3(ratio,0),r4(0,-ratio),r5(0,1);for(int i=0;i<=(sz>>1);i++){int j=(sz-i)&(sz-1);num a1=(fa[i]+conj(fa[j]));num a2=(fa[i]-conj(fa[j]))*r2;num b1=(fb[i]+conj(fb[j]))*r3;num b2=(fb[i]-conj(fb[j]))*r4;if(i!=j){num c1=(fa[j]+conj(fa[i]));num c2=(fa[j]-conj(fa[i]))*r2;num d1=(fb[j]+conj(fb[i]))*r3;num d2=(fb[j]-conj(fb[i]))*r4;fa[i]=c1*d1+c2*d2*r5;fb[i]=c1*d2+c2*d1;}fa[j]=a1*b1+a2*b2*r5;fb[j]=a1*b2+a2*b1;}fft(fa,sz);fft(fb,sz);vector<int> res(need);for(int i=0;i<need;i++){int64_t aa=fa[i].x+0.5;int64_t bb=fb[i].x+0.5;int64_t cc=fa[i].y+0.5;res[i]=(aa+((bb%m)<<15)+((cc%m)<<30))%m;}return res;}vector<int> square_mod(vector<int> &a,int m){return multiply_mod(a,a,m,1);}
};vector<int> ksm(vector<int>&x , int64_t p, int n){if(p == 1) {return x;}vector<int> z = ksm(x, p/2, n);z = fft::multiply_mod(z,z,mod);while(z.size() > 12*n+4)z.pop_back();if(p%2 == 0){return z;}z = fft::multiply_mod(z,x,mod);while(z.size() > 12*n+4)z.pop_back();return z;
}int64_t ksm(int64_t x, int64_t p, int64_t mod){int64_t r = 1;while (p > 0) {if (p % 2==1) {r = r * x % mod;}p /= 2;x = x * x % mod;}return r;
}int main() {int64_t inv100 = ksm(100, mod-2, mod);int t;cin >> t;while (t--) {int n, k;cin >> n >> k;vector<int> p(13);for (int& x: p) cin >> x;vector<int> b;for(int i=0 ; i <=12; i ++){b.push_back((p[i]*inv100)%mod);}b = ksm(b, n, n);cout << b[k] << '\n';}return 0;
}

注: 这边FFT/NTT 板子来自于官解