2024第三届大学生算法大赛真题训练2 解题报告 | 珂学家

前言

在这里插入图片描述

题解

D是FFT板子题，这么来看，其实处于ACM入门题，哭了T_T.

D. 行走之谜

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

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

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

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

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;
}