Problem H Rock Paper Scissors，FFT

topic

topic link

meaning of the title

Give the order of two pieces of rock-paper-scissors$S$and$T$,in$T$To be shorter, now let you put$T$Past$S$lean on a certain position, so that after leaning,$T$can win$S$The maximum number of subsections.
As shown in the figure:

answer

This problem is very typical FFT, first of all we put$T$Change the sequence to a sequence it can win$T'$, that is,$T$in sequence$R、S、P$corresponding to$S、P、R$form$T'$。

In this case, we only need to$S$It is enough to find a paragraph with the greatest degree of matching, and the greatest degree of matching is the answer.

To match, we put$T'$Inverted, denoted as$rT'$, we imagine the process of doing convolution and
find the first number in the new convolution sequence$k$The value of the positions is equal to$S[k-lenT,k-1]$and$T'$The sum of the products of the corresponding positions of the sequence.

To solve this problem using convolution, we split the problem into 3 parts, i.e. consider separately$P、S、R$At the same time, the maximum matching degree, and finally the matching degree of the same position is added up.

code

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
double pi = acos(-1.0);
struct complex{
    double re,im;
    complex(double r = 0.0,double i = 0.0):re(r),im(i){};
    complex operator+(complex com){
        return complex(re+com.re,im+com.im);
    }
    complex operator-(complex com){
        return complex(re-com.re,im-com.im);
    }
    complex operator*(complex com){
        return complex(re*com.re-im*com.im,re*com.im+im*com.re);
    }
};
complex wn,wntmp;
void rader(complex arr[],int n){
    int num = n-1;
    for(int i = 0;i < n;++i){
        int tn = n>>1;
        while(num && num >= tn) num ^= tn,tn >>= 1;
        num |= tn;
        if(num > i) swap(arr[i],arr[num]);
    }
}
void FFT(complex cs[],int n,int f){
    rader(cs,n);
    for(int s = 1;s < n;s <<= 1){
        wn = complex(cos(f*2*pi/(s*2)),sin(f*2*pi/(s*2)));
        for(int offset = 0;offset < n;offset += s<<1){
            wntmp = complex(1.0,0.0);
            for(int i = 0;i < s;++i){
                complex u = cs[offset+i],v = cs[offset+i+s]*wntmp;
                cs[offset+i] = u + v;
                cs[offset+i+s] = u - v;
                wntmp = wntmp * wn;
            }
        }
    }
    if(f == -1)
        for(int i = 0;i < n;++i)
            cs[i].re /= n;
}
int n,m;
const int maxn = 1e5+7;
char S[maxn],T[maxn];
int ans[maxn*4];
complex csA[maxn*4],csB[maxn*4];
#define pr(x) cout<<#x<<":"<<x<<endl
void solve(char c){
    memset(csA,0,sizeof(csA));
    memset(csB,0,sizeof(csB));
    for(int i = 0;i < n;++i) csA[i] = complex(S[i]==c?1.0:0);
    for(int i = 0;i < m;++i) csB[i] = complex(T[i]==c?1.0:0);
    int len = 1;
    while(len < n) len<<=1;
    len <<= 1;
    FFT(csA,len,1);
    FFT(csB,len,1);
    for(int i = 0;i < len;++i) csA[i] = csA[i]*csB[i];
    FFT(csA,len,-1);
    for(int i = m-1;i < len;++i) {
        ans[i] += int(csA[i].re+0.5);
    };

}
char big(char c){
    if(c == 'R') return 'S';
    if(c == 'S') return 'P';
    if(c == 'P') return 'R';
}
int main(){
    cin>>n>>m>>S>>T;
    for(int i = 0;i < m/2;++i) swap(T[i],T[m-1-i]);
    for(int i = 0;i < m;++i) T[i] = big(T[i]);
    solve('P');
    solve('S');
    solve('R');

    int mx = 0;
    for(int i = 0;i < n+m+1;++i) mx = max(mx,ans[i]);
    cout<<mx<<endl;

    return 0;
}