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[题目通道](【L&K R-03】密码串匹配 - 洛谷)
一道神题。
如果没有修改操作，翻转A数组或B数组后就是裸的FFT了
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如果每次操作都暴力修改+FFT时间复杂度显然爆炸
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如果每次操作都不修改，记下修改序列，询问时加上修改序列的贡献，复杂度仍然爆炸
于是考虑把两种方法结合起来，对操作序列分块
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#define M 1000005
#define mod 998244353
#define g 3
using namespace std;
int n,ll,m,len=1,L,power3[M],power_inv3[M],rev[M],H,cnt;
int X[M],Y[M],E[M];
long long A[M],B[M],C[M],tmp_A[M],tmp_B[M],len_inv,sum;
long long add(long long u,long long v){return (u+=v)>=mod?u-mod:u;}
long long inv(long long x){return x==1?1:(mod-mod/x)*inv(mod%x)%mod;}
long long power(long long x,long long y){long long ans=1,now=x;for (register long long i=y;i;i>>=1,now=now*now%mod)if (i&1) ans=ans*now%mod;return ans;
}
char get_char(){//快读（字母）char c=getchar();while (c<'a'||c>'z') c=getchar();return c-'a';
}
int read(){//快读（整数)char c=getchar();int ans=0;while (c<'0'||c>'9') c=getchar();while (c>='0'&&c<='9') ans=ans*10+c-'0',c=getchar();return ans;
}
void Write(long long x){//快速输出if (x<10) putchar(x^48);else Write(x/10),putchar((x%10)^48);return;
}
void NTT(long long *A,int flag){//NTT板子for (register int i=0;i<len;i++)if (i<rev[i]) swap(A[i],A[rev[i]]);for (register int l=1;l<len;l<<=1){long long T=(flag==1?power3[l]:power_inv3[l]);for (register int i=0;i<len;i+=(l<<1)){long long t=1;for (register int j=0;j<l;j++,t=t*T%mod){long long u=A[i+j],v=A[i+j+l]*t%mod;A[i+j]=add(u,v),A[i+j+l]=add(u,mod-v);}}}return;
}
void mul(){//多项式乘法for (register int i=0;i<len;i++) tmp_B[i]=B[i];NTT(tmp_B,1);for (register int i=0;i<len;i++) C[i]=tmp_A[i]*tmp_B[i]%mod;NTT(C,-1);for (register int i=0;i<len;i++) C[i]=C[i]*len_inv%mod;return;
}
long long query(int x){//处理询问操作long long now=C[x+ll-1];x+=ll-1;for (register int i=1;i<=X[0];i++) now+=Y[i]*A[x-X[i]];//暴力加上修改块内修改操作贡献return sum+E[x]-(x>=ll?E[x-ll]:0)-now*2;
}
int main(){n=read(),ll=read(),m=read();H=(int)(pow(n*log2(n),0.5))*3;while (len<=n+ll) len<<=1,++L;for (register int i=0;i<len;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<L-1);for (register int i=0;i<n;i++) tmp_A[i]=A[i]=get_char(),E[i]=E[i-1]+A[i]*A[i];for (register int i=0;i<ll;i++) B[i]=get_char(),sum+=B[i]*B[i];for (register int i=0,j=ll-1;i<j;i++,j--) swap(B[i],B[j]);for (register int l=1;l<len;l<<=1) power3[l]=power(g,(mod-1)/(l<<1)),power_inv3[l]=power(inv(g),(mod-1)/(l<<1));len_inv=inv(len);NTT(tmp_A,1);mul();//对密码串T只需要做一次FFTfor (register int i=1,opt;i<=m;i++){opt=read();if (opt==1) Write(query(read()-1)),putchar('\n');else {X[++X[0]]=ll-read(),Y[++Y[0]]=get_char();sum-=B[X[X[0]]]*B[X[X[0]]];int tmp=B[X[X[0]]];B[X[X[0]]]=Y[Y[0]],Y[Y[0]]-=tmp;sum+=B[X[X[0]]]*B[X[X[0]]];}if (X[0]==H) mul(),X[0]=Y[0]=0;//块的末尾暴力做NTT}return 0;
}