BZOJ 4259 残缺的字符串

思路

同样是FFT进行字符串匹配
只不过两个都有通配符
匹配函数再乘一个\(A_i\)即可

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#define int long long
using namespace std;
const int MAXN = 1200000;
const int MOD = 998244353;
const int G = 3;
const int invG = 332748118;
int rev[MAXN];
void cal_rev(int n,int lim){
    for(int i=0;i<n;i++)
        rev[i]=(rev[i>>1]>>1)|((i&1)<<(lim-1));
}
int pow(int a,int b){
    int ans=1;
    while(b){
        if(b&1)
            ans=(1LL*ans*a)%MOD;
        a=(1LL*a*a)%MOD;
        b>>=1;
    }
    return ans;
}
void NTT(int *a,int opt,int n,int lim){
    for(int i=0;i<n;i++)
        if(i<rev[i])
            swap(a[i],a[rev[i]]);
    for(int i=2;i<=n;i<<=1){
        int len=i/2,tmp=pow((opt)?G:invG,(MOD-1)/i);
        for(int j=0;j<n;j+=i){
            int arr=1;
            for(int k=j;k<j+len;k++){
                int t=(1LL*a[k+len]*arr)%MOD;
                a[k+len]=(a[k]-t+MOD)%MOD;
                a[k]=(a[k]+t)%MOD;
                arr=(1LL*arr*tmp)%MOD;
            }
        }
    }
    if(!opt){
        int invN = pow(n,MOD-2);
        for(int i=0;i<n;i++)
            a[i]=(1LL*a[i]*invN)%MOD;
    }
}
int n,m,s[MAXN],t[MAXN],a[MAXN],b[MAXN],c[MAXN],ans[MAXN],cnt;
char S[MAXN];
signed main(){
    freopen("test.in","r",stdin);
    freopen("test.out","w",stdout);
    scanf("%lld %lld",&m,&n);
    scanf("%s",S);
    for(int i=0;i<m;i++)
        t[i]=(S[i]=='*')?0:S[i]-'a'+1;
    reverse(t,t+m);
    scanf("%s",S);
    for(int i=0;i<n;i++)
        s[i]=(S[i]=='*')?0:S[i]-'a'+1;
    int midlen=1,midlim=0;
    while(midlen<(n+m))
        midlen<<=1,midlim++;
    cal_rev(midlen,midlim);
    
    for(int i=0;i<midlen;i++)
        a[i]=(s[i]*s[i]*s[i])%MOD,b[i]=t[i];
    NTT(a,1,midlen,midlim);
    NTT(b,1,midlen,midlim);
    for(int i=0;i<midlen;i++)
        c[i]=(a[i]*b[i])%MOD;

    for(int i=0;i<midlen;i++)
        a[i]=(2*s[i]*s[i])%MOD,b[i]=(t[i]*t[i])%MOD;
    NTT(a,1,midlen,midlim);
    NTT(b,1,midlen,midlim);
    for(int i=0;i<midlen;i++)
        c[i]=(c[i]-a[i]*b[i]+MOD)%MOD;
    
    for(int i=0;i<midlen;i++)
        a[i]=s[i],b[i]=(t[i]*t[i]*t[i])%MOD;
    NTT(a,1,midlen,midlim);
    NTT(b,1,midlen,midlim);
    for(int i=0;i<midlen;i++)
        c[i]=(c[i]+a[i]*b[i])%MOD;
    
    NTT(c,0,midlen,midlim);
    // for(int i=0;i<midlen;i++)
    //     printf("!%lld\n",c[i]);
    for(int i=m-1;i<n;i++)
        if(!c[i])
            ans[++cnt]=i-m+1;
    printf("%lld\n",cnt);
    for(int i=1;i<=cnt;i++)
        printf("%lld ",ans[i]+1);
    return 0;
}
posted @ 2019-04-23 18:49  dreagonm  阅读(90)  评论(0编辑  收藏  举报