BZOJ2160 拉拉队排练

来源：互联网发布：wps数据透视表使用方法编辑：程序博客网时间：2024/04/29 23:20

傻逼题……manacher求出每个长度的极长回文串都有多少个，然后把长度相等的快速幂一起乘，然后把长度-2

#include<iostream>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<algorithm>#include<iomanip>#include<vector>#include<stack>#include<queue>#include<map>#include<set>#include<bitset>using namespace std;#define MAXN 2000010#define MAXM 1010#define ll long long#define INF 1000000000#define MOD 19930726#define eps 1e-8char s[MAXN];int n;ll k;int f[MAXN],mx,now;int v[MAXN];ll ans;ll mi(ll x,ll y){    ll re=1;    while(y){        if(y&1){            (re*=x)%=MOD;        }        (x*=x)%=MOD;        y>>=1;    }    return re;}int main(){    int i;    scanf("%d%lld",&n,&k);    scanf("%s",s+1);    for(i=n;i;i--){        s[i<<1]=s[i];        s[i<<1|1]='*';    }    s[1]='*';    s[0]='$';    n=n<<1|1;    for(i=1;i<=n;i++){        if(i<=mx){            f[i]=min(f[now*2-i],mx-i);        }        for(;s[i+f[i]+1]==s[i-f[i]-1];f[i]++){                     }        if(i+f[i]>mx){            mx=i+f[i];            now=i;        }    }    for(i=2;i<=n;i+=2){        f[i>>1]=f[i];    }    n>>=1;    for(i=1;i<=n;i++){        v[f[i]]++;    }    ans=1;    for(i=n;i;i--){        (ans*=mi(i,min((ll)v[i],k)))%=MOD;        k-=v[i];        if(i!=1){            v[i-2]+=v[i];        }        if(k<=0){            break;        }    }    printf("%lld\n",ans);    return 0;} /* */

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