ACM-ICPC 2018 焦作赛区网络预赛 H、L

https://nanti.jisuanke.com/t/31721

题意 n个位置 有几个限制相邻的三个怎么怎么样，直接从3开始 矩阵快速幂进行递推就可以了

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const ll mod=1e9+,maxn=;
 struct Matrix
 {
     ll m[maxn][maxn];
     Matrix()
     {
         memset(m,,sizeof(m));
     }
     void init()
     {
         for(int i=; i<maxn; i++)
             for(int j=; j<maxn; j++)
                 m[i][j]=(i==j);
     }
     Matrix  operator +(const Matrix &b)const
     {
         Matrix c;
         for(int i=; i<maxn; i++)
         {
             for(int j=; j<maxn; j++)
             {
                 c.m[i][j]=(m[i][j]+b.m[i][j])%mod;
             }
         }
         return c;
     }
     Matrix  operator *(const Matrix &b)const
     {
         Matrix c;
         for(int i=; i<maxn; i++)
         {
             for(int j=; j<maxn; j++)
             {
                 for(int k=; k<maxn; k++)
                 {
                     c.m[i][j]=(c.m[i][j]+(m[i][k]*b.m[k][j])%mod)%mod;
                 }
             }
         }
         return c;
     }
     Matrix  operator^(const ll &t)const
     {
         Matrix ans,a=(*this);
         ans.init();
         ll n=t;
         while(n)
         {
             if(n&) ans=ans*a;
             a=a*a;
             n>>=;
         }
         return ans;
     }
 };
 int xishu[][]={
 ,,,,,,,,,
 ,,,,,,,,,
 ,,,,,,,,,
 ,,,,,,,,,
 ,,,,,,,,,
 ,,,,,,,,,
 ,,,,,,,,,
 ,,,,,,,,,
 ,,,,,,,,};
 int main()
 {
     int t;
     ll n,m,a,b;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%lld",&n);
         if(n==)
         {
             printf("3\n");
             continue;
         }
         if(n==)
         {
             printf("9\n");
             continue;
         }
         Matrix temp,aa;
         for(int i=;i<maxn;i++)
         {
             for(int j=;j<maxn;j++)
             {
                 temp.m[i][j]=xishu[i][j];
             }
         }
         for(int i=;i<maxn;i++)
             aa.m[i][]=;
         temp=temp^(n-);
         aa=temp*aa;
         ll ans=;
         for(int i=;i<maxn;i++)
             ans=(ans+aa.m[i][])%mod;
         cout<<ans<<endl;
     }
     return ;
 }

https://nanti.jisuanke.com/t/31717

解析 傻逼题，数据范围应该是2e5，真是傻逼，被安排的明明白白。存个后缀自动机的板子。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+;
typedef long long ll;
char s[maxn];
int len,k,n,m;
char temp[];
struct SAM{
    int last,cnt,nxt[maxn*][],fa[maxn*],l[maxn*],num[maxn*];
    ll ans;
    void init(){
        last = cnt=;
        memset(nxt[],,sizeof nxt[]);
        fa[]=l[]=num[]=;
        ans=;
    }
    int inline newnode(){
        cnt++;
        memset(nxt[cnt],,sizeof nxt[cnt]);
        fa[cnt]=l[cnt]=num[cnt]=;
        return cnt;
    }
    void add(int c){
        int p = last;
        int np = newnode();
        last = np;
        l[np] =l[p]+;
        while (p&&!nxt[p][c]){
            nxt[p][c] = np;
            p = fa[p];
        }
        if (!p){
            fa[np] =;
        }else{
            int q = nxt[p][c];
            if (l[q]==l[p]+){
                fa[np] =q;
            }else{
                int nq = newnode();
                memcpy(nxt[nq],nxt[q],sizeof nxt[q]);
                fa[nq] =fa[q];
                num[nq] = num[q];
                l[nq] = l[p]+;
                fa[np] =fa[q] =nq;
                while (nxt[p][c]==q){
                    nxt[p][c]=nq;
                    p = fa[p];
                }
            }
        }
        int temp = last;
        while (temp){
            if (num[temp]>=k){
                break;
            }
            num[temp]++;
            if (num[temp]==k){
                ans+=l[temp]-l[fa[temp]];
            }
            temp = fa[temp];
        }
    }
}sam;
int main(){
    while (scanf("%s",s)!=EOF){
        scanf("%d%d",&m,&k);k++;
        n=len = strlen(s);
        sam.init();
        for (int i=;i<len;i++){
            sam.add(s[i]-'A');
        }
        ll ans1=sam.ans;
        k=m;
        sam.init();
        for (int i=;i<len;i++){
            sam.add(s[i]-'A');
        }
        ll ans2=sam.ans;
        printf("%lld\n",ans2-ans1);
    }
    return ;
}