uva1358 - Generator

链接

https://vjudge.net/problem/UVA-1358

题解

借助$kmp$写出递推关系
然后用高斯消元去解
这题的答案貌似一定是整数（我也不知道为啥）

代码

#include <bits/stdc++.h>
#define cl(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;
struct matrix
{
    ll n, m, a[15][15];
    ll* operator[](ll index){return a[index];}
    void clear(){n=m=0;cl(a);}
    void gauss()
    {
        ll i, j, k, r;
        for(i=1;i<=n;i++)
        {
            r=i;
            for(j=i+1;j<=n;j++)if(a[j][i])r=j;
            for(j=1;j<=n+1;j++)swap(a[i][j],a[r][j]);
            for(j=i+1;j<=n;j++)
            {
                while(a[j][i])
                {
                    ll t(a[i][i]/a[j][i]), tmp;
                    for(k=i;k<=n+1;k++)
                    {
                        tmp=a[j][k];
                        a[j][k]=a[i][k]-t*a[j][k];
                        a[i][k]=tmp;
                    }
                }
            }
        }
    }
    ll getval(ll index)
    {
        return a[index][n+1]/a[index][index];
    }
    void show()
    {
        ll i, j;
        for(i=1;i<=n;i++)for(j=1;j<=m;j++)printf("%lld",a[i][j]), putchar(j==m?10:32);
    }
}mat;
struct KMP  //所有传进的字符串都要封尾(len+1的位置放一个0)
{
    int n, next[15], t[15];
    void clear()
    {
        cl(next), n=0;
    }
    void build(int *r, int len)
    {
        int i, j=0;
        n=len;
        for(i=1;i<=len;i++)t[i]=r[i];
        for(i=2;i<=len;i++)
        {
            for(;j and t[j+1]!=t[i];j=next[j]);
            next[i] = t[j+1]==t[i]?++j:0;
        }
    }
    int move(int pos, int x)
    {
        for(;pos and t[pos+1]!=x;pos=next[pos]);
        return t[pos+1]==x ? pos+1 : 0;
    }
}kmp;
int read(int x=0)
{
    int c, f(1ll);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-48;
    return f*x;
}
int main()
{
    auto T=read();
    int kase(0);
    while(T--)
    {
        auto n=read();
        char s[15];
        int r[15], i, len;
        scanf("%s",s+1);
        len=strlen(s+1);
        for(auto i=1;i<=len;i++)r[i]=s[i]-'A'+1;
        r[len+1]=0;
        kmp.clear();
        kmp.build(r,len);
        mat.clear();
        mat.n=len+1, mat.m=len+2;
        for(i=1;i<len;i++)
        {
            mat[i][i]=n;
            for(auto alpha=1;alpha<=n;alpha++)
            {
                auto to=kmp.move(i,alpha);
                if(to==0)to=len+1;
                mat[i][to]-=1;
            }
            mat[i][len+2]=n;
        }
        mat[len+1][len+1]=n;                //处理0
        for(auto alpha=1;alpha<=n;alpha++)
        {
            auto to=kmp.move(0,alpha);
            if(to==0)to=len+1;
            mat[len+1][to]-=1;
        }
        mat[len+1][len+2]=n;
        mat[len][len]=1;                //结尾
        mat[len][len+2]=0;
        mat.gauss();
        kase++;
        printf("Case %d:\n%lld\n",kase,mat.getval(len+1) );
        if(T)putchar(10);
    }
    return 0;
}