bzoj1009 [HNOI2008]GT考试
用a[i][j]表示 匹配到i 转移到 匹配到j 的方案数
用矩阵快速幂求解
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cstdio> 6 #include<string> 7 #include<cmath> 8 #include<ctime> 9 #include<queue> 10 #include<stack> 11 #include<map> 12 #include<set> 13 #define rre(i,r,l) for(int i=(r);i>=(l);i--) 14 #define re(i,l,r) for(int i=(l);i<=(r);i++) 15 #define Clear(a,b) memset(a,b,sizeof(a)) 16 #define inout(x) printf("%d",(x)) 17 #define douin(x) scanf("%lf",&x) 18 #define strin(x) scanf("%s",(x)) 19 #define LLin(x) scanf("%lld",&x) 20 #define op operator 21 #define CSC main 22 typedef unsigned long long ULL; 23 typedef const int cint; 24 typedef long long LL; 25 using namespace std; 26 void inin(int &ret) 27 { 28 ret=0;int f=0;char ch=getchar(); 29 while(ch<'0'||ch>'9'){if(ch=='-')f=1;ch=getchar();} 30 while(ch>='0'&&ch<='9')ret*=10,ret+=ch-'0',ch=getchar(); 31 ret=f?-ret:ret; 32 } 33 int n,m,mod,pre[100]; 34 struct M 35 { 36 int a[25][25]; 37 int* op [] (int x){return x[a];} 38 M(){Clear(a,0);} 39 void in(int x) 40 { 41 re(i,0,25)re(j,0,25)if(i==j)a[i][j]=1; 42 } 43 M op * (M &rhs) 44 { 45 M c; 46 re(i,0,m-1)re(j,0,m-1)re(k,0,m-1) 47 (c[i][j]+=a[i][k]*rhs[k][j])%=mod; 48 return c; 49 } 50 }; 51 char s[100]; 52 M qpow(M a,int b) 53 { 54 M ret;ret.in(1); 55 while(b) 56 { 57 if(b&1)ret=ret*a; 58 b>>=1;a=a*a; 59 } 60 return ret; 61 } 62 M a; 63 void kmp() 64 { 65 int k=0; 66 re(i,2,m) 67 { 68 while(k&&s[i]!=s[k+1])k=pre[k]; 69 if(s[k+1]==s[i])k++; 70 pre[i]=k; 71 } 72 re(i,0,m-1)re(j,0,9) 73 { 74 int x=i; 75 while(x&&s[x+1]-'0'!=j)x=pre[x]; 76 if(j==s[x+1]-'0')a[i][x+1]++; 77 else a[i][0]++; 78 } 79 } 80 int CSC() 81 { 82 inin(n),inin(m),inin(mod); 83 strin(s+1); 84 kmp(); 85 M f=qpow(a,n); 86 int ans=0; 87 re(i,0,m-1)ans=(ans+f[0][i])%mod; 88 printf("%d",ans); 89 return 0; 90 }
