阿申准备报名参加GT考试,准考证号为N位数X1X2....Xn(0<=Xi<=9),他不希望准考证号上出现不吉利的数字。
他的不吉利数学A1A2...Am(0<=Ai<=9)有M位,不出现是指X1X2...Xn中没有恰好一段等于A1A2...Am. A1和X1可以为0
标签:hnoi names desc 考试 lex lag 矩阵 ++i eof
阿申准备报名参加GT考试,准考证号为N位数X1X2....Xn(0<=Xi<=9),他不希望准考证号上出现不吉利的数字。
他的不吉利数学A1A2...Am(0<=Ai<=9)有M位,不出现是指X1X2...Xn中没有恰好一段等于A1A2...Am. A1和X1可以为0
第一行输入N,M,K.接下来一行输入M位的数。 N<=10^9,M<=20,K<=1000
阿申想知道不出现不吉利数字的号码有多少种,输出模K取余的结果.
/* Stay hungry, stay foolish. */ #include<iostream> #include<algorithm> #include<queue> #include<string> #include<map> #include<vector> #include<set> #include<sstream> #include<stack> #include<ctime> #include<cmath> #include<cctype> #include<climits> #include<cstring> #include<cstdio> #include<cstdlib> #include<iomanip> #include<bitset> #include<complex> using namespace std; /* #define getchar() getc() char buf[1<<15],*fs,*ft; inline char getc() {return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin)),fs==ft)?0:*fs++;} */ template <class _T> inline void read(_T &_x) { int _t; bool _flag=false; while((_t=getchar())!=‘-‘&&(_t<‘0‘||_t>‘9‘)) ; if(_t==‘-‘) _flag=true,_t=getchar(); _x=_t-‘0‘; while((_t=getchar())>=‘0‘&&_t<=‘9‘) _x=_x*10+_t-‘0‘; if(_flag) _x=-_x; } typedef long long LL; const int maxm = 30; int n, m, mod; struct Matrix { int v[maxm][maxm], n, m; Matrix() {memset(v, 0, sizeof v); } Matrix(int a, int b):n(a), m(b) {memset(v, 0, sizeof v); } void init() {for (int i = 0; i <= n && i <= m; ++i) v[i][i] = 1; } Matrix operator * (Matrix B)const { Matrix C(n, B.m); for (int i = 0; i <= n; ++i) for (int j = 0; j <= B.m; ++j) for (int k = 0; k <= m; ++k) (C.v[i][j] += v[i][k] * B.v[k][j]) %= mod; return C; } Matrix operator ^ (int t) { Matrix ans(n, m), x = *this; ans.init(); for ( ; t; t >>= 1, x = x * x) if (t & 1) ans = ans * x; return ans; } inline void print() { for (int i = 0; i <= n; ++i) { for (int j = 0; j <= m; ++j) { printf("%d ", v[i][j]); } puts(""); } } }; int nxt[maxm]; int main() { //freopen("test.in","r",stdin); //freopen(".out","w",stdout); read(n), read(m), read(mod); char s[maxm]; scanf("%s", s + 1); nxt[1] = 0; for (int i = 2, j; i <= m; ++i) { j = nxt[i - 1]; while (j && s[j + 1] != s[i]) j = nxt[j]; if (s[j + 1] == s[i]) ++j; nxt[i] = j; } Matrix x(m, m); for (int i = 0; i < m; ++i) { for (int j = 0, k; j <= 9; ++j) { k = i; while (k && s[k + 1] - ‘0‘ != j) k = nxt[k]; if (s[k + 1] - ‘0‘ == j) ++k; if (k != m) (++x.v[k][i]) %= mod; } } x = x ^ n; int sum = 0; for (int i = 0; i < m; ++i) (sum += x.v[i][0]) %= mod; cout << sum << endl; return 0; }
标签:hnoi names desc 考试 lex lag 矩阵 ++i eof
原文地址:http://www.cnblogs.com/akhpl/p/7116136.html