写了一个早上...就因为把长度为m的也算进去了...
dp(i, j)表示准考证号前i个字符匹配了不吉利数字前j个的方案数. kmp预处理, 然后对于j进行枚举, 对数字0~9也枚举算出f(i, j)表示dp(x-1, j)对dp(x, i)的贡献.然后用矩阵快速幂就可以了. 时间复杂度O(M3logN + M)
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#include<bits/stdc++.h>
using namespace std;
const int maxn = 25;
int fail[maxn], S[maxn], N, M, MOD;
struct matrix {
int n, m;
int a[maxn][maxn];
matrix(int _n = 0, int _m = 0):n(_n), m(_m) {
memset(a, 0, sizeof a);
}
void unit() {
for(int i = 0; i < n; i++)
a[i][i] = 1;
}
matrix operator * (matrix o) {
matrix ret(n, o.m);
for(int i = 0; i < n; i++)
for(int k = 0; k < m; k++)
for(int j = 0; j < o.m; j++)
ret.a[i][j] = (ret.a[i][j] + a[i][k] * o.a[k][j]) % MOD;
return ret;
}
matrix operator = (matrix o) {
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
a[i][j] = o.a[i][j];
return *this;
}
matrix operator ^ (int k) {
matrix ret(n, m), t = *this; ret.unit();
for(; k; k >>= 1) {
if(k & 1) ret = ret * t;
t = t * t;
}
return ret;
}
};
void kmp() {
fail[0] = fail[1] = 0;
for(int i = 1; i < M; i++) {
int p = fail[i];
while(p && S[i] != S[p]) p = fail[p];
fail[i + 1] = S[i] == S[p] ? p + 1 : 0;
}
}
int main() {
scanf("%d%d%d", &N, &M, &MOD);
for(int i = 0; i < M; i++) {
char c = getchar();
for(; !isdigit(c); c = getchar());
S[i] = c - '0';
}
kmp();
matrix Q(M, M);
for(int i = 0; i < M; i++)
for(int j = 0; j < 10; j++) {
int p = i;
while(p && S[p] != j) p = fail[p];
if(S[p] == j) p++;
Q.a[p][i]++;
}
matrix ans(M, 1);
ans.a[0][0] = 1;
ans = (Q ^ N) * ans;
int tot = 0;
for(int i = 0; i < M; i++)
if((tot += ans.a[i][0]) >= MOD) tot -= MOD;
printf("%d\n", tot);
return 0;
}
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1009: [HNOI2008]GT考试
Time Limit: 1 Sec Memory Limit: 162 MB
Submit: 2236 Solved: 1368
[Submit][Status][Discuss]
Description
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
Input
第一行输入N,M,K.接下来一行输入M位的数。 100%数据N<=10^9,M<=20,K<=1000 40%数据N<=1000 10%数据N<=6
Output
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.
Sample Input
4 3 100
111
111
Sample Output
81