Problem Description
As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:
Yuta has 100110.
Yuta has 100110.
Input
The first line contains a number ).
Output
For each testcase, print a single line with a single number -- the answer modulo 998244353.
Sample Input
2
2 2
011
001
2 3
011
001
Sample Output
1
4
题解：
要是能在比赛中A掉就爽了
和题解做法一样
#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define ls i<<1 #define rs ls | 1 #define mid ((ll+rr)>>1) #define pii pair<int,int> #define MP make_pair typedef long long LL; const long long INF = 1e18+1LL; const double pi = acos(-1.0); const int N = 1e4+10, M = 1e3+20,inf = 2e9; const LL mod = 998244353LL; int dp[2][121][121][80],sum[2][N]; int nex[2][N][2],cnt0,cnt1,head1,tail1,head0,tail0,q[2][N],fail[2][N]; void insert(char *s,int p) { int now = 1,len = strlen(s); for(int i = 0; i < len; ++i) { int index = s[i] - '0'; if(!nex[0][now][index]) nex[0][now][index] = ++cnt0; sum[0][nex[0][now][index]] |= sum[0][now]; now = nex[0][now][index]; //cout<<now<<" "<<index<<endl; } sum[0][now] |= (1<<p); now = 1; for(int i = len-1; i >= 0; --i) { int index = s[i] - '0'; if(!nex[1][now][index]) nex[1][now][index] = ++cnt1; sum[1][nex[1][now][index]] |= sum[1][now]; now = nex[1][now][index]; //cout<<now<<" "<<index<<endl; } sum[1][now] |= (1<<p); } void build_fail() { head0 = 0, tail0 = 0;head1 = 0, tail1 = 0; for(int i = 0; i < 2; ++i) nex[0][0][i] = 1,nex[1][0][i] = 1; fail[0][1] = 0,fail[1][1] = 0; q[0][tail0++] = 1;q[1][tail1++] = 1; while(head0 != tail0) { int now = q[0][head0++]; sum[0][now] |= sum[0][fail[0][now]]; for(int i = 0; i < 2; ++i) { int p = fail[0][now]; if(!nex[0][now][i]) { nex[0][now][i] = nex[0][p][i];continue; } fail[0][nex[0][now][i]] = nex[0][p][i]; q[0][tail0++] = nex[0][now][i]; } } while(head1 != tail1) { int now = q[1][head1++]; sum[1][now] |= sum[1][fail[1][now]]; for(int i = 0; i < 2; ++i) { int p = fail[1][now]; if(!nex[1][now][i]) { nex[1][now][i] = nex[1][p][i];continue; } fail[1][nex[1][now][i]] = nex[1][p][i]; q[1][tail1++] = nex[1][now][i]; } } } int len[N],mx,n,L; char a[N]; int dfs() { int now = 1; int ret = 0; for(int i = 1; i <= 2*mx; ++i) { now = nex[0][now][len[i]]; ret |= sum[0][now]; } return ret; } int ma(int p) { int now = 1; if(p) for(int i = mx; i >= 1; --i) now = nex[1][now][len[i]]; else for(int i = mx+1; i <= 2*mx; ++i) now = nex[0][now][len[i]]; return now; } void init() { memset(dp,0,sizeof(dp)); memset(nex,0,sizeof(nex)); cnt0 = 1;mx = -1;cnt1 = 1; memset(fail,0,sizeof(fail)); memset(sum,0,sizeof(sum)); } int main() { int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&L); init(); for(int i = 1; i <= n; ++i) { scanf("%s",a); insert(a,i-1); mx = max(mx,(int)strlen(a)); } int ff = 0; mx-=1; build_fail(); for(int i = 0; i < (1<<mx); ++i) { for(int j = 1; j <= mx; ++j) len[j] = ((i>>(j-1))&1); for(int j = mx+1; j <= 2*mx; ++j) len[j] = 1^(len[2*mx - j + 1]); int now = dfs(); int z = ma(1),f = ma(0); dp[ff][z][f][now] += 1; dp[ff][z][f][now] %= mod; // cout<<i<<" "<<now<<" "<<z<<" "<<f<<endl; } for(int i = mx; i < L; i++) { memset(dp[ff^1],0,sizeof(dp[ff^1])); for(int j = 0; j < tail1; ++j) { for(int k = 0; k < tail0; ++k) { for(int h = 0; h < (1<<n); ++h) { if(!dp[ff][q[1][j]][q[0][k]][h]) continue; int p = nex[1][q[1][j]][0],np = nex[0][q[0][k]][1]; int tmp = (h|sum[1][p]); tmp |= sum[0][np]; dp[ff^1][p][np][tmp] += dp[ff][q[1][j]][q[0][k]][h]; dp[ff^1][p][np][tmp] %= mod; p = nex[1][q[1][j]][1],np = nex[0][q[0][k]][0]; tmp = (h|sum[1][p]); tmp |= sum[0][np]; dp[ff^1][p][np][tmp] += dp[ff][q[1][j]][q[0][k]][h]; dp[ff^1][p][np][tmp] %= mod; } } } ff^=1; } LL ans = 0; for(int i = 0; i < tail1; ++i) for(int j = 0; j < tail0; ++j) ans = ( ans + dp[ff][q[1][i]][q[0][j]][(1<<n)-1]) % mod; printf("%lld\n",ans); } return 0; }