题目链接
题解
可以预处理出(S)每个位置能匹配哪些字符
对每种字符
构造两个序列
如果(S[i])可以匹配该字符，则该位置为(0)，否则为(1)
如果(T[i])可以匹配该字符，则该位置为(1)，否则为(0)
将(T)翻转一下做卷积
如果某个字符意义下的某个位置为(1)，就说明出现了(T)能匹配而(S)不能的情况，此时(T)不匹配(S)
否则(T)匹配(S)
即寻找有多少位置都为(0)
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 800005,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
return flag ? out : -out;
}
const int G = 3,P = 998244353;
inline int qpow(int a,int b){
int re = 1;
for (; b; b >>= 1,a = 1ll * a * a % P)
if (b & 1) re = 1ll * re * a % P;
return re;
}
int R[maxn],c[maxn];
void NTT(int* a,int n,int f){
for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
for (int i = 1; i < n; i <<= 1){
int gn = qpow(G,(P - 1) / (i << 1));
for (int j = 0; j < n; j += (i << 1)){
int g = 1,x,y;
for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
x = a[j + k],y = 1ll * g * a[j + k + i] % P;
a[j + k] = (x + y) % P,a[j + k + i] = (x + P - y) % P;
}
}
}
if (f == 1) return;
int nv = qpow(n,P - 2); reverse(a + 1,a + n);
for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
void conv(int* a,int* b,int deg1,int deg2){
int n = 1,L = 0;
while (n <= (deg1 + deg2)) n <<= 1,L++;
for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
for (int i = 1; i <= deg2; i++) c[i] = b[i];
for (int i = deg2 + 1; i < n; i++) c[i] = 0; c[0] = 0;
NTT(a,n,1); NTT(c,n,1);
for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * c[i] % P;
NTT(a,n,-1);
}
int A[4][maxn],B[4][maxn];
char ss[maxn],tt[maxn];
int N,M,K,last[4],S[maxn],T[maxn];
int id(char c){
if (c == 'A') return 0;
if (c == 'C') return 1;
if (c == 'G') return 2;
return 3;
}
int main(){
N = read(); M = read(); K = read();
scanf("%s%s",ss + 1,tt + 1); reverse(tt + 1,tt + 1 + M);
REP(i,N) S[i] = id(ss[i]);
REP(i,M) T[i] = id(tt[i]);
REP(i,N) for (int j = 0; j < 4; j++) A[j][i] = 1;
for (int i = 1; i <= N; i++){
last[S[i]] = i;
for (int j = 0; j < 4; j++){
if (last[j] && i - last[j] <= K) A[j][i] = 0;
}
}
for (int j = 0; j < 4; j++) last[j] = 0;
for (int i = N; i; i--){
last[S[i]] = i;
for (int j = 0; j < 4; j++){
if (last[j] && last[j] - i <= K) A[j][i] = 0;
}
}
for (int i = 1; i <= M; i++){
for (int j = 0; j < 4; j++){
if (T[i] == j) B[j][i] = 1;
else B[j][i] = 0;
}
}
for (int j = 0; j < 4; j++) conv(A[j],B[j],N,M);
int ans = 0;
for (int i = 1 + M; i <= N + 1; i++){
int flag = 1;
for (int j = 0; j < 4; j++) if (A[j][i]){flag = 0; break;}
ans += flag;
}
printf("%d
",ans);
return 0;
}