题意：
给定一个模式串和目标串按下图方式匹配，错开位置不多于k
解题思路：
总共只有(A C G T)四个字符，那么我们可以按照各个字符进行匹配，比如按照(A)进行匹配时,当(k=1)时，我们将目标串
(ACAT)化作
(1~0~1~0)
模式串
(AGCAATTCAT)化作
(1~1~1~1~1~1~0~1~1~1)
同样是反置目标串
可以得到以x为匹配终点的位置的匹配函数(p(X)=sum_{i+j=x}A(i)B(j))
如此进行4次FFT，最后如果目标位置贡献等于目标串长度，则说明匹配成功
#include <bits/stdc++.h>
using namespace std;
/* freopen("k.in", "r", stdin);
freopen("k.out", "w", stdout); */
//clock_t c1 = clock();
//std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 1e6 + 7;
const ll MAXM = 1e6 + 7;
const ll MOD = 998244353;
const double eps = 1e-6;
const double pi = acos(-1.0);
template <class T>
inline void in(T &x)
{
static char ch;
static bool neg;
for (ch = neg = 0; ch < '0' || '9' < ch; neg |= ch == '-', ch = getchar())
;
for (x = 0; '0' <= ch && ch <= '9'; (x *= 10) += ch - '0', ch = getchar())
;
x = neg ? -x : x;
}
struct Complex
{
double x, y;
Complex(double xx = 0, double yy = 0) { x = xx, y = yy; }
} a[MAXN], b[MAXN], c[MAXN], ans[MAXN];
Complex operator+(Complex a, Complex b) { return Complex(a.x + b.x, a.y + b.y); }
Complex operator-(Complex a, Complex b) { return Complex(a.x - b.x, a.y - b.y); }
Complex operator*(Complex a, Complex b) { return Complex(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); } //不懂的看复数的运算那部分
int N, M;
int l, r[MAXN];
int limit = 1;
void FFT(Complex *A, int type)
{
for (int i = 0; i < limit; i++)
if (i < r[i])
swap(A[i], A[r[i]]); //求出要迭代的序列
for (int mid = 1; mid < limit; mid <<= 1)
{ //待合并区间的长度的一半
Complex Wn(cos(pi / mid), type * sin(pi / mid)); //单位根
for (int R = mid << 1, j = 0; j < limit; j += R)
{ //R是区间的长度，j表示前已经到哪个位置了
Complex w(1, 0); //幂
for (int k = 0; k < mid; k++, w = w * Wn)
{ //枚举左半部分
Complex x = A[j + k], y = w * A[j + mid + k]; //蝴蝶效应
A[j + k] = x + y;
A[j + mid + k] = x - y;
}
}
}
/*if (type == -1)
for (int i = 0; i < limit; ++i)
a[i].x /= limit;//我们推过的公式里面有一个1/n这一项*/
}
char s[MAXN], t[MAXN];
void init(int N, int M)
{
while (limit <= N + M)
limit <<= 1, l++;
for (int i = 0; i < limit; i++)
r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
}
int change(char str)
{
if (str == 'A')
return 1;
else if (str == 'T')
return 2;
else if (str == 'G')
return 3;
else
return 4;
}
int pre[MAXN], cnt;
int main()
{
int n, m, k;
scanf("%d%d%d %s %s", &n, &m, &k, s, t);
reverse(t, t + m);
init(n, m);
for (int ca = 1; ca <= 4; ca++)
{
cnt = -1;
memset(pre, 0, sizeof(pre));
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
for (int i = 0; i < n; i++)
{
if (change(s[i]) == ca)
pre[++cnt] = i;
a[i].x = change(s[i]) == ca ? 1 : 0, a[i].y = 0;
}
for (int i = 0; i < m; i++)
b[i].x = change(t[i]) == ca ? 1 : 0, b[i].y = 0;
int now = -1;
for (int i = 0; i <= cnt; i++)
{
int L = max(pre[i] - k, 0);
int R = min(pre[i] + k, n - 1);
if (now > R)
continue;
now = max(L, now);
for (; now <= R; now++)
a[now].x = 1;
now--;
}
FFT(a, 1);
FFT(b, 1);
for (int i = 0; i < limit; i++)
a[i] = b[i] * a[i];
FFT(a, -1);
for (int i = 0; i < limit; i++)
c[i] = c[i] + a[i];
}
int ans = 0;
for (int i = m - 1; i < limit; i++)
if (int(c[i].x / limit + 0.5) == m)
ans++;
printf("%d
", ans);
return 0;
}