题解：

将字符串A、B中的a和b分别以1和-1表示，对字符串B进行反转。
将A和B看成多项式，求卷积，这样的话从结果区间的$[lenB-1,lenA)$中的每一个点的值$val$，$(lenB-val)/2$代表当前位置的字串与B串的距离，然后对字串进行字符串hash，去重就是答案。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <set>
using namespace std;
double pi = acos(-1.0);
struct complex{double re,im;complex(double r = 0.0,double i = 0.0):re(r),im(i){};complex operator+(complex com){return complex(re+com.re,im+com.im);}complex operator-(complex com){return complex(re-com.re,im-com.im);}complex operator*(complex com){return complex(re*com.re-im*com.im,re*com.im+im*com.re);}
};
complex wn,wntmp;
void rader(complex arr[],int n){int num = n-1;for(int i = 0;i < n;++i){int tn = n>>1;while(num && num >= tn) num ^= tn,tn >>= 1;num |= tn;if(num > i) swap(arr[i],arr[num]);}
}
void FFT(complex cs[],int n,int f){rader(cs,n);for(int s = 1;s < n;s <<= 1){wn = complex(cos(f*2*pi/(s*2)),sin(f*2*pi/(s*2)));for(int offset = 0;offset < n;offset += s<<1){wntmp = complex(1.0,0.0);for(int i = 0;i < s;++i){complex u = cs[offset+i],v = cs[offset+i+s]*wntmp;cs[offset+i] = u + v;cs[offset+i+s] = u - v;wntmp = wntmp * wn;}}}if(f == -1)for(int i = 0;i < n;++i)cs[i].re /= n;
}
int K;
const int maxn = 600007;
char A[maxn],B[maxn];
complex csA[maxn],csB[maxn];
unsigned long long fac = 9973;
unsigned long long pow(int x){unsigned long long ans = 1,base = fac;while(x){if(x & 1)ans *= base;base *= base;x >>= 1;}return ans;
}
int main(){int cas = 0;while(cin>>K && K != -1){memset(csA,0,sizeof(csA)),memset(csB,0,sizeof(csB));cin>>A>>B;int lenA = strlen(A),lenB = strlen(B);for(int i = 0;i < lenB/2;++i) swap(B[i],B[lenB-i-1]);int len = 1;while(len < lenA || len < lenB) len <<= 1;len <<= 1;for(int i = 0;i < lenA;++i) csA[i].re = A[i] == 'a'?1:-1;FFT(csA,len,1);for(int i = 0;i < lenB;++i) csB[i].re = B[i] == 'a'?1:-1;FFT(csB,len,1);for(int i = 0;i < len;++i) csA[i] = csA[i]*csB[i];FFT(csA,len,-1);set<unsigned long long> st;unsigned long long hash = 0,base = pow(lenB-1);for(int i = 0;i < lenB;++i) hash = hash*fac + (A[i] == 'a');long long ans = 0; for(int i = lenB-1;i < lenA;++i){int dis= (lenB - int(csA[i].re+100000.5) + 100000)/2;if(dis <= K) st.insert(hash),ans++;hash = (hash - base * (A[i-lenB+1] == 'a'))*fac+(A[i+1] == 'a');}printf("Case %d: %d\n",++cas,st.size());}return 0;
}