UVALive

题意:海明距离的定义:两个相同长度的字符串中不同的字符数.现给出母串A和模式串B,求A中有多少与B海明距离<=k的不同子串
分析:将字符a视作1,b视作0.则A与B中都是a的位置乘积是1.现将B逆置,并设B的长度为n,令(C(n+k-1)= sum_{i=0}^{n-1}A_{i+k}*B_{n-i-1}),表示母串A中从位置k开始,长度为n的子串与B中字符都是'a'的位置的数目,可以通过FFT运算得到.再对字符'b'做一次同样的运算,(ans[i])统计母串A中以i结尾的子串与B相同字符的个数.
设A的长度为m,则一共有(m-n+1)个子串,若(n-ans[i] leq k),则盖子串符合条件,但本题需要求不同的子串,所以将母串哈希掉,将符合要求的哈希值存在集合中去重.

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 4e5 + 10;
const double PI = acos(-1.0);
struct Complex{
    double x, y;
    inline Complex operator+(const Complex b) const {
        return (Complex){x +b.x,y + b.y};
    }
    inline Complex operator-(const Complex b) const {
        return (Complex){x -b.x,y - b.y};
    }
    inline Complex operator*(const Complex b) const {
        return (Complex){x *b.x -y * b.y,x * b.y + y * b.x};
    }
} va[MAXN * 2 + MAXN / 2], vb[MAXN * 2 + MAXN / 2];
int lenth = 1, rev[MAXN * 2 + MAXN / 2];
int N, M;   // f 和 g 的数量
    //f g和 的系数
    // 卷积结果
    // 大数乘积
int f[MAXN],g[MAXN];
vector<LL> conv;
vector<LL> multi;
//f g
void init()
{
    int tim = 0;
    lenth = 1;
    conv.clear(), multi.clear();
    memset(va, 0, sizeof va);
    memset(vb, 0, sizeof vb);
    while (lenth <= N + M - 2)
        lenth <<= 1, tim++;
    for (int i = 0; i < lenth; i++)
        rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (tim - 1));
}
void FFT(Complex *A, const int fla)
{
    for (int i = 0; i < lenth; i++){
        if (i < rev[i]){
            swap(A[i], A[rev[i]]);
        }
    }
    for (int i = 1; i < lenth; i <<= 1){
        const Complex w = (Complex){cos(PI / i), fla * sin(PI / i)};
        for (int j = 0; j < lenth; j += (i << 1)){
            Complex K = (Complex){1, 0};
            for (int k = 0; k < i; k++, K = K * w){
                const Complex x = A[j + k], y = K * A[j + k + i];
                A[j + k] = x + y;
                A[j + k + i] = x - y;
            }
        }
    }
}
void getConv(){             //求多项式
    init();
    for (int i = 0; i < N; i++)
        va[i].x = f[i];
    for (int i = 0; i < M; i++)
        vb[i].x = g[i];
    FFT(va, 1), FFT(vb, 1);
    for (int i = 0; i < lenth; i++)
        va[i] = va[i] * vb[i];
    FFT(va, -1);
    for (int i = 0; i <= N + M - 2; i++)
        conv.push_back((LL)(va[i].x / lenth + 0.5));
}

char s1[100005],s2[100005];
LL res[MAXN];

const int seed = 3;
LL dig[MAXN],Hash[MAXN];
set<LL> dp;

void pre(){
    dig[0] =1;
    for(int i=1;i<=100005;++i){
        dig[i] = dig[i-1]*seed;
    }
}

LL getHash(int L ,int R){
    if(L==0) return Hash[R];
    return Hash[R] - Hash[L-1]*dig[R-L+1];
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    pre();
    int k,cas=1;
    while(scanf("%d",&k)==1){
        if(k==-1) break;
        memset(res,0,sizeof(res));
        dp.clear();
        scanf("%s",s1);
        scanf("%s",s2);
        int len1 = strlen(s1), len2 = strlen(s2);
        N = len1, M  = len2;
        for(int i=0;i<len1;++i){
            if(s1[i]=='a') f[i] = 1;
            else f[i] = 0;
        }
        for(int i=0;i<len2;++i){
            if(s2[len2-i-1]=='a') g[i] = 1;
            else g[i] = 0;
        }
        getConv();
        int sz =conv.size();
        for(int i=len2-1;i<sz;++i){
            res[i] += conv[i];
        }

        for(int i=0;i<len1;++i){
            if(s1[i]=='b') f[i] = 1;
            else f[i] = 0;
        }
        for(int i=0;i<len2;++i){
            if(s2[len2-i-1]=='b') g[i] = 1;
            else g[i] = 0;
        }
        getConv();
        sz =conv.size();
        for(int i=len2-1;i<sz;++i){
            res[i] += conv[i];
        }

        //Hash
        Hash[0] = s1[0]-'a'+1;
        for(int i=1;i<len1;++i){
            Hash[i] = Hash[i-1]*seed + s1[i]-'a'+1;
        }

        for(int i=len2-1;i<len1;++i){
            LL now = getHash(i-len2+1,i);
            if(len2-res[i]<=k){
                dp.insert(now);
            }
        }
        printf("Case %d: %d
",cas++,(int)dp.size());
    }
    return 0;
}