Codeforces 528D

题意：给定一个主串和一个模式（都只含有ATCG）,求模式在主串中出现的次数

但是匹配的时候有一个偏移量K，即在用t[i]匹配s[j]的时候  只要s[j-k]---s[j+k]中有字符t[i]就算匹配；

这里用了FFT

参考：http://blog.csdn.net/u013368721/article/details/45565729

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;

const double PI = acos(-1.0);

struct complex
{
double r,i;
complex(double _r = 0.0,double _i = 0.0)
{
r = _r; i = _i;
}
complex operator +(const complex &b)
{
return complex(r+b.r,i+b.i);
}
complex operator -(const complex &b)
{
return complex(r-b.r,i-b.i);
}
complex operator *(const complex &b)
{
return complex(r*b.r-i*b.i,r*b.i+i*b.r);
}
};

void change(complex y[],int len)
{
int i,j,k;
for(i = 1, j = len/2;i < len-1; i++)
{
if(i < j)swap(y[i],y[j]);
k = len/2;
while( j >= k)
{
j -= k;
k /= 2;
}
if(j < k) j += k;
}
}

void fft(complex y[],int len,int on)
{
change(y,len);
for(int h = 2; h <= len; h <<= 1)
{
complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
for(int j = 0;j < len;j+=h)
{
complex w(1,0);
for(int k = j;k < j+h/2;k++)
{
complex u = y[k];
complex t = w*y[k+h/2];
y[k] = u+t;
y[k+h/2] = u-t;
w = w*wn;
}
}
}
if(on == -1)
for(int i = 0;i < len;i++)
y[i].r /= len;
}
const int N = 300005;
char s
, t
, ss[5] = "ATCG";
complex x1[N<<1],x2[N<<1], res[N<<1];
int sum[N<<1];
int ls, lt, k;
void solve(char c, int len){
memset(sum,0,sizeof(int)*len);
memset(x1,0,sizeof(complex)*len);
memset(x2,0,sizeof(complex)*len);
for(int i = 0; i < ls; i++){
if(i == 0){
sum[i] = (s[i] == c);
}
else {
sum[i] = sum[i-1] + (s[i] == c);
}
}
for(int i = 0; i < ls; i++){
int l = max(0,i-k);
int r = min(ls-1,i+k);
if(l == 0){
x1[i] = complex(sum[r]>0,0);
}
else {
x1[i] = complex((sum[r]-sum[l-1])>0,0);
}
}
for(int i = 0; i < lt; i++){
x2[i] = complex(t[i] == c,0);
}
fft(x1,len,1);
fft(x2,len,1);
for(int i = 0; i < len; i++)res[i] = res[i] + x1[i] * x2[i];
}
int main(){
while(~scanf("%d%d%d",&ls, <, &k)){
scanf("%s%s",s,t);
int len = 1;
while(len < (ls+lt+1))len<<=1;
reverse(t, t+lt);
for(int i = 0; i < 4; i++){
solve(ss[i],len);
}
fft(res, len, -1);
int ans = 0;
for(int i = 0; i < len; i++){
ans += (res[i].r+0.5) >= lt;
}
printf("%d\n",ans);
}
return 0;
}