bzoj 4503: 两个串

和上一题一样的，这不过这个去掉一个就可以，大概是酱紫 ∑(aj-bi)*bi 就好（此处a，b可以和原题不同）

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define LL long long
#define pi acos(-1)
using namespace std;

const int N=100005;

struct complex
{
    double r,i;
    complex (double x=0, double y=0) {r=x; i=y;}
    complex operator + (const complex a) {return complex(r+a.r,i+a.i);}
    complex operator - (const complex a) {return complex(r-a.r,i-a.i);}
    complex operator * (const complex a) {return complex(r*a.r-i*a.i,r*a.i+i*a.r);}
}a[N<<2],b[N<<2],c1[N<<2],c2[N<<2],c3[N<<2];

int n,len,lem,m,qwq;
int ans[N],ans1[N],ans2[N],ans3[N],A[N],B[N];
int rev[N<<2];
char s1[N],s2[N];

void FFT(complex *a, int f)
{
    for (int i=0; i<n; i++) if (rev[i]>i) swap(a[i],a[rev[i]]);
    for (int h=2; h<=n; h<<=1)
    {
        complex wn(cos(2*pi*f/h),sin(2*pi*f/h));
        for (int i=0; i<n; i+=h)
        {
            complex w(1,0);
            for (int j=0; j<(h>>1); j++,w=w*wn)
            {
                complex x=a[i+j],y=w*a[i+j+(h>>1)];
                a[i+j]=x+y;
                a[i+j+(h>>1)]=x-y;
            }
        }
    }
    if (f==-1) for (int i=0; i<n; i++) a[i].r/=(double)n;
}

int main(int argc, char const *argv[])
{
    scanf("%s",s1); scanf("%s",s2);
    n=strlen(s1); m=strlen(s2); qwq=n;
    for (int i=0; i<n; i++) A[i]=s1[i]-'a'+1;
    for (int i=0; i<m; i++) B[i]=(s2[m-i-1]=='?')?0:s2[m-i-1]-'a'+1;
    int orz=n<<1,L=0;
    for (n=1; n<orz; n<<=1) L++;
    for (int i=0; i<n; i++)
        rev[i]=(rev[i>>1]>>1)|((i&1)?n>>1:0);

    for (int i=0; i<qwq; i++) a[i]=A[i]*A[i];
    for (int i=0; i<m; i++) b[i]=B[i];
    FFT(a,1); FFT(b,1);
    for (int i=0; i<n; i++) c1[i]=a[i]*b[i];
    FFT(c1,-1);
    for (int i=0; i<qwq; i++) ans1[i]=(int)(c1[i].r+0.1);

    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    for (int i=0; i<qwq; i++) a[i]=1;
    for (int i=0; i<m; i++) b[i]=B[i]*B[i]*B[i];
    FFT(a,1); FFT(b,1);
    for (int i=0; i<n; i++) c2[i]=a[i]*b[i];
    FFT(c2,-1);
    for (int i=0; i<qwq; i++) ans2[i]=(int)(c2[i].r+0.1);

    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    for (int i=0; i<qwq; i++) a[i]=A[i];
    for (int i=0; i<m; i++) b[i]=B[i]*B[i];
    FFT(a,1); FFT(b,1);
    for (int i=0; i<n; i++) c3[i]=a[i]*b[i];
    FFT(c3,-1);
    for (int i=0; i<qwq; i++) ans3[i]=(int)(c3[i].r+0.1);

    for (int i=m-1; i<qwq; i++)
            if (ans1[i]+ans2[i]-2*ans3[i]==0) ans[++ans[0]]=i;
    printf("%d
",ans[0]);
    for (int i=1; i<=ans[0]; i++) printf("%d
",ans[i]-m+1);
    return 0;
}