bzoj3160万径人踪灭

具体见http://blog.csdn.net/PoPoQQQ/article/details/42193259

code：

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<cmath>

 4 #include<cstring>

 5 #include<algorithm>

 6 #define maxn 262144

 7 #define pi 3.14159265358979323846

 8 #define mod 1000000007

 9 using namespace std;

10 char s[maxn],ss[maxn];

11 bool ok;

12 int ans,ans1,len,m,n,id,pow2[maxn],f[maxn],g[maxn];

13 struct comp{

14     double rea,ima;

15     void clear(){rea=ima=0;}

16     comp operator +(const comp &x){return (comp){rea+x.rea,ima+x.ima};}

17     comp operator -(const comp &x){return (comp){rea-x.rea,ima-x.ima};}

18     comp operator *(const comp &x){return (comp){rea*x.rea-ima*x.ima,rea*x.ima+ima*x.rea};}

19 }a[maxn],c[maxn],tmp[maxn],w,wn;

20 void fft(comp *a,int st,int siz,int step,int op){

21     if (siz==1) return;

22     fft(a,st,siz>>1,step<<1,op),fft(a,st+step,siz>>1,step<<1,op);

23     int x=st,x1=st,x2=st+step;

24     w=(comp){1,0},wn=(comp){cos(op*2*pi/siz),sin(op*2*pi/siz)};

25     for (int i=0;i<(siz>>1);i++,x+=step,x1+=(step<<1),x2+=(step<<1),w=w*wn)

26         tmp[x]=a[x1]+(w*a[x2]),tmp[x+(siz>>1)*step]=a[x1]-(w*a[x2]);

27     for (int i=st;siz;i+=step,siz--) a[i]=tmp[i];

28 }

29 void manacher(){

30     m=(len<<1)+1,ss[1]='#',id=0;

31     for (int i=1;i<=len;i++) ss[i<<1]=s[i],ss[(i<<1)+1]='#';

32     for (int i=1;i<=m;i++){

33         f[i]=(id+f[id]>=i)?min(f[(id<<1)-i],id+f[id]-i):0;

34         while (i-f[i]-1>=1&&i+f[i]+1<=m&&ss[i-f[i]-1]==ss[i+f[i]+1]) f[i]++;

35         if (i+f[i]>id+f[id]) id=i;

36         ans+=((f[i]+1)>>1),ans%=mod;

37     }

38 }

39 int main(){

40     scanf("%s",s+1); n=1,pow2[0]=1;

41     len=strlen(s+1),manacher();

42     while (n<(len<<1)) n<<=1;

43     for (int i=1;i<n;i++) pow2[i]=(pow2[i-1]<<1)%mod;

44     for (int i=1;i<=len;i++) a[i].rea=(s[i]=='a');

45     fft(a,0,n,1,1);

46     for (int i=0;i<n;i++) c[i]=a[i]*a[i];

47     for (int i=0;i<n;i++) a[i].clear();

48     for (int i=1;i<=len;i++) a[i].rea=(s[i]=='b');

49     fft(a,0,n,1,1);

50     for (int i=0;i<n;i++) c[i]=c[i]+a[i]*a[i];

51     fft(c,0,n,1,-1);

52     for (int i=0;i<n;i++) g[i]+=(int)round(c[i].rea/n);

53     for (int i=0;i<n;i++) ans1+=pow2[(g[i]+1)>>1]-1,ans1%=mod;

54     ans1-=ans,ans1%=mod,ans1+=mod,ans1%=mod;

55     printf("%d\n",ans1);

56     system("pause");

57     return 0;

58 }