标签:int swap can ace ons inf i++ oid 技术
1 #include <cstdio> 2 #include <istream> 3 #define ll long long 4 using namespace std; 5 const ll N=200010,mo=786433; 6 int T,n,m,k,x,y,sum,num[N],f[N],len,L,rev[N],v[10][N]; 7 ll a[10][N],inv[mo],fac[N],ny[N],ans[N],p[N],P[N]; 8 char s[N]; 9 ll ksm(ll a,ll b) { ll r=1; for (;b;b>>=1,a=a*a%mo) if (b&1) r=r*a%mo; return r; } 10 void ntt(ll *a,int len,int f) 11 { 12 for (int i=0;i<len;i++) if (rev[i]>i) swap(a[i],a[rev[i]]); 13 for (int i=2;i<=len;i<<=1) 14 { 15 ll r,m=i>>1; 16 if (f==1) r=p[i]; else r=P[i]; 17 for (int j=0;j<len;j+=i) 18 { 19 ll R=1; 20 for (int k=0;k<m;k++,R=R*r%mo) 21 { 22 ll x=a[j+k],y=a[j+k+m]*R%mo; 23 a[j+k]=(x+y)%mo,a[j+k+m]=(x-y+mo)%mo; 24 } 25 } 26 } 27 if (f==-1) { ll r=ksm(len,mo-2); for (int i=0;i<len;i++) a[i]=a[i]*r%mo; } 28 } 29 int main() 30 { 31 32 fac[0]=ny[0]=1; 33 for (int i=1;i<N;i++) p[i]=ksm(10,(mo-1)/i),P[i]=ksm(p[i],mo-2); 34 for (int i=0;i<mo;i++) inv[i]=ksm(i,mo-2); 35 for (int i=1;i<N;i++) fac[i]=fac[i-1]*(ll)i%mo; 36 ny[N-1]=ksm(fac[N-1],mo-2); 37 for (int i=N-2;i;i--) ny[i]=ny[i+1]*(ll)(i+1)%mo; 38 scanf("%d",&T); 39 while (T--) 40 { 41 scanf("%d%d%d",&n,&m,&k),n--,sum=0; 42 for (int i=0;i<n;i++) 43 { 44 scanf("%s",s); 45 for (int j=0;j<=m;j++) v[i][j]=s[j]-‘0‘; 46 } 47 for (len=1,L=-1;len<=n*m;len*=2) L++; 48 for (int i=0;i<len;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<L); 49 for (int i=0;i<len;i++) num[i]=0,f[i]=1; 50 for (int i=0;i<n;i++) 51 { 52 for (int j=0;j<len;j++) a[i][j]=0; 53 for (int j=0;j<=m;j++) if (v[i][j]) a[i][j]=ny[j]; 54 ntt(a[i],len,1); 55 for (int j=0;j<len;j++) if (a[i][j]) f[j]=f[j]*a[i][j]%mo; else num[j]++; 56 } 57 for (int i=0;i<len;i++) ans[i]=num[i]?0:f[i]; 58 for (int x,y;k;k--) 59 { 60 scanf("%d%d",&x,&y),x--; 61 for (int i=0;i<len;i++) if (a[x][i]) f[i]=f[i]*inv[a[x][i]]%mo; else num[i]--; 62 ll flag; 63 if (v[x][y]==1) flag=-1; else flag=1; 64 v[x][y]^=1; 65 ll r=1,R=ksm(p[len],y); 66 for (int i=0;i<len;i++,r=r*R%mo) a[x][i]=(a[x][i]+flag*inv[fac[y]]*r%mo+mo)%mo; 67 for (int i=0;i<len;i++) if (a[x][i]) f[i]=f[i]*a[x][i]%mo; else num[i]++; 68 for (int i=0;i<len;i++) ans[i]=(ans[i]+(num[i]?0:f[i]))%mo; 69 } 70 ntt(ans,len,-1); 71 for (int i=1;i<len;i++) sum=(sum+ans[i]*fac[i]%mo)%mo; 72 printf("%lld\n",sum); 73 } 74 }
[生成函数][DFT][NTT] Hdu P6067 Big Integer
标签:int swap can ace ons inf i++ oid 技术
原文地址:https://www.cnblogs.com/Comfortable/p/11364236.html