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2020-2021:teams:legal_string:jxm2001:contest:edu_102 [2021/01/16 16:16] jxm2001 [题解] |
2020-2021:teams:legal_string:jxm2001:contest:edu_102 [2021/01/17 13:59] (当前版本) jxm2001 [G. Tiles] |
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} | } | ||
enter(ans-solver.Maxflow(s,t)); | enter(ans-solver.Maxflow(s,t)); | ||
+ | return 0; | ||
+ | } | ||
+ | </code> | ||
+ | </hidden> | ||
+ | |||
+ | ===== G. Tiles ===== | ||
+ | |||
+ | 快速傅里叶变换好题 | ||
+ | |||
+ | <hidden 查看代码> | ||
+ | <code cpp> | ||
+ | const int MAXN=5005,MAXV=2e5+5,Mod=998244353; | ||
+ | int quick_pow(int a,int k){ | ||
+ | int ans=1; | ||
+ | while(k){ | ||
+ | if(k&1)ans=1LL*ans*a%Mod; | ||
+ | a=1LL*a*a%Mod; | ||
+ | k>>=1; | ||
+ | } | ||
+ | return ans; | ||
+ | } | ||
+ | namespace Poly{ | ||
+ | const int G=3; | ||
+ | int rev[MAXN<<2],Pool[MAXN<<3],*Wn[30]; | ||
+ | void init(){ | ||
+ | int lg2=0,*pos=Pool,n,w; | ||
+ | while((1<<lg2)<MAXN*2)lg2++; | ||
+ | n=1<<lg2,w=quick_pow(G,(Mod-1)/(1<<lg2)); | ||
+ | while(~lg2){ | ||
+ | Wn[lg2]=pos,pos+=n; | ||
+ | Wn[lg2][0]=1; | ||
+ | _for(i,1,n)Wn[lg2][i]=1LL*Wn[lg2][i-1]*w%Mod; | ||
+ | w=1LL*w*w%Mod; | ||
+ | lg2--;n>>=1; | ||
+ | } | ||
+ | } | ||
+ | int build(int k){ | ||
+ | int n,pos=0; | ||
+ | while((1<<pos)<=k)pos++; | ||
+ | n=1<<pos; | ||
+ | _for(i,0,n)rev[i]=(rev[i>>1]>>1)|((i&1)<<(pos-1)); | ||
+ | return n; | ||
+ | } | ||
+ | void NTT(int *f,int n,bool type){ | ||
+ | _for(i,0,n)if(i<rev[i]) | ||
+ | swap(f[i],f[rev[i]]); | ||
+ | int t1,t2; | ||
+ | for(int i=1,lg2=1;i<n;i<<=1,lg2++){ | ||
+ | for(int j=0;j<n;j+=(i<<1)){ | ||
+ | _for(k,j,j+i){ | ||
+ | t1=f[k],t2=1LL*Wn[lg2][k-j]*f[k+i]%Mod; | ||
+ | f[k]=(t1+t2)%Mod,f[k+i]=(t1-t2)%Mod; | ||
+ | } | ||
+ | } | ||
+ | } | ||
+ | if(!type){ | ||
+ | reverse(f+1,f+n); | ||
+ | int div=quick_pow(n,Mod-2); | ||
+ | _for(i,0,n)f[i]=(1LL*f[i]*div%Mod+Mod)%Mod; | ||
+ | } | ||
+ | } | ||
+ | void Mul(int *f,int _n,int *g,int _m,int xmod=0){ | ||
+ | int n=build(_n+_m-2); | ||
+ | _for(i,_n,n)f[i]=0;_for(i,_m,n)g[i]=0; | ||
+ | NTT(f,n,true);NTT(g,n,true); | ||
+ | _for(i,0,n)f[i]=1LL*f[i]*g[i]%Mod; | ||
+ | NTT(f,n,false); | ||
+ | if(xmod)_for(i,xmod,n)f[i]=0; | ||
+ | } | ||
+ | } | ||
+ | int frac[MAXV],invfrac[MAXV],a[MAXN],b[MAXN]; | ||
+ | int C(int n,int m){ | ||
+ | if(m<0||m>n)return 0; | ||
+ | return 1LL*frac[n]*invfrac[m]%Mod*invfrac[n-m]%Mod; | ||
+ | } | ||
+ | int dp[MAXN<<2],temp[MAXN<<2],c[MAXN<<2]; | ||
+ | int main() | ||
+ | { | ||
+ | Poly::init(); | ||
+ | int n=read_int(); | ||
+ | _for(i,0,n)a[i]=read_int(),b[i]=read_int(); | ||
+ | frac[0]=1; | ||
+ | _for(i,1,MAXV) | ||
+ | frac[i]=1LL*frac[i-1]*i%Mod; | ||
+ | invfrac[MAXV-1]=quick_pow(frac[MAXV-1],Mod-2); | ||
+ | for(int i=MAXV-1;i;i--) | ||
+ | invfrac[i-1]=1LL*invfrac[i]*i%Mod; | ||
+ | dp[0]=1; | ||
+ | int len=1; | ||
+ | _for(i,0,n){ | ||
+ | _for(j,0,2*len+a[i]-b[i]-1) | ||
+ | c[j]=C(a[i]+b[i],j+b[i]-len+1); | ||
+ | Poly::Mul(dp,len,c,2*len+a[i]-b[i]-1); | ||
+ | _for(j,0,len+a[i]-b[i]) | ||
+ | dp[j]=dp[j+len-1]; | ||
+ | len+=a[i]-b[i]; | ||
+ | } | ||
+ | int ans=0; | ||
+ | _for(i,0,len) | ||
+ | ans=(ans+dp[i])%Mod; | ||
+ | enter(ans); | ||
return 0; | return 0; | ||
} | } | ||
</code> | </code> | ||
</hidden> | </hidden> |