⑨ to write (Codevs 1697)

Title Description Description

Qi Cirno (Ice Goblin) has the ability to control the air-conditioning. It is more dangerous to freeze a small thing in an instant than a normal goblin. had been releasing the air-conditioner around her always very cold.

Due to the following three points reason ...

• Qi Cirno's Fuca symbol "Icicle Fall"-easy of the bomb screen is enough stupid, as long as standing in front of her there is no play will touch you;

• Zun in "Red Devil Township" introduced her when she said she was a bit stupid;

• In the Zun release "Oriental Phantasmagoria Tomb" Introduction map, in the picture put the Qi Lulu in the position of ⑨, and with "⑨ idiot" simple take, from "⑨" and "idiot" became her alias ...

So Kiki Cirno got the nickname of "idiot".

One day, Kiki Cirno again 2 ...

She wrote n letters to be loaded into N envelopes, but all were wrong ... Now you want to know how many kinds of possibilities are wrong.

Enter a description input Description

Number of letters and envelopes N.

Outputs description Output Description

The number of possibilities to mount the error.

Sample input to sample

Input Example 1

2

Input Example 2

4

Sample output Sample Outputs

Output Example 1

1

Output Example 2

9

Data Size & Hint

1≤n≤100

/*recursion + high precision.  The first time got 60 points, the B, C array to the main function outside it is over, long memory ...    There are n letters, the first letter is loaded wrong words have i-1 kind of possibility, set i-1 one is K, then K has two kinds of loading method: ①: Into the first envelope, at this time the remaining I-2 letter to be installed in i-2 envelope, there is f[i-2];   ②: Loaded into an envelope other than I, then the i-1 function and I, the rest of the f[i-1]; To sum up, get the state transfer equation: f[i]= (i-1) * (F[i-1]+f[i-2])*/#include<cstdio>#include<iostream>#include<cstring>#defineM 210using namespacestd;structnode{intA[m],len;}; Node F[m];intB[m],c[m];intMain () {intN; scanf ("%d",&N); f[1].a[1]=0; f[2].a[1]=1; f[1].len=1; f[2].len=1;  for(intI=3; i<=n;i++)    {        //High-precision addition        intlena=f[i-1].len,lenb=f[i-2].len,lenc=1, x=0;  while(lenc<=lena| | lenc<=LenB) {F[i].a[lenc]=f[i-1].a[lenc]+f[i-2].a[lenc]+x; X=f[i].a[lenc]/Ten; F[I].A[LENC]%=Ten; Lenc++; } F[i].a[lenc]=Y; if(f[i].a[lenc]==0) lenc--; F[i].len=Lenc; //high-precision multiplication        intla=f[i].len,zh=i-1, lb=0; memset (b,0,sizeof(b)); Memset (c,0,sizeof(c));  while(ZH) {b[++lb]=zh%Ten; ZH/=Ten; }         for(intj=1; j<=la;j++)        {            intx=0;  for(intk=1; k<=lb;k++) {c[j+k-1]+=f[i].a[j]*b[k]+x; X=c[j+k-1]/Ten; C[j+k-1]%=Ten; } c[j+lb]=x; }        intlc=la+lb; F[i].len=LC;  while(c[lc]==0&&lc>1) lc--;  for(intj=lc;j>=1; j--) F[i].a[j]=C[j]; }    intflag=0;  for(inti=f[n].len;i>=1; i--)      if(!flag&&!f[n].a[i])Continue; Else{printf ("%d", F[n].a[i]); Flag=1; }    return 0;}

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⑨ to write (Codevs 1697)