Codeforces 100548F (XI ' an field game f): the principle of tolerance and repulsion

Main topic:

Color the n rows of items in a row, available in M colors.

Requires different colors for adjacent items, and there is a total of K colors, asking all possible scenarios

Analysis:

Choose K species from M color, there are C (m,k) methods, then we only consider the k color of the coating method can be

Obviously the first item has K-type coating method, the back because can not be the same as the previous only k-1 seed coating method

So it's easy to think of a formula: k* (k-1) ^ (n-1)

But this formula is not more than the k color of the coating method, the question must be K-type, how to do?

Consider a simplified version of the issue first:

The number of programs that use and use up to 5 different colors to coat unrelated five items

Using factorial method can calculate ans=120, change a way of thinking about this problem, it is easy to think

Ans (Take Five colors) =5^5 (take no more than 5 colors)-C (5,4) *4^5 (take no more than 4 colors)

However, it turns out that ans is less than 0, how is this going? Easy to find in fact, less than four colors of the scheme is reduced repetition

So I thought of the need for a rebuke

Ans=c (5,5) *5^5-c (5,4) *4^5+c (5,3) *3^5-c (5,2) *2^5+c (5,1) *1^5 =120

This problem has been solved. The original problem is almost the same.

Code:

#include <iostream>#include<stdio.h>#include<string.h>#include<algorithm>#include<string>#include<ctype.h>using namespacestd;Const Long LongMod=1000000007;Const Long Longny=500000004;Long Longn,m,k;Long Longcm[1000010];Long Longcn[1000010];Long Longck[1000010];Long Longinv[1000010];Long LongMoLong Longx) {     while(x<0) x+=MoD; returnX%MoD;}Long LongEXGCD (Long LongALong LongBLong Long&x,Long Long&y) {    if(a==0&&b==0)return-1; if(b==0) {x=1; y=0;returnA;} Long LongD=EXGCD (b,a%b,y,x); Y-=a/b*x; returnD;}Long LongINV (Long LongALong LongN) {    Long Longx, y; Long LongD=EXGCD (a,n,x,y); if(d==1)return(x%n+n)%N; Else return-1;}Long LongQuickmod (Long LongALong LongBLong LongM//a^b%m{    Long Longres=1;  while(b) {if(b&1) Res=res*a%MoD; b>>=1; A=a*a%MoD; }    returnRes;}voidIni () {cn[0]=cm[0]=1; memset (CM,0,sizeof(cm)); cm[0]=1; intTmp=min (m/2, K);  for(intI=1; i<=tmp;i++) {Cm[i]= (cm[i-1]* (m+1-i)%mod*inv[i])%MoD; }    if(cm[k]==0) Cm[k]=cm[m-K]; ck[0]=ck[k]=1;  for(intI=1; i<=k/2; i++) {Ck[i]= (ck[i-1]* (k +1-i)%mod*inv[i])%MoD; Ck[k-i]=Ck[i]; }}intMain () {//freopen ("In.txt", "R", stdin); //freopen ("OUT.txt", "w", stdout);inv[1]=1;  for(intI=2; i<=1000000; i++) {Inv[i]=INV (I,MOD); }    intT; scanf ("%d",&t); intcas=0;  while(t--) {CAs++; scanf ("%i64d%i64d%i64d",&n,&m,&k);        INI (); Long Longans=0; Long Longp=1;  for(inti=k;i>=1; i--) {ans= (ans+p* (ck[k-i]) *i%mod*quickmod (i-1, N-1, MoD)%mod) +mod)%MoD; P=-p; } ans= (Ans*cm[k])%MoD; printf ("Case #%d:%c%i64d\n"Cas' ', ans); }    return 0;}

Codeforces 100548F (XI ' an field game f): the principle of tolerance and repulsion