2014 Xian live game F Q Uvala 7040

Address

Test instructions: To choose K colors in M colors, there are several ways to paint n flowers.

Analysis: Drawing can be found that the basic formula is KX (K-1) ^ (n-1). However, this only guarantees that the adjacent color, the total number of colors not more than k, and there is no guarantee that there is a k color; Then there is a repulsion problem, the above calculation method contains only 2, 3 、...、 (k-1) Color of the case, need to be removed by the principle of repulsion. Assuming P (2 <= P <= k-1) color, choose p species from the k color to paint, the scheme number is C (k,p) xpx (p-1) ^ (n-1); The total number of scenarios is C (M,K) x (KX (K-1) ^ (n-1) +∑ (( -1) ^px C (k, p) xpx (p-1) ^ (n-1)).

Diagram Method:

There are 8 kinds of cases, but there are two types that are only two colors and need to be lost.

TLE Reason: 1) The inverse element does not hit the table; 2) The parameter is redundant, the mod is a global variable, but the template has this item, when the parameters are passed on, the result is tle.

This is our last year in Xi ' an live game when the topic, on the card in this topic, has been tle, has not been able to change, today the topic reproduced once, still tle. It is not a solid thing when learning new knowledge.

1#include <cstdio>2#include <iostream>3#include <sstream>4#include <cmath>5#include <cstring>6#include <cstdlib>7#include <string>8#include <vector>9#include <map>Ten#include <Set> One#include <queue> A#include <stack> -#include <algorithm> - using namespacestd; the #definell Long Long - #define_cle (M, a) memset (M, A, sizeof (m)) - #defineRepu (I, A, b) for (int i = A; I < b; i++) - #defineREPD (I, A, b) for (int i = b; i >= A; i--) + #defineSFI (n) scanf ("%d", &n) - #definePFI (n) printf ("%d\n", N) + #defineSfi2 (n, m) scanf ("%d%d", &n, &m) A #definePfi2 (n, m) printf ("%d%d\n", N, m) at #definePfi3 (A, B, c) printf ("%d%d%d\n", A, B, c) - #defineMAXN 1000005 - #defineMOD 1000000007 - Const intINF =0x3f3f3f3f; - ll INV[MAXN]; - ll Quickpow (ll M, ll N) in { -ll ans =1; to      while(n) +     { -         if(N &1)///if n is an odd number theAns = (ans * m)%MOD; *n = n >>1;///bitwise operation "Right shift 1 similar to 2" $m = (M * m)%MOD;Panax Notoginseng     } -     returnans; the } + ll C (ll N, ll M) A { the     if(M > N)return 0; +ll ans =1; -      for(inti =1; I <= m; i++) $     { $ll a = (n-m + i)%MOD; -ll B = i%MOD; -Ans = ans * (A * QUICKPOW (b, MOD-2)% MOD)%MOD; the     } -     returnans;Wuyi } the ll Lucas (ll N, ll M) - { Wu     if(M = =0)return 1; -     Else About         return(C (n mod, m% MoD) * Lucas (N/mod, m/mod))%MOD; $ } - void Get() - { -Repu (I,1, MAXN) AInv[i] = Quickpow (i, MOD-2); + } thell cc[1010100]; - intMain () $ { the     Get(); the     intT; the SFI (T); the     intKase =1; - ll N, M, K; in      while(t--) the     { thescanf"%lld%lld%lld", &n, &m, &k); About         if(k==1) the         { theprintf"Case #%d:", kase++); the             if(n==1) printf ("%d\n", m); +             Elseprintf"0\n"); -             Continue ; the         }Bayill CMK =Lucas (M, k); thell t = Quickpow (K-1N1); thet = (t * k)%MOD; -         intFlag =0; -ll cc =1; the          for(ll p = k-1; P >=2; p--) the         { theCC = (((CC * (P +1)) (% MOD) * Inv[k-p])%MOD; the             ///The hand residue added a mod also will be tle ....  -ll q = (((CC * p)% MOD) * QUICKPOW (P-1N1)) %MOD; the             if(flag) T = (t + q)%MOD; the             ElseT = (t-q + MOD)%MOD; theFlag =!Flag;94         } theprintf"Case #%d:%lld\n", kase++, (t * cmk)%MOD); the     } the     return 0;98}

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2014 Xian live game F Q Uvala 7040