Gym101194H(Great Cells)-思維
阿新 • • 發佈:2019-02-08
題意:有一張N*M的格子紙,每個格子可以填1到K之間的數。如果一個格子裡的數嚴格大於本行的其他格子裡的數,並且嚴格大於本列的的其他格子裡的數,則這個格子叫做Great Cell。Ag表示有Ag種填法使得格子紙中恰有g個Great Cell。
求:
思路:
選一個坑,然後這一行一列的數就要受到這個坑的限制,就有n-1+m-1受到限制,剩下的就有(n-1)*(m-1)沒限制。
上述受到限制的結果就是(i-1)^(n-1+m-1) 沒限制的結果就是k^(n-1)*(m-1),然後n行m列選一個的話有n*m種選擇。
//china no.1 #pragma comment(linker, "/STACK:1024000000,1024000000") #include <vector> #include <iostream> #include <string> #include <map> #include <stack> #include <cstring> #include <queue> #include <list> #include <stdio.h> #include <set> #include <algorithm> #include <cstdlib> #include <cmath> #include <iomanip> #include <cctype> #include <sstream> #include <functional> #include <stdlib.h> #include <time.h> #include <bitset> using namespace std; #define pi acos(-1) #define PI acos(-1) #define endl '\n' #define srand() srand(time(0)); #define me(x,y) memset(x,y,sizeof(x)); #define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++) #define close() ios::sync_with_stdio(0); cin.tie(0); #define FOR(x,n,i) for(int i=x;i<=n;i++) #define FOr(x,n,i) for(int i=x;i<n;i++) #define fOR(n,x,i) for(int i=n;i>=x;i--) #define fOr(n,x,i) for(int i=n;i>x;i--) #define W while #define sgn(x) ((x) < 0 ? -1 : (x) > 0) #define bug printf("***********\n"); #define db double #define ll long long typedef long long LL; const int INF=0x3f3f3f3f; const LL LINF=0x3f3f3f3f3f3f3f3fLL; const int dx[]={-1,0,1,0,1,-1,-1,1}; const int dy[]={0,1,0,-1,-1,1,-1,1}; const int maxn=2e5+10; const int maxx=1e6+100; const double EPS=1e-8; const double eps=1e-8; const int mod=1e9+7; template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);} template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);} template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));} template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));} template <class T> inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;} while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;} inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false; while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;} else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}} if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}} if(IsN) num=-num;return true;} void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');} void print(LL a){ Out(a),puts("");} //freopen( "in.txt" , "r" , stdin ); //freopen( "data.txt" , "w" , stdout ); //cerr << "run time is " << clock() << endl; int _pow(LL a,LL n) { LL ret=1; W(n) { if(n&1) ret=ret*a%mod; a=a*a%mod; n>>=1; } return ret; } int n,m,k; int main() { int t; scan_d(t); for(int cas=1;cas<=t;cas++) { LL ans=0; scan_d(n),scan_d(m),scan_d(k); ans+=_pow(k,n*m); ans%=mod; FOR(1,k,i) { ans+=(LL)n*m%mod*_pow(i-1,n-1+m-1)%mod*_pow(k,(n-1)*(m-1)); ans%=mod; } printf("Case #%d: %d\n",cas,ans); } }