124 - ZOJ Monthly, March 2013 - E Choosing number 选数的方法数

http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=4963

Choosing number

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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There are n people standing in a row. And There are m numbers, 1.2...m. Every one should choose a number. But if two persons standing adjacent to each other choose the same number, the number shouldn't equal or less thank. Apart from this rule, there are no more limiting conditions.

And you need to calculate how many ways they can choose the numbers obeying the rule.

Input

There are multiple test cases. Each case contain a line, containing three integer n (2 ≤ n ≤ 10⁸), m (2 ≤ m ≤ 30000), k(0 ≤ k ≤ m).

Output

One line for each case. The number of ways module 1000000007.

Sample Input

4 4 1

Sample Output

216

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Author: GU, Shenlong

题意：  输入  n m k

有n个人 选择m个数      每个人可以任意选数 ，   当2个相邻的人选择相同的数 当且仅当这个数大于k  问总共有多少种方法选择数

思路：

设x0 为上一个人选的时候选择的小于k的数的方法数， x1为上一个人选的时候选择的大于k的数的方法数

y0为当前人选择的小于k的数的方法数，y1为当前人选择的大于k的数的方法数

则有  y0=x0*(k-1)+x1*k

          y1=x0*(m-k)+x1*(m-k)

这样可以构造一个矩阵了

;  k-1      k       ;         ;x0;                   ;y0

;                       ;    *   ;   ;          =        ;

; m-k   m-k     ;         ;x1;                   ;

#include <stdio.h>#define N 2#define mod 1000000007struct mat{long long mar[2][2];};mat a,b,c,init,temp;mat res=  {  1,0,          0,1     };mat mul(mat a1,mat b1)  {      long long i,j,l;      mat c1;      for (i=0;i<N;i++)      {          for (j=0;j<N;j++)          {              c1.mar[i][j]=0;              for (l=0;l<N;l++)              {                  c1.mar[i][j]+=(a1.mar[i][l]*b1.mar[l][j])%mod;                  c1.mar[i][j]%=mod;              }          }      }      return c1;  }  mat er_fun(mat e,long long x){mat tp;tp=e;e=res;while(x){if(x&1)e=mul(e,tp);tp=mul(tp,tp);x>>=1;}return e;}long long mi(long long a,long long k,long long M){    long long b=1;    while(k>=1){        if(k%2==1){            b=a*b%M;        }        a=(a%M)*(a%M)%M;        k/=2;    }    return b;}int main(){long long i,j;long long x1,x2;long long dp[2];long long ss;long long n,m,k;while(scanf("%lld%lld%lld",&n,&m,&k)!=EOF){if(k!=0){init.mar[0][0]=k-1;init.mar[0][1]=k;init.mar[1][0]=m-k;init.mar[1][1]=m-k;a=init;            b=er_fun(a,n-1);dp[0]=k;dp[1]=m-k;ss=b.mar[0][0]*dp[0]%mod+b.mar[0][1]*dp[1]%mod+b.mar[1][0]*dp[0]%mod+b.mar[1][1]*dp[1]%mod;ss=ss%mod;printf("%lld\n",ss);}else{printf("%lld\n",mi(m,n,1000000007));}}return 0;}