蓝桥杯-组合公式求值（java）

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                算法提高 组合公式求值              时间限制：1.0s   内存限制：256.0MB            问题描述            　　给定n, m，求：            输入格式            　　输入一行，包含两个整数n, m。            输出格式            　　输出一行，包含求得的值，由于答案可能非常大，请输出此公式除以987654321的余数。            样例输入            3 1            样例输出            162            数据规模和约定            　　1<=m<=n<=10^7。

    package com.sihai.improve;    import java.math.BigInteger;      import java.util.Scanner;      class Number {          long x,y,dd;          /**          * @param x          * @param y          * @param dd          */          public Number(long x, long y, long dd) {              super();              this.x = x;              this.y = y;              this.dd = dd;          }          public Number() {              super();          }      }      public class Main {          static BigInteger n, m;          static int k;          static long monum = 999101;          static BigInteger mobig = new BigInteger("999101");          static BigInteger big2 = new BigInteger("2");          static long ansnum1,ans;          static long dp[][], bignum[], subnum[];          static long fact[];          private static Number gcd(long a,long b) {            if (b==0) return new Number(1,0,a);            Number number=gcd(b, a%b);            long x=number.y;            long y=number.x-(a/b)*number.y;            long dd=number.dd;            return new Number(x,y,dd);          }          private static long mod_inverse(long num) {              if (num==0) return 0;              Number number=gcd(num, monum);              long x=(number.x+monum)%monum;              return x;          }          private static long cal(BigInteger num) {              if (num.equals(BigInteger.ZERO)) return 1;              if (num.equals(BigInteger.ONE)) return 2;              long mnum = cal(num.divide(big2));              mnum = mnum*mnum%monum;              BigInteger mo = num.mod(big2);              if (mo.equals(BigInteger.ONE))                  mnum=mnum*2%monum;              return mnum;          }          private static void init() {              fact = new long[(int) (monum + 1)];              fact[0] = 1;              for (int i = 1; i <= monum; i++)                   fact[i]=(fact[i-1]*(long)i)%monum;          }          private static long calc(int n,int m) {              if (n<m) return 0;              long mo=fact[m]*fact[n-m]%monum;              long divnum=mod_inverse(mo);              long res=fact[n]*divnum%monum;              return res;          }          private static long lucas(BigInteger n,BigInteger m){              if (m.equals(BigInteger.ZERO)) return 1;              int nmo=n.mod(mobig).intValue();              int mmo=m.mod(mobig).intValue();              return calc(nmo, mmo)*lucas(n.divide(mobig),m.divide(mobig))%monum;          }          public static void main(String[] args) {              // TODO Auto-generated method stub              Scanner reader = new Scanner(System.in);              n = reader.nextBigInteger();              m = reader.nextBigInteger();              k = reader.nextInt();              BigInteger kbig=new BigInteger(String.valueOf(k));              dp = new long[k + 1][k + 1];              dp[0][0]=1;              long mon=n.mod(mobig).longValue();              for (int i = 0; i <= k - 1; i++)                  for (int j = 0; j <= i; j++) {                      dp[i + 1][j] = (dp[i+1][j]+(long)j*dp[i][j])%monum;                      dp[i + 1][j + 1] =(dp[i+1][j+1]+(long)(mon-j+monum)*dp[i][j])%monum;                  }              long mulnum =cal(n.subtract(kbig));              for (int i = k; i >= 0; i--) {                  ansnum1 =(ansnum1+dp[k][i]*mulnum)%monum;                  mulnum = (mulnum*2)%monum;              }              init();              ans=ansnum1*lucas(n, m)%monum;              System.out.println(ans);          }      }

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