演算法提高 組合公式求值  
            時間限制：1.0s   記憶體限制：256.0MB

            問題描述
            　　給定n, m，求：

            輸入格式
            　　輸入一行，包含兩個整數n, m。
            輸出格式
            　　輸出一行，包含求得的值，由於答案可能非常大，請輸出此公式除以987654321的餘數。
            樣例輸入
            3 1
            樣例輸出
            162
            資料規模和約定
            　　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);  
        }  

    }