ZOJ3411 Special Special Judge

问题描述：

The norm of a vector v = (v₁, v₂, ... , v_n) is define as norm(v) = |v₁| + |v₂| + ... + |v_n|. And in Problem 0000, there are n cases, the standard output is recorded as a vector x = (x₁, x₂, ... , x_n). The output of a submitted solution is also records as a vector y = (y₁, y₂, ... , y_n). The submitted solution will be accepted if and only if norm(x - y) ≤ m (a given tolerance).

    Knowing that all the output x_i are in range [a, b], an incorrent solution that outputs integer y_i in range [a, b] with equal probability may also get accepted. Given the standard output x_i, you are supposed to calculate the possibility of getting accepted using such solution.

n, m, a, b (1 ≤ n, m ≤ 50, -50 ≤ a < b ≤ 50)

Accepted

3411

Java

380MS

9938K

import java.util.*;
import java.math.*;

public class Main{
    public static void main(String[] args){
        int n,m,a,b,x[]=new int[51];
        BigInteger d[][]=new BigInteger[51][51],f1,f2,f3;
        Scanner cin=new Scanner(System.in);
        while(cin.hasNext()) {
            n=cin.nextInt();
            m=cin.nextInt();
            a=cin.nextInt();
            b=cin.nextInt();
            //System.out.println(n+" "+m+" "+a+" "+b);
            int i,j,k,t;
            for(i=1;i<=n;i++)
                x[i]=cin.nextInt();
            for(i=0;i<=n;i++)
                for(j=0;j<=m;j++)
                    d[i][j]=BigInteger.ZERO;
            for(i=a;i<=b;i++){
                t=Math.abs(i-x[1]);
                //System.out.println(t);
                if(t<=m) {
                    d[1][t]=d[1][t].add(BigInteger.ONE);
                    //System.out.println("d[1]["+t+"]="+d[1][t]);
                }
            }
            
            for(i=2;i<=n;i++){
                for(j=a;j<=b;j++){
                    t=Math.abs(j-x[i]);
                    if(t<=m){
                        for(k=t;k<=m;k++){
                            d[i][k]=d[i][k].add(d[i-1][k-t]);
                            //System.out.println("d["+i+"]["+k+"]="+d[i][k]);
                        }
                    }
                }
            }
            f1=BigInteger.ZERO;
            for(i=0;i<=m;i++)
                f1=f1.add(d[n][i]);
            f2=BigInteger.ONE;
            for(i=1;i<=n;i++)
                f2=f2.multiply(BigInteger.valueOf(b-a+1));
            //System.out.println(f1+" "+f2);
            f3=f1.gcd(f2);
            f1=f1.divide(f3);
            f2=f2.divide(f3);
            System.out.println(f1+"/"+f2);
        }
    }
}
/*
1 1 0 2
1
4 2 2 5
2 3 2 4
4 3 2 5
2 3 2 4

1/1
3/32
7/32
*/

代码的核心部分是Lines 30-40。简单吧，嘻嘻…