Zoj3411 special judge

Problem description:

The norm of a vectorV= (V1, V2,..., vn)Is defineNorm (V)= |V1| + |V2| +... + |Vn|. And in problem 0000, there areNCases, the standard output is recorded as a vectorX= (X1, x2,..., xn). The output of a submitted solution is also records as a vectorY= (Y1, Y2,..., yn). The submitted solution will be accepted if and only ifNorm (x-y)≤M(A given tolerance ).

Knowing that all the outputXIAre in range [A,B], An incorrent solution that outputs integerYiIn range [A,B] With equal probability may also get accepted. Given the standard outputXI, 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

380 Ms

9938 K

  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  Using (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
  */ 

CodeThe core part is lines 30-40. That's simple...