POJ1837

题意:给出N个挂钩,M个砝码,每个挂钩有一个权值,可正可负(相当于天平的左右边).

求用M个砝码使天平平衡的方案数(砝码必须全用上).

分析:f[i,j]表示用前i个砝码,左右差值为j的方案数.转移见代码.

ans=f[m,0].

code:

const range=8000;
      r=7500;
var   f:array[0..21,-range..range] of longint;
      n,m,k,i,j:longint;
      p,w:array[0..21] of longint;



begin
      readln(n,m);
      for i:=1 to n do read(p[i]);
      readln;
      for i:=1 to m do read(w[i]);

      for i:=1 to n do
      inc(f[1,w[1]*p[i]]);

      for i:=2 to m do
         for j:=1 to n do
            for k:=-r to r do
            f[i,k+w[i]*p[j]]:=f[i,k+w[i]*p[j]]+f[i-1,k];

      writeln(f[m,0]);
end.