作业帮n*m的矩形,用1*2的砖块铺,求方案数.这个公式:Prod[j=1..m/2,Prod[k=1..n/2,4cos(j*Pi/(m+1))^2 + 4cos(k*Pi/(n+1))^2 ]].Pascal程序:ans:=1;for i:=1 to n shr 1 dofor j:=1 to m shr 1 dobegintemp:=sqr(cos(pi*i/(n+1)))+sqr(cos(pi*j/(m+1)));ans:
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![n*m的矩形,用1*2的砖块铺,求方案数.这个公式:Prod[j=1..m/2,Prod[k=1..n/2,4cos(j*Pi/(m+1))^2 + 4cos(k*Pi/(n+1))^2 ]].Pascal程序:ans:=1;for i:=1 to n shr 1 dofor j:=1 to m shr 1 dobegintemp:=sqr(cos(pi*i/(n+1)))+sqr(cos(pi*j/(m+1)));ans:](/uploads/image/z/5023251-27-1.jpg?t=n%2Am%E7%9A%84%E7%9F%A9%E5%BD%A2%2C%E7%94%A81%2A2%E7%9A%84%E7%A0%96%E5%9D%97%E9%93%BA%2C%E6%B1%82%E6%96%B9%E6%A1%88%E6%95%B0.%E8%BF%99%E4%B8%AA%E5%85%AC%E5%BC%8F%3AProd%5Bj%3D1..m%2F2%2CProd%5Bk%3D1..n%2F2%2C4cos%28j%2APi%2F%28m%2B1%29%29%5E2+%2B+4cos%28k%2APi%2F%28n%2B1%29%29%5E2+%5D%5D.Pascal%E7%A8%8B%E5%BA%8F%3Aans%3A%3D1%3Bfor+i%3A%3D1+to+n+shr+1+dofor+j%3A%3D1+to+m+shr+1+dobegintemp%3A%3Dsqr%28cos%28pi%2Ai%2F%28n%2B1%29%29%29%2Bsqr%28cos%28pi%2Aj%2F%28m%2B1%29%29%29%3Bans%3A)
n*m的矩形,用1*2的砖块铺,求方案数.这个公式:Prod[j=1..m/2,Prod[k=1..n/2,4cos(j*Pi/(m+1))^2 + 4cos(k*Pi/(n+1))^2 ]].Pascal程序:ans:=1;for i:=1 to n shr 1 dofor j:=1 to m shr 1 dobegintemp:=sqr(cos(pi*i/(n+1)))+sqr(cos(pi*j/(m+1)));ans:
n*m的矩形,用1*2的砖块铺,求方案数.
这个公式:
Prod[j=1..m/2,Prod[k=1..n/2,4cos(j*Pi/(m+1))^2 + 4cos(k*Pi/(n+1))^2 ]].
Pascal程序:
ans:=1;
for i:=1 to n shr 1 do
for j:=1 to m shr 1 do
begin
temp:=sqr(cos(pi*i/(n+1)))+sqr(cos(pi*j/(m+1)));
ans:=ans*temp*4;
end;
writeln(ans:0:0);
c++程序:
ans=1
for (i=1;i
n*m的矩形,用1*2的砖块铺,求方案数.这个公式:Prod[j=1..m/2,Prod[k=1..n/2,4cos(j*Pi/(m+1))^2 + 4cos(k*Pi/(n+1))^2 ]].Pascal程序:ans:=1;for i:=1 to n shr 1 dofor j:=1 to m shr 1 dobegintemp:=sqr(cos(pi*i/(n+1)))+sqr(cos(pi*j/(m+1)));ans:
killnow5788 拌得很厉害...lz不要理他.
参考M.E.Fischer的一篇论文吧
Statistcal Mechanics of Dimers on a Plane Lattice
你写的那两个程序就是把公式给表达出来而已。
你是不是公式没理解啊？
而且这两个程序用了最冗余的办法，算法都没优化。
哦，那你觉得好就好吧~