[noip2006]能量项链 区间DP

program tt;
var n,i,l,j,k,ans:longint;
    a:array[1..300]of longint;
    f:array[1..300,1..300]of longint;
function max(a,b:longint):longint;
begin
 if a<b then max:=b else max:=a;
end;
begin
 read(n);
 for i:=1 to n do
  begin
   read(a[i]);
   a[n+i]:=a[i];
   f[i,i]:=0;
   f[n+i,n+i]:=0;
  end;
 for i:=2*n-1 downto 1 do//起始位置，倒着循环可以让求f[i,j]所要用到的状态都用到
  for j:=i+1 to i+n-1 do //终止位置
   for k:=i to j-1 do  //注意这里是到j-1，因为f[i,j]最多i=j 为0
    f[i,j]:=max(f[i,j],f[i,k]+f[k+1,j]+a[i]*a[k+1]*a[j+1]);
 ans:=0;
 for i:=1 to n do
  ans:=max(ans,f[i,n+i-1]);
 writeln(ans);
end.

f[i,j]=max{f[i,k]+f[k+1,j]+a[i]*a[k+1]*a[j+1]}

循环有两种写法：

 

program tt;
var n,i,l,j,k,ans:longint;
    a:array[1..300]of longint;
    f:array[1..300,1..300]of longint;
function max(a,b:longint):longint;
begin
 if a<b then max:=b else max:=a;
end;
begin
 read(n);
 for i:=1 to n do
  begin
   read(a[i]);
   a[n+i]:=a[i];
   f[i,i]:=0;
   f[n+i,n+i]:=0;
  end;
 for l:=2 to n do
  for i:=1 to 2*n do
   begin
    j:=i+l-1;
    if j<2*n then
     for k:=i to j-1 do
      f[i,j]:=max(f[i,j],f[i,k]+f[k+1,j]+a[i]*a[k+1]*a[j+1]);
   end;
 for i:=1 to n do
  ans:=max(ans,f[i,i+n-1]);
 writeln(ans);
 readln;readln;
end.

第二种：