暴力出奇迹,这种题,看着有点难其实用两重循环就行了。
题目步骤:
1st:先用一个快排,从小到大
2nd:循环
for i:=n downto 1 do
begin
for j:=1 to i-1 do
if (a[i]+a[j])>s then break
else
if i<> j then inc(ans);
end;
3rd:输出
4th:没了!!
源代码(福利)var
n,s,i,j,ans:longint;
a:array[1..20000] of longint;
procedure sort(l,r: longint);
var
i,j,x,y: longint;
begin
i:=l;
j:=r;
x:=a[(l+r) div 2];
repeat
while a[i]<x do
inc(i);
while x<a[j] do
dec(j);
if not(i>j) then
begin
y:=a[i];
a[i]:=a[j];
a[j]:=y;
inc(i);
j:=j-1;
end;
until i>