一道防AK的好题
不得不吐槽这个题目背景和算法。这就是一脑筋急转弯啊。
Code
program hard;
CONST FileName='hard';
var n,i,tot:longint;
x,a,b,c,ans:array[1..500000] of int64;
left,right:int64;
function calc(a,b,c:int64):int64;
var i:longint;
t:int64;
begin
for i:=1 to n do
begin
t:=a*(i+1)*x[i]*x[i]+(b+1)*i*x[i]+c+i;
write(t:5);
if(t=0)then calc:=i;
end;
end;
BEGIN
readln(n);
for i:=1 to n do read(x[i]);
while not eof do
begin
inc(tot);
readln(a[tot],b[tot],c[tot]);
end;
ans[tot-1]:=-a[tot];
for i:=tot-1 downto 2 do
begin
left:=a[i]*(ans[i]+1)*x[ans[i]]*x[ans[i]]+(b[i]+1)*ans[i]*x[ans[i]]+c[i]+ans[i];
right:=(ans[i]+1)*x[ans[i]]*x[ans[i]]+ans[i]*x[ans[i]]+1;
ans[i-1]:=-left div right
end;
for i:=1 to tot-1 do writeln(ans[i]);
END.
天空中的繁星
四十分做法得了70分，好开心。
Code
CONST dx:array[1..4] of longint=(1,-1,0,0);
dy:array[1..4] of longint=(0,0,1,-1);
maxn=500;
var Q:array[1..maxn*maxn,1..2] of longint;
vis,A,map:array[0..maxn+1,0..maxn+1] of boolean;
sum:array[0..maxn+1,0..maxn+1] of longint;
n,m,p,i,j,k,l,ans:longint;
s:ansistring;
procedure BFS(x,y:longint);
var h,t,i,nx,ny:longint;
isstar:boolean;
begin
h:=0;t:=1;
Q[1][1]:=x;
Q[1][2]:=y;
vis[x][y]:=true;
while h<t do
begin
inc(h);isstar:=true;
for i:=1 to 4 do
begin
nx:=Q[h][1]+dx[i];
ny:=Q[h][2]+dy[i];
if(map[nx][ny])and(not vis[nx][ny]) then
begin
vis[nx][ny]:=true;inc(t);
Q[t][1]:=nx;Q[t][2]:=ny;
end
else if not map[nx][ny]
then isstar:=false;
end;
if isstar then A[Q[h][1]][Q[h][2]]:=true;
end;
end;
BEGIN
readln(n,m,p);
for i:=1 to n do
begin
readln(s);
for j:=1 to m do
if s[j]='*'
then map[i][j]:=true;
end;
for i:=1 to n do
for j:=1 to m do
if(not vis[i][j])and(map[i][j])
then BFS(i,j);
for i:=1 to n do
for j:=1 to m do
begin
sum[i][j]:=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];
if A[i][j] then inc(sum[i][j]);
for k:=1 to i-2 do
for l:=1 to j-2 do
if sum[i-1][j-1]-sum[i-1][l]-
sum[k][j-1]+sum[k][l]>=p
then inc(ans);
end;
writeln(ans);
END.
应该注意到，对于给定的上、下、右三个边界，左边界能否满足条件满足一个单调性，是可以二分的，如果继续优化发现右边界增大，最大的满足条件的左边界不会减小，通过左边界计算方案数。滑窗应该算是新知识。
Code
program star;
CONST dx:array[1..4] of longint=(1,-1,0,0);
dy:array[1..4] of longint=(0,0,1,-1);
FileName='star';
maxn=500;
var Q:array[1..maxn*maxn,1..2] of longint;
vis,A,map:array[0..maxn+1,0..maxn+1] of boolean;
sum:array[0..maxn+1,0..maxn+1] of longint;
n,m,p,i,j,k,r,mid,l:longint;
s:ansistring;
find:boolean;
ans:int64;
function calc(a,b,c,d:longint):longint;
begin calc:=sum[c][d]-sum[c][b-1]-sum[a-1][d]+sum[a-1][b-1];end;
procedure BFS(x,y:longint);
var h,t,i,nx,ny:longint;
isstar:boolean;
begin
h:=0;t:=1;
Q[1][1]:=x;
Q[1][2]:=y;
vis[x][y]:=true;
while h<t do
begin
inc(h);isstar:=true;
for i:=1 to 4 do
begin
nx:=Q[h][1]+dx[i];
ny:=Q[h][2]+dy[i];
if(map[nx][ny])and(not vis[nx][ny]) then
begin
vis[nx][ny]:=true;inc(t);
Q[t][1]:=nx;Q[t][2]:=ny;
end
else if not map[nx][ny]
then isstar:=false;
end;
if isstar then A[Q[h][1]][Q[h][2]]:=true;
end;
end;
BEGIN
Assign(input,FileName+'.in');
Assign(output,FileName+'.out');
Reset(input);Rewrite(output);
readln(n,m,p);
for i:=1 to n do
begin
readln(s);
for j:=1 to m do
if s[j]='*'
then map[i][j]:=true;
end;
for i:=1 to n do
for j:=1 to m do
if(not vis[i][j])and(map[i][j])
then BFS(i,j);
for i:=1 to n do
for j:=1 to m do
begin
sum[i][j]:=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];
if A[i][j] then inc(sum[i][j]);
l:=1;
end;
for i:=1 to n do
for j:=i+2 to n do
begin
l:=0;//滑了个窗
for k:=3 to m do
begin
while calc(i+1,l+2,j-1,k-1)>=p do inc(l);
inc(ans,l);
end;
{
二了个分
for k:=3 to m do
begin
l:=0;r:=k;
while l<r-1 do
begin
mid:=(l+r)>>1;
if calc(i+1,mid+1,j-1,k-1)>=p
then l:=mid
else r:=mid;
end;
inc(ans,l);
end;
}
end;
writeln(ans);
Close(input);Close(output);
END.
柠檬的密码
考试没来得及交，按二十分写的暴力居然过了6个点。
Code
//写的乱七八糟。。。
program password;
var s,p:array[1..2000] of char;
a,b,l,r,maxi,n,ans,ans2,temp,len,i,j:longint;
begin
readln(n);
for i:=1 to n do
begin
read(s[i]);
p[n-i+1]:=s[i];
end;
len:=0;
for i:=1 to n do
begin
temp:=0;
for j:=1 to n do
if(p[j]=s[i+j-1])and(i+j-1<n-j+1)
then inc(temp)
else break;
if temp>len
then begin
len:=temp;
maxi:=i;
end;
end;
ans:=len<<1;
l:=maxi+len-1;r:=n-len;
for i:=l to r do
for j:=i+1 to r do
begin
len:=(j-i+1);temp:=0;
if(len and 1)=1
then begin
a:=i-1+(len+1)>>1-1;
b:=a+2;
temp:=1;
end
else begin
a:=i+-1+len>>1;
b:=i+-1+a+1;
end;
while(l<=a)and(b<=r)and(s[a]=s[b])do
begin
inc(temp,2);
dec(a);inc(b);
end;
if temp>ans2 then ans2:=temp;
end;
inc(ans,ans2);
writeln(ans);
end.
全部拿暴力分，排到51名，显然发现第一题是个脑筋急转弯的人不多。手快点就好了，第三题交上就能rank24了，要坚定敢于写暴力的信心~~~~~
