303 条题解
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0
wangyk LV 6 @ 2009-05-10 00:06:09
经验错误
先是看成保留多少人
错误一次
身高比较当成导弹拦截了
加上了=
又错了一次 -
0
abrahamlee LV 8 @ 2009-05-08 08:41:20
本问题可以分解成:
1 从前向后递推最大升序列,可以利用f[0]=0;
2 从后向前递推最大降序列,可以利用g[n+1]=0;
3 枚举所有学生以其为最高点的最大升序和最大降序之和,求最大;
4 从总数中减去,注意自身多重复一次,需要做加一减一处理. -
0
var
a,f:array[0..200] of longint;
n,ans,tot,sum:longint;
procedure init;
var i:longint;
begin
readln(n);
for i:=1 to n do read(a[i]);
end;
procedure main;
var i,j,k:longint;
begin
for k:=1 to n do
begin
tot:=0;
for i:=1 to n do f[i]:=1;
for i:=k downto 1 do
for j:=i-1 downto 1 do
if (a[i]>a[j]) and (f[j] -
0
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Accepted 有效得分：100 有效耗时：0ms
var q,f,a:array[1..1000] of integer;
min,i,j,max,k,n,m,x:integer;
begin
readln(n);min:=maxint;
for i:=1 to n do read(a[i]);
for k:=1 to n do
begin
m:=k;max:=0;
for i:=1 to n do q[i]:=1;
for i:=2 to k do
for j:=1 to i-1 do
if (a[i]max then max:=q[i];
m:=m-max;
for i:=1 to n do q[i]:=1;
x:=n-k+1;
max:=0;
for i:=n-1 downto k do
for j:=n downto i+1 do
if (a[i]max then max:=q[i];
x:=x-max;
if x+m -
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noname01.pas LV 7 @ 2009-04-04 16:02:51
一定要记住：
不能只找最高点
而要把每个人当作最高点试一遍！
交了两次
刚学DP，练练手 请高人给几道经典题 -
0
var a,b,c:array[1..100] of longint;
i,j,k,n,num,ml,mr,ii,jj:longint;procedure left(x2:longint);
begin
ml:=0;
for ii:=1 to x2-1 do b[ii]:=1;
for ii:=2 to x2-1 do
for jj:=1 to ii-1 do
begin
if (a[ii]a[jj]) and (b[jj]+1>b[ii]) then b[ii]:=b[jj]+1;
if b[ii]>ml then ml:=b[ii];
end;
end;procedure right(x3:longint);
begin
mr:=0;
for ii:=x3+2 to n do c[ii]:=1;
for ii:=x3+2 to n do
for jj:=x3+1 to ii-1 do
begin
if (a[ii]mr then mr:=c[ii];
end;
end;begin
readln(n);
for i:=1 to n do read(a[i]);
for i:=1 to n do
begin
left(i); right(i);
if ml+mr>num then num:=ml+mr;
end;
left(n);
writeln(n-num-1);
readln;
readln;
end. -
0
不知道怎么反过来~ 只好弄2个~
function dd(k:integer):integer;
var max,i,j:integer;
begin
max:=0;
for i:= 1 to k do f[i]:=1;
for i:= k downto 2 do
for j:= i downto 1 do
begin
if ( a[i] > a[j] ) and ( f[i] + 1 > f[j] ) then
f[j]:=f[i]+1;
if f[j] > max then max:=f[j];
end;
exit(max);
end;function bb(k:integer):integer;
var max,i,j:integer;
begin
max:=0;
for i:= k to n do f[i]:=1;
for i:= k to n-1 do
for j:= i to n do
begin
if ( a[i] > a[j] ) and ( f[i] + 1 > f[j] ) then
f[j]:=f[i]+1;
if f[j] > max then max:=f[j];
end;
exit(max);
end; -
0
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Accepted 有效得分：100 有效耗时：0ms -
0
lwq7153910 LV 7 @ 2009-03-15 13:00:51
#include
int main()
{
int a[100],i,j,t=0,s=0,p=0,q=0,number;
scanf("%d",&number);
for(i=0;i=0;j--)
{
if(a[i] -
0
var
a:array[0..101,1..3]of longint;
i,j,n,max,maxn:longint;
begin
readln(n);
a[0,1]:=-maxlongint;
a[n+1,1]:=-maxlongint;
for i:=1 to n do
begin
read(a);
a:=1;
a:=1;
end;
for i:=1 to n do
begin
max:=0;
for j:=i-1 downto 0 do
if(a[j,1]max)then max:=a[j,2];
inc(a,max);
end;
for i:=n downto 1 do
begin
max:=0;
for j:=i+1 to n+1 do
if(a[j,1]max)then max:=a[j,3];
inc(a,max);
end;
maxn:=0;
for i:=1 to n do
if a+a>maxn then maxn:=a+a-1;
writeln(n-maxn);
end. -
0
var
t,f1,f2:array[1..100]of byte;
i,j,n,max:integer;begin
readln(n);
for i:=1 to n do begin
read(t[i]);f1[i]:=1;f2[i]:=1;
end;
max:=0;
for i:=2 to n do
for j:=1 to i-1 do
begin
if (t[i]>t[j])and(f1[j]>=f1[i]) then f1[i]:=f1[j]+1;
if (t[n-i+1]>t[n-j+1])and(f2[n-j+1]>=f2[n-i+1]) then
f2[n-i+1]:=f2[n-j+1]+1;
end;
for i:=1 to n do if max -
0
chinese7004 LV 4 @ 2009-02-05 16:30:44
#include
using namespace std;
int a[10000],b[10000],c[10000],i,j,k,n;
int main()
{
cin>>n;
for (i=1;i>a[i];
for (i=1;ii;j--)
if (a[j]c[i])
c[i]=c[j]+1;
}
int maxn=1;
for (i=1;imaxn)
maxn=c[i]+b[i];
cout -
0
victorygod LV 5 @ 2009-02-04 19:32:23
#include
using namespace std;int p[102],a[102],b[102];
int l1[102],l2[102],n;int find(int x,int len)
{
int l,r;
l=1;r=len;
while (l=x) r=mid-1; else l=mid+1;
}
return l;
}main()
{
cin>>n;
for (int i=1;i>a[i];
memset(p,0,sizeof(p));l1[0]=0;p[1]=30000;
for (int i=1;il1[i]) {p[j+1]=30000;l1[i]+=1;}
p[j]=a[i];
}
memset(p,0,sizeof(p));l2[n+1]=0;p[1]=30000;
for (int i=n;i>=1;i--)
{
l2[i]=l2;
int j=find(a[i],l2[i]);
if (j>l2[i]) {p[j+1]=30000;l2[i]+=1;}
p[j]=a[i];
}
int m=0;
for (int i=1;im) m=l1[i]+l2[i]-1;
cout
wangyk
LV 6
@ 2009-05-10 00:06:09
