303 条题解
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刀刀 LV 6 @ 2012-08-01 14:39:17
#include
using namespace std;
int main()
{
int n,i,j,max,kk;
int a[100],h[100],f[100];cin>>n;
for(i=0;i>a[i];
f[i]=1;
h[i]=1;
}for(i=1;i
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编译通过...
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├ 测试数据 10：答案正确... 0ms
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Accepted 有效得分：100 有效耗时：0ms点击查看代码
动态规划——最长上升子序列和最长下降子序列
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编译通过...
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Accepted 有效得分：100 有效耗时：0ms程序代码：
program p1098;
var
a,b,c:array[1..100] of integer;
i,j,n,ans:integer;
begin
readln(n);
for i:=1 to n do read(a[i]);
fillchar(b,sizeof(b),0);
fillchar(c,sizeof(c),0);
for i:=1 to n do
begin
b[i]:=1;
for j:=1 to i-1 do
if (a[i]>a[j]) and (b[j]+1>b[i]) then b[i]:=b[j]+1;
end;
for i:=n downto 1 do
begin
c[i]:=1;
for j:=i+1 to n do
if (a[i]>a[j]) and (c[j]+1>c[i]) then c[i]:=c[j]+1;
end;
ans:=0;
for i:=1 to n do
if (b[i]+c[i]>ans) then ans:=b[i]+c[i];
ans:=n-ans+1;
write(ans);
end. -
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program chorus;
var
a:array[0..100]of integer;
n,i,j,k:integer;
l,r:array[0..100]of integer;
begin
assign(input,'chorus.in');
assign(output,'chorus.out');
reset(input);
rewrite(output);
readln(n);
l[1]:=1;
r[n]:=1;
for i:=1 to n do
read(a[i]);
for i:=2 to n do
for j:=n-1 downto 0 do
begin
if (a[j]l[i])
then l[i]:=l[j]+1;
end;
for i:=n downto 1 do
for j:=i+1 to n+1 do
begin
if (a[j]r[i])
then r[i]:=r[j]+1;
end;
for i:=1 to n do
begin
k:=l[i]+r[i];
if l[i]+r[i]>k
then k:=l[i]+r[i];
end;
writeln(n-k+1);
close(input);
close(output);
end. -
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编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
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├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
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Accepted 有效得分：100 有效耗时：0ms
program p1098;{hechangdui}
var n:2..100;i,j:1..100;c:array[1..100,1..3] of 0..230;
begin
readln(n);
for i:=1 to n do begin read(c);c:=0;c:=0 end;
for i:=2 to n do
for j:=1 to i-1 do
if (c[j,1]=c) then c:=c[j,2]+1;
for i:=n-1 downto 1 do
for j:=n downto i+1 do
if (c[j,1]=c) then c:=c[j,3]+1;
j:=c[1,2]+c[1,3];
for i:=2 to n do
if j -
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majiamajiaco LV 6 @ 2009-11-10 10:31:47
#include
int main()
{
int n,i,j,a[110],f[110]={0},g[110]={0},max;
scanf("%d",&n);
for(i=1;i -
0
AC的第一道DP题
program P1098;
var a,t,b:array[0..100] of integer;
i,j,n,total:integer;
begin
readln(n);
for i:=1 to n do
read(t[i]);
for i:=1 to n do begin
a[i]:=1;
for j:=0 to i-1 do
if (t[i]>t[j])and(a[i]t[j])and(b[i]total then total:=a[i]+b[i];
writeln(n-total+1);
end. -
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mingyuan1240 LV 5 @ 2009-11-09 20:12:15
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
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Accepted 有效得分：100 有效耗时：0ms
怎么全是pascal？，来个c的吧！
int n,t[101]={0},min[101],max[101];
void Init()
{
int i;
scanf("%d",&n);
for(i=1;ik)
k=max[j];
max[i]+=k;
}
}
int DP_min()
{
int i,j,k;
for(i=1;i=1;i--)
{
k=0;
for(j=i+1;jt[j]&&min[j]>k)
k=min[j];
min[i]+=k;
}
}
void process()
{
int m=0,i;
for(i=1;im)
m=min[i]+max[i];
printf("%d",n-m+1);
}
int main()
{
Init();
DP_max();
DP_min();
process();
return 0;
} -
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同样的方法，都是正向反向各求一次最长不降序列，但求法略有不同
因为粗心“n-i+1”写成了“i-n+1”
悲剧~~~program p1098;
var n,i,j,x:integer;
t,h,a,b:array[1..100] of integer;
f1,f2:text;
begin
assign(f1,'in.txt');
assign(f2,'out.txt');
reset(f1);
rewrite(F2);
readln(f1,n);
for i:=1 to n do begin read(f1,t[i]); a[i]:=1; b[i]:=1; end;
for i:=n downto 1 do h[n-i+1]:=t[i];
for i:=2 to n do
for j:=1 to i-1 do
if (t[i]>t[j]) and (a[j]+1>a[i]) then a[i]:=a[j]+1;
for i:=2 to n do
for j:=1 to i-1 do
if (h[i]>h[j]) and (b[j]+1>b[i]) then b[i]:=b[j]+1;
x:=n-b[n];
for i:=2 to n do
if (n-a[i]-b[n-i+1]+1) -
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编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
++++++++++++++++++++++++++++++++++++++++++++++++++++++++
TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Accepted 有效得分：100 有效耗时：0ms+++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++++++++
TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
++++++++++++++++++++++++++++++++++++++++++++++++++++++++program ex;
var n,i,j,max:longint;
a,b,c:array[1..100] of longint;
begin
readln(n);
for i:=1 to n do read(a[i]);
for i:=1 to n do
begin
b[i]:=1;
for j:=1 to i-1 do
if (a[i]>a[j])and(b[j]+1>b[i]) then b[i]:=b[j]+1;
end;
for i:=n downto 1 do
begin
c[i]:=1;
for j:=n downto i+1 do
if (a[i]>a[j])and(c[j]+1>c[i]) then c[i]:=c[j]+1;
end;
for i:=1 to n do
if b[i]+c[i]>max then max:=b[i]+c[i];
writeln(n-max+1);
end. -
0
刀刀
LV 6
@ 2012-08-01 14:39:17
