C++博客-powerwater-最新评论http://www.cppblog.com/powerwater/CommentsRSS.aspxzh-cnMon, 02 Sep 2024 09:59:42 GMTMon, 02 Sep 2024 09:59:42 GMTcnblogsre: NOIP1999第1题 拦截导弹http://www.cppblog.com/powerwater/archive/2016/05/17/200197.html#213531卢卢Tue, 17 May 2016 03:54:00 GMThttp://www.cppblog.com/powerwater/archive/2016/05/17/200197.html#213531#include<sstream>
#include<algorithm>
using namespace std;
const int N(1000);
int h[N],f[N],g[N];
int main()
{
int n(0),x;
string s;
getline(cin,s);
istringstream is(s);
while (is>>x) h[++n]=x;
fill(f+1,f+n+1,1);
fill(g+1,g+n+1,1);
for (int i=1;i<=n;i++)
for (int j=1;j<i;j++)
{
if (h[j]>=h[i])
f[i]=max(f[j]+1,f[i]);
if (h[j]<h[i])
g[i]=max(g[j]+1,g[i]);
}
int ans1(1),ans2(1);
for (int i=1;i<=n;i++)
ans1=max(ans1,f[i]),ans2=max(ans2,g[i]);
cout<<ans1<<" "<<ans2<<endl;
system("pause");
return 0;
}

卢 2016-05-17 11:54 发表评论
]]>re: NOIP2005提高组第2题 过河http://www.cppblog.com/powerwater/archive/2016/05/14/187391.html#213514高飞高飞Sat, 14 May 2016 00:49:00 GMThttp://www.cppblog.com/powerwater/archive/2016/05/14/187391.html#213514

高飞 2016-05-14 08:49 发表评论
]]>re: NOIP2005提高组第2题 过河http://www.cppblog.com/powerwater/archive/2016/05/14/187391.html#213513王靖琰王靖琰Sat, 14 May 2016 00:49:00 GMThttp://www.cppblog.com/powerwater/archive/2016/05/14/187391.html#213513

王靖琰 2016-05-14 08:49 发表评论
]]>re: NOIP1999第1题 拦截导弹http://www.cppblog.com/powerwater/archive/2016/05/10/200197.html#213483小犇小犇Tue, 10 May 2016 03:44:00 GMThttp://www.cppblog.com/powerwater/archive/2016/05/10/200197.html#213483发的题目无大问题；
有没有关于高精的题目

小犇 2016-05-10 11:44 发表评论
]]>re: NOIP2012 普及组 第3题 摆花[未登录]http://www.cppblog.com/powerwater/archive/2015/11/05/195733.html#212196yyyyThu, 05 Nov 2015 11:54:00 GMThttp://www.cppblog.com/powerwater/archive/2015/11/05/195733.html#212196#include<cmath>
#include<cstdio>
#include<fstream>
using namespace std;
int a[110],f[110][110]={0};
int main()
{
ifstream cin("flower.in");
ofstream cout("flower.out");
int n,m;
cin>>n>>m;
for (int i=1;i<=n;i++) cin>>a[i];
for (int i=0;i<=a[1];i++) f[1][i]=1;
for (int i=2;i<=n;i++)
{
f[i][0]=1;
for (int j=1;j<=m;j++)
for (int k=0;k<=a[i];k++)
if (j-k>=0)
f[i][j]=(f[i][j]+f[i-1][j-k])%1000007;
}
cout<<f[n][m]<<endl;
return 0;
}

yy 2015-11-05 19:54 发表评论
]]>