";
print " 아래 점수는 참여한 모든 분에 대한 결과입니다.
다음에 준비중인 모의고사에도 참가해 주시기 바랍니다.
";
print "
범례 : o - Accepted, x - Wrong Answer, s - signal out(runtime error), t - Time Limit Exceeded
";
print "2회 나는 코더다!! 최종결과
";
print "우승(1st place Gold) : dijkstra(715pts)
";
print "준우승(2nd place Silver) : ryanch1(560pts)
";
print "3등상(3rd place Bronze) : zlzmsrhak56(475pts), gs12059(460pts), makesource(460pts), gs12065(455pts), vega4792(390pts)
";
$dir="/var/www/bbs/moi/2nd/";
$pnum = 4;
$files1 = scandir($dir);
print 'Result for All Participant
|
user
|
Sum(200)
|
Ship Travelling(200)
|
Cutting Confectionery(300)
|
Comeback vault 101(300)
|
Score
|
';
$number=1;
for( $i=0 ; $files1[$i] != NULL ; $i++ )
{
if($files1[$i]!="admin" && $files1[$i]!="." && $files1[$i]!=".." && is_dir($files1[$i])==true)
{
$number++;
$score=0;
if( $number%2 == 1 ) $bcolor = "ffffff";
else $bcolor = "ececff";
print '
| '.$files1[$i].' | ';
for( $j=1 ; $j <= $pnum ; $j++ )
{
if( $j>2 ) $sz = 30;
else $sz = 15;
print ' ';
$prob_result = $dir.$files1[$i].'/'.$files1[$i].'.res'.$j;
if(file_exists($prob_result)==true)
{
$fp = fopen($prob_result, "r");
$str = fgets($fp, filesize($prob_result)+1);
//print $str."";
for( $k=0; $k <= strlen($str) ; $k++ )
if(SUBSTR($str, $k, 1)== "o"){
print ''.SUBSTR($str, $k, 1).'';
switch($j)
{
case 1: case 2: $score = $score + 20; break;
case 3: case 4: $score = $score + 15; break;
}
}
else if(SUBSTR($str, $k, 1)=="x")
print ''.SUBSTR($str, $k, 1).'';
else if(SUBSTR($str, $k, 1)=="t")
print ''.SUBSTR($str, $k, 1).'';
else if(SUBSTR($str, $k, 1)=="s")
print ''.SUBSTR($str, $k, 1).'';
}
else print 'Not submited!!';
print ' | ';
}
print ''.$score.' | ';
print "
";
}
}
print "
";
print "이번 대회에 참가하신 ".$number."분 모두 수고하셨습니다!! 다음 경연을 기대하세요~ (1,2주 내에 실시 예정(공지참조))
";
print "앞으로 기회가 된다면 사은품도 준비예정 ^^;";
print "
Here is some User's perfect Solution!!
1. sum (by vega4792)";
print '
#include <stdio.h> // this solution is dynamic programming using memory cutting!
int n, i, front, cnt, sum;
main()
{
scanf("%d",&n);
for(i=1;i<=n;i++)
{
sum+=i;
while(sum>n)
sum-= ++front;
if(sum==n) cnt++;
}
printf("%d",cnt);
}
';
print "
Here is another solution 1. sum (by k5888200)";
print '
#include <stdio.h> // This is methmatical skill
long long N;
int count;
int main(){
long long k;
scanf("%lld",&N);
for(k=1;k<=N;k++){
if( (2*N-k*k+k)%(2*k) ==0 && (2*N-k*k+k)/(2*k)>=1) count++;
}
printf("%d",count);
return 0;
}
';
print "
Here is solution 2. Ship Travelling(by zlzmsrhak56)";
print '
#include<stdio.h> // This is SCC(Strongly Connected Component). Try seach SCC (using googling!!)
#include<vector>
using namespace std;
vector<int> way[10005];
vector<int> bay[10005];
vector<int> nay[10005];
vector<int> stack;
int num[10005],ad[10005],dy[10005],n,m,st,en,cnt=0;
bool check[10005],sch,ech;
void input()
{
int i,a,b;
scanf("%d%d%d%d",&n,&m,&st,&en);
for(i=0;i<m;i++){
scanf("%d%d",&a,&b);
way[a].push_back(b);
bay[b].push_back(a);
}
}
void DFS(int x)
{
int i;
for(i=0;i<way[x].size();i++){
if(check[way[x][i]])continue;
check[way[x][i]]=true;
DFS(way[x][i]);
}
stack.push_back(x);
}
void nDFS(int x)
{
if(x==st&&!sch){ st=cnt; sch=true; }
if(x==en&&!ech){ en=cnt; ech=true; }
int i;
for(i=0;i<bay[x].size();i++){
if(ad[bay[x][i]]!=0&&ad[bay[x][i]]!=cnt&&(nay[ad[bay[x][i]]].empty()||nay[ad[bay[x][i]]].back()!=cnt)){
nay[ad[bay[x][i]]].push_back(cnt);
}
if(check[bay[x][i]])continue;
check[bay[x][i]]=true;
nDFS(bay[x][i]);
}
num[cnt]++;
ad[x]=cnt;
}
void scc()
{
int i;
for(i=1;i<=n;i++){
if(check[i])continue;
check[i]=true;
DFS(i);
}
for(i=1;i<=n;i++)check[i]=false;
for(i=stack.size()-1;i>=0;i--){
if(check[stack[i]])continue;
cnt++;
check[stack[i]]=true;
nDFS(stack[i]);
}
}
void solve()
{
int i,j;
dy[st]=num[st];
for(i=st;i<=en;i++){
for(j=0;j<nay[i].size();j++){
if(dy[nay[i][j]]<num[nay[i][j]]+dy[i])dy[nay[i][j]]=num[nay[i][j]]+dy[i];
}
}
printf("%d",dy[en]);
}
int main()
{
input();
scc();
solve();
return 0;
}
';
print "
Here is solution 4. Comeback vault 101(by dijkstra)";
print '
#include<stdio.h> // This is Dynamic Programming (named Bitonic Tour) best solution!!
#define max(a,b) (a>b?(a):(b))
int d[220][110][110],n,m;
int s[110][110];
main()
{
int i,j,k;
scanf("%d%d",&m,&n);
for(i=1;i<=n;i++) for(j=1;j<=m;j++) scanf("%1d",&s[i][j]);
for(i=1;i<=n;i++) for(j=1;j<=m;j++) s[i][j]=s[i][j]==1?1:s[i][j]==0?0:-1;
for(k=2,d[1][1][1]=s[1][1]+1;k<n+m;k++) for(i=1;i<=n;i++)
{
if(s[i][k-i+1]==-1) continue;
for(j=i;j<=n;j++)
{
int score=0;
if(i!=j) score = s[i][k-i+1]+s[j][k-j+1];
else score = s[i][k-i+1];
if(s[j][k-j+1]==-1) continue;
d[k][i][j] = max(d[k-1][i-1][j],max(d[k-1][i][j-1],max(d[k-1][i][j],d[k-1][i-1][j-1])));
d[k][i][j] = d[k][i][j]?d[k][i][j]+score:0;
}
}
printf("%d\n",d[n+m-1][n][n]?d[n+m-1][n][n]-1:0);
}
';
print "위 해법을 분석하여 열심히 공부합시다. 이번에 3번은 만점자가 없습니다. 문제 등록되면 도전하세요!!";
?>