";
print " 아래 점수는 참여한 모든 분에 대한 결과입니다.
수상자들 모두 축하드립니다. 수고하셨습니다. 다음에 준비중인 모의고사에도 참가해 주시기 바랍니다.
";
print "
범례 : o - Accepted, x - Wrong Answer, s - signal out(runtime error), t - Time Limit Exceeded
";
print "3회 나는 코더다(Eagle)!! 결과
";
print "Gold Medal : august14, gs10107, gs10087
";
print "Silver Medal : myungwoo, wclee2265, gs12065, zlzmsrhak56, teram
";
print "Bronze Medal : pl0892029, ryanch1, shinism, kk1401, commaster0, cjmp1
";
$dir="/var/www/bbs/moi/3st/";
$pnum = 4;
$files1 = scandir($dir);
$number=1;
$iuser = array();
$ip1 = array();
$ip2 = array();
$ip3 = array();
$ip4 = array();
$isc = array();
$rank = array();
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";
$id_name="";
for($ttt=0;$ttt<= strlen($files1[$i]) ;$ttt++)
if(1<=$ttt&&$ttt<=-2) $id_name.= "*";
else $id_name.=SUBSTR($files1[$i],$ttt,1);
$iuser[$number] = $id_name;
for( $j=1 ; $j <= $pnum ; $j++ )
{
$sz = 20;
$prob_result = $dir.$files1[$i].'/'.$files1[$i].'.res'.$j;
if(file_exists($prob_result)==true)
{
$score+=($j*0.001);
$fp = fopen($prob_result, "r");
$str = fgets($fp, filesize($prob_result)+1);
$resres = "";
for( $k=0; $k <= strlen($str) ; $k++ )
if(SUBSTR($str, $k, 1)== "o"){
$resres.=''.SUBSTR($str, $k, 1).'';
$score += 25+($j*0.001);
}
else if(SUBSTR($str, $k, 1)=="x")
$resres.=''.SUBSTR($str, $k, 1).'';
else if(SUBSTR($str, $k, 1)=="t")
$resres.=''.SUBSTR($str, $k, 1).'';
else if(SUBSTR($str, $k, 1)=="s")
$resres.=''.SUBSTR($str, $k, 1).'';
}
else $resres='Not submited!!';
if($j==1) $ip1[$number] = $resres;
else if($j==2) $ip2[$number] = $resres;
else if($j==3) $ip3[$number] = $resres;
else $ip4[$number] = $resres;
}
$isc[$number]=$score;
}
}
for($i=2;$i<=$number;$i++)
for($j=2;$j<=$number;$j++)
if( $isc[$i] < $isc[$j] ) $rank[$i]++;
for($i=2;$i<=$number-1;$i++)
for($j=$i+1;$j<=$number;$j++)
if($isc[$i] < $isc[$j])
{
$tttt=$iuser[$i];$iuser[$i]=$iuser[$j];$iuser[$j]=$tttt;
$tttt=$ip1[$i];$ip1[$i]=$ip1[$j];$ip1[$j]=$tttt;
$tttt=$ip2[$i];$ip2[$i]=$ip2[$j];$ip2[$j]=$tttt;
$tttt=$ip3[$i];$ip3[$i]=$ip3[$j];$ip3[$j]=$tttt;
$tttt=$ip4[$i];$ip4[$i]=$ip4[$j];$ip4[$j]=$tttt;
$ttttt=$isc[$i];$isc[$i]=$isc[$j];$isc[$j]=$ttttt;
$ttttt=$rank[$i];$rank[$i]=$rank[$j];$rank[$j]=$ttttt;
}
print 'Result for All Participant (동점일 경우에 4, 3, 2, 1번 문제의 점수 순으로 우선순위가 정해집니다.)
|
Rank
|
user
|
Number Card(250)
|
Commuting Road(250)
|
Direction Deaf(250)
|
Good Patten(250)
|
Score
|
';
for($i=2; $i <= $number ; $i++ )
{
if( $i%2 == 1 ) $bcolor = "ffffff";
else $bcolor = "ececff";
print ' | '.($rank[$i]+1).' | ';
print ' '.$iuser[$i].' | ';
print ' '.$ip1[$i].' | ';
print ' '.$ip2[$i].' | ';
print ' '.$ip3[$i].' | ';
print ' '.$ip4[$i].' | ';
print ''.(floor($isc[$i])).' | ';
print "
";
}
print "
";
print "이번 대회에 참가하신 ".($number)."분 모두 수고하셨습니다!! 다음 경연을 기대하세요~
";
print "앞으로 기회가 된다면 사은품도 준비예정 ^^;";
print "
Here is some User's perfect Solution!!
1. Number Card
CHICK 4번 해법을 참고하세요";
print "
2. Commuting Road (by commaster0)";
print '
#include<cstdio>
long long int a[102][102],b[102][102],c[102][102],d[102][102];
int main()
{
long long int w,h,i,j;
scanf("%lld%lld",&w,&h);
for(i=1;i<=100;i++)
{
b[1][i]=1;
d[i][1]=1;
}
for(i=4;i<=100;i++)
{
for(j=2;j<=i-2;j++)
{
a[j][i-j]=d[j][i-j-1]%100000;
b[j][i-j]=a[j][i-j-1]%100000+b[j][i-j-1]%100000;
c[j][i-j]=b[j-1][i-j]%100000;
d[j][i-j]=c[j-1][i-j]%100000+d[j-1][i-j]%100000;
}
}
for(j=2;j<=99;j++)
{
a[j][101-j]=d[j][101-j-1]%100000;
b[j][101-j]=a[j][101-j-1]%100000+b[j][101-j-1]%100000;
c[j][101-j]=b[j-1][101-j]%100000;
d[j][101-j]=c[j-1][101-j]%100000+d[j-1][101-j]%100000;
}
for(i=102;i<=200;i++)
{
for(j=i-100;j<=100;j++)
{
a[j][i-j]=d[j][i-j-1]%100000;
b[j][i-j]=a[j][i-j-1]+b[j][i-j-1]%100000;
c[j][i-j]=b[j-1][i-j]%100000;
d[j][i-j]=c[j-1][i-j]%100000+d[j-1][i-j]%100000;
}
}
printf("%lld",(a[w][h]+b[w][h]+c[w][h]+d[w][h])%100000);
return 0;
}
';
print "
Here is another solution 2. Commuting Road(by teram)";
print '
#include<stdio.h>
#define mod 100000
struct ROAD {
int left;
int up;
} road[110][110];
int main() {
int w,h;
int i,j;
scanf("%d %d", &w, &h);
for(i=1 ; i<=w ; i++)
road[1][i].left=1;
for(i=1 ; i<=h ; i++)
road[i][1].up=1;
for(i=2 ; i<=h ; i++) {
for(j=2 ; j<=w ; j++) {
road[i][j].left=road[i][j-1].left+road[i-1][j-1].up;
road[i][j].up=road[i-1][j].up+road[i-1][j-1].left;
road[i][j].left %= mod, road[i][j].up %= mod;
}
}
printf("%d\n", (road[h][w].left+road[h][w].up)%mod);
return 0;
}
';
print "
3, 4번 문항은 만점자가 없습니다. 사실 이 두 문항은 TLE제한 상 만점 받기 어려운 문제였습니다. TLE는 걸렸지만 가장 고득점 한 소스코드를 공개합니다. 시간만 충분하다면 모든 답을 구할 수 있는 solution들입니다.
";
print "
3. Direction Deaf(by wclee2265)";
print '
#include <stdio.h> //Dynamic Programming Solution
#define N 15
#define M 15
FILE *in=stdin;
FILE *out=stdout;
int n,m,map[N][M],ans,cnt,mat[30][30],dy[1200000][22];
struct data
{
int x,y;
}a[30];
void process()
{
int i,j,k,x;
for(i=1;i<(1<<cnt);i++)
{
for(j=1;j<=cnt;j++)
{
if((i>>(j-1))&1)
{
x=i^(1<<(j-1));
if(x==0)
{
if(mat[0][j]!=-1) dy[i][j]=1;
}
else
{
for(k=1;k<=cnt;k++)
{
if((x>>(k-1))&1 && mat[j][k]!=-1 && (mat[j][k]&x)==0)
dy[i][j]+=dy[x][k];
}
}
}
}
}
for(i=1;i<=cnt;i++)
{
if(mat[i][0]==0)
ans+=dy[(1<<cnt)-1][i];
}
}
bool check_in(int l, int r, int t)
{
if(l<r)
{
if(l<t && t<r) return true;
return false;
}
if(r<t && t<l) return true;
return false;
}
void get_house()
{
int i,j,k;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(map[i][j]==1)
{
cnt++;
a[cnt].y=i;
a[cnt].x=j;
}
else if(map[i][j]==2)
{
a[0].y=i;
a[0].x=j;
}
}
}
for(i=0;i<=cnt;i++)
{
for(j=i+1;j<=cnt;j++)
{
if(a[i].x!=a[j].x && a[i].y!=a[j].y) mat[i][j]=mat[j][i]=-1;
if(a[i].x==a[j].x)
{
for(k=1;k<=cnt;k++)
{
if(a[k].x==a[i].x && check_in(a[i].y,a[j].y,a[k].y))
{
mat[i][j]|=(1<<(k-1));
mat[j][i]|=(1<<(k-1));
}
}
}
else
{
for(k=1;k<=cnt;k++)
{
if(a[k].y==a[i].y && check_in(a[i].x,a[j].x,a[k].x))
{
mat[i][j]|=(1<<(k-1));
mat[j][i]|=(1<<(k-1));
}
}
}
}
}
}
int main()
{
int i,j;
fscanf(in,"%d %d",&m,&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++) fscanf(in,"%d",&map[i][j]);
}
get_house();
process();
fprintf(out,"%d",ans);
fclose(in);
fclose(out);
return 0;
}
';
print "
3. Here is another solution Direction Deaf(by gs12065)";
print '
#include <stdio.h>
#include <vector>
struct pos{
int x,y;
};
int n, m;
int map[12][12], numHouse, res;
bool chk[12][12];
pos church;
std::vector < pos > dir[5][12][12];
void fly(int x, int y, int cnt){
if(numHouse == cnt){
if(x == church.x || y == church.y) res++;
return;
}
int i, j;
for(i=0; i<4; i++){
for(j=0; j<dir[i][y][x].size(); j++){
if(!chk[dir[i][y][x][j].y][dir[i][y][x][j].x]){
chk[dir[i][y][x][j].y][dir[i][y][x][j].x] = 1;
fly(dir[i][y][x][j].x, dir[i][y][x][j].y, cnt+1);
chk[dir[i][y][x][j].y][dir[i][y][x][j].x] = 0;
break;
}
}
}
}
int main(){
scanf("%d%d", &m, &n);
int i, j, k;
for(i=1; i<=n; i++){
for(j=1; j<=m; j++){
scanf("%d", &map[i][j]);
if(map[i][j] == 2) church.x = j, church.y = i;
else if(map[i][j] == 1) numHouse++;
}
}
pos tmp;
for(i=1; i<=n; i++){
tmp.y = i;
for(j=1; j<=m; j++){
if(map[i][j] == 1){
tmp.x = j;
for(k=j-1; k>=1; k--)
if(map[i][k] > 0) dir[0][i][k].push_back(tmp);
}
}
for(j=m; j>=1; j--){
if(map[i][j] == 1){
tmp.x = j;
for(k=j+1; k<=m; k++)
if(map[i][k] > 0) dir[1][i][k].push_back(tmp);
}
}
}
for(j=1; j<=m; j++){
tmp.x = j;
for(i=1; i<=n; i++){
if(map[i][j] == 1){
tmp.y = i;
for(k=i-1; k>=1; k--)
if(map[k][j] > 0) dir[2][k][j].push_back(tmp);
}
}
for(i=n; i>=1; i--){
if(map[i][j] == 1){
tmp.y = i;
for(k=i+1; k<=n; k++)
if(map[k][j] > 0) dir[3][k][j].push_back(tmp);
}
}
}
fly(church.x, church.y, 0);
printf("%d\n", res);
return 0;
}
';
print "
4. Good Pattern(by august14)";
print '
#include <stdio.h> // Best solution for Good Pattern By Winner!!
// using mem cutting 2^21*2^21 -> 2*2^21
const int mod = 100000;
int N,M,C; char S[22][22];
int D[2][3][1<<21],f;
void sum(int& a, int b){a = (a + b) % mod;}
int main()
{
int i,j,k,c;
scanf ("%d %d",&N,&M); C = 1;
for (i=0;i<N;i++){
scanf ("%s",S[i]);
for (j=0;j<M;j++){
if (S[i][j] == '?') C = (C * 3) % mod;
}
}
D[1][0][0] = 1;
for (i=0;i<N;i++) for (j=0;j<M;j++){
for (k=0;k<(1<<M);k++) D[f][0][k] = D[f][1][k] = D[f][2][k] = 0;
for (k=0;k<(1<<M);k++){
c = (D[!f][0][k] + D[!f][1][k] + D[!f][2][k]) % mod;
if (S[i][j] == 'K' || S[i][j] == '?') sum(D[f][0][k>>1],c);
if ((S[i][j] == 'I' || S[i][j] == '?') && (k & 1) == 0) sum(D[f][2][k>>1],c);
if (j == 0){
if (S[i][j] == 'O' || S[i][j] == '?') sum(D[f][1][k>>1],c);
}
else{
if(S[i][j] == 'O' || S[i][j] == '?'){
sum(D[f][1][k>>1],D[!f][1][k]+D[!f][2][k]);
sum(D[f][1][(k>>1)+(1<<(M-2))],D[!f][0][k]);
}
}
}
f = !f;
}
c = 0;
for (k=0;k<(1<<M);k++){
c = (c + D[!f][0][k] + D[!f][1][k] + D[!f][2][k]) % mod;
}
printf ("%d",(C-c+mod)%mod);
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 3, 4번은 만점자가 없습니다. 게시된 솔루션의 주인이 설명을 Discuss Board에 작성해 주시면 등록해 드립니다.";
print '';
?>