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MMA

鶴亀算

Before:

   1 #include <stdio.h>
   2 
   3 int main(void)
   4 {
   5   int num,leg,tsuru,kame;
   6   printf("鶴と亀の数の合計 ? ");
   7   scanf("%d",&num);
   8   printf("鶴と亀の足の本数の合計 ? ");
   9   scanf("%d",&leg);
  10   kame = (leg - 2 * num) / 2;
  11   tsuru = num - kame;
  12   printf("鶴と亀の数の合計 %d\n",num);
  13   printf("鶴と亀の足の本数の合計 %d\n",leg);
  14   printf("鶴 %d 羽, 亀 %d 匹\n",tsuru,kame);
  15 
  16   return 0;
  17 }

After:

   1 #define P;printf("%s%s "
   2 #define S;scanf("%d",
   3 s="鶴と亀の",y="足の本数の合計";main(n,l){P"? ",s,y+6)S&n)P"? ",s,y)S&l)P"%d\n%s%s %d\n鶴 %d 羽, 亀 %d 匹\n",s,y+6,n,s,y,l,n*2-l/2,l/2-n);}

(187B)

Revised (ytoku):

   1 #define P;printf("%s%s "
   2 #define S y);scanf("%d",
   3 s="鶴と亀の",y="足の本数の合計";main(n,l){P"? ",s,6+S&n)P"? ",s,S&l)P"%d\n",s,y+6,n)P"%d\n鶴 %d 羽, 亀 %d 匹\n",s,y,l,n*2-l/2,l/2-n);}

(185B)

PKU 1006

http://poj.org/problem?id=1006

Before:

   1 #include <stdio.h>
   2 
   3 #define C_PHY 23
   4 #define C_EMO 28
   5 #define C_INT 33
   6 
   7 int findpeak(int a, int ca, int b, int cb)
   8 {
   9     a -= b;
  10     while (a % cb != 0) {
  11         a += ca;
  12     }
  13     return a+b;
  14 }
  15 
  16 int main()
  17 {
  18     int n;
  19     for (n = 1; ; n++) {
  20         int p, e, i, d;
  21         scanf("%d %d %d %d", &p, &e, &i, &d);
  22         if (p == -1) break;
  23 
  24         int tmp;
  25         tmp = findpeak(e, C_EMO, i, C_INT);
  26         tmp = findpeak(tmp, C_EMO*C_INT, p, C_PHY);
  27 
  28         tmp -= d;
  29         if (tmp <= 0) tmp += C_EMO*C_INT*C_PHY;
  30         else if (tmp > C_EMO*C_INT*C_PHY) tmp -= C_EMO*C_INT*C_PHY;
  31 
  32         printf("Case %d: the next triple peak occurs in %d days.\n", n, tmp);
  33     }
  34     return 0;
  35 }

After:

   1 main(n,p,e,i,d){for(;scanf("%d%d%d%d",&p,&e,&i,&d),p+1;printf("Case %d: the next triple peak occurs in %d days.\n",n++,(i-d+21251)%21252+1))while((i-p)%23|(i-e)%28)i+=33;}

(171B)

四則演算と剰余

Before:

   1 #include <stdio.h>
   2 
   3 int main(void)
   4 {
   5     int a, b;
   6 
   7     printf("input two numbers\n");
   8     scanf("%d%d", &a, &b);
   9     printf("%d+%d=%d\n", a, b, a + b);
  10     printf("%d-%d=%d\n", a, b, a - b);
  11     printf("%d*%d=%d\n", a, b, a * b);
  12     if (b) {
  13         printf("%d/%d=%d\n", a, b, a / b);
  14         printf("%d%%%d=%d\n", a, b, a % b);
  15     }
  16 
  17     return 0;
  18 }

After:

   1 #define p(x) printf("%d%s%d=%d\n",a,#x,b,a x b);
   2 a;main(b){scanf("%d%d",&a,&b);p(+)p(-)p(*)b&&p(/)b&&p(%)}

(107B)

Revised (renda):

   1 #define p(x)printf("%d%s%d=%d\n",a,#x,b,a x b);
   2 a;main(b){scanf("%d%d",&a,&b);p(+)p(-)p(*)b&&p(/)b&&p(%)}

(106B)

Revised (ytoku):

   1 #define p(x)&&printf("%d%s%d=%d\n",a,#x,b,a x b)
   2 a;main(b){scanf("%d%d",&a,&b)p(+)p(-)p(*),b p(/)p(%);}

(104B)

ICPC2010 インターネット予選 A問題

Before:

あとで (1103B)

After:

   1 #define K scanf("%d",
   2 #define I(x)for(x=N;--x;)
   3 #define L(a,b,c)I(i)I(j)if(k[i][j])b=b>a?b:a,c=a<c?a:c;b-=c-1;
   4 N=404;main(c,i,j,l,e,f,g,h,n,d){int k[N][N],m[N],o[N];for(;c;){I(i)I(j)l=k[i][j]=m[j]=o[j]=0;e=g=m[0]=o[0]=(f=h=N)/2;K&c);if(k[e][e]=c){for(;--c;K&d),d/2?i=3-d,j=2-d:(i=d-1,j=d),d=m[n]+i,n=o[n]+j,k[d][n]=m[++l]=d,o[l]=n)K&n);L(i,e,f)L(j,g,h)c=printf("%d %d\n",e,g);}}}

(380B)

Revised (nomeaning):

   1 n,x[200],y[200],a,s,d,f,k,l,q[]={1,0,-1,0};main(i){for(;scanf("%d",&n),n;a=s=d=f=!printf("%d %d\n",s-a+1,f-d+1))for(i=0;++i<n;a=fmin(a,x[i]),s=fmax(s,x[i]),d=fmin(d,y[i]),f=fmax(f,y[i]))scanf("%d%d",&l,&k),x[i]=x[l]+q[k],y[i]=y[l]+q[k^1];}

(239B)

3文字の文字列の反転

Before:

   1 #include <stdio.h>
   2 
   3 int main(void)
   4 {
   5     char str[4], rts[4];
   6 
   7     printf("文字列を入力してください ");
   8     scanf("%3s", str);
   9 
  10     rts[0] = str[2];
  11     rts[1] = str[1];
  12     rts[2] = str[0];
  13     rts[3] = str[3];
  14 
  15     printf("%s %s %s %s \n ", str, rts, str, rts);
  16 
  17     return 0;
  18 }

(299B)

After:

   1 #define P(x);printf("%s ",x);
   2 #define S(i);r[i]=s[2-i]
   3 char str[4],rts[4],*s,*r=rts;main(){P("文字列を入力してください")s=gets(str)S(0)S(1)S(2)P(s)P(r)P(s)P(r)P("\n")}

(168B)

Revised (ytoku):

   1 #define P(x)printf("%s ",x);
   2 #define S(i)r[i]=s[2-i];
   3 char str[4],rts[4],*s,*r=rts;main(){P("文字列を入力してください")P(s=gets(str))S(0)S(1)S(2)P(r)P(s)P(r)P("\n")}

(166B)

Revised (ytoku on the dark side):

   1 #define P(x)printf("%s ",x);
   2 #define S r[i]=s[2-i++]
   3 char str[4],rts[4],*s,*r=rts;i;main(){P("文字列を入力してください")P(s=gets(str))S;S;S;P(r)P(s)P(r)P("\n")}

(161B)

Revised (ytoku):

   1 #define P(x)printf("%s ",x);
   2 char str[4],rts[4],*s,*r=rts+3;main(){P("文字列を入力してください")for(P(s=gets(str))*s;)*--r=*s++;P(r)P(str)P(r)P("\n")}

(151B)

Revised (nomeaning):

   1 char a[9],b[];i;main(){printf("文字列を入力してください ");for(gets(a);i<3;i++)b[i]=a[2-i];printf("%s %s %s %s \n ",a,b,a,b);}

(126B)


ツリーの表示

コマンドラインから整数n(n>=0を仮定してよい)を受け取り、n段のツリーを表示する。nが与えられなかったとき、およびn=0のときは何も表示しない。次に示す例はn=5のときである:

    *
   * *
  * * *
 * * * *
* * * * *

Before

   1 #include <stdio.h>
   2 #include <stdlib.h>
   3 
   4 int main(int argc, char **argv)
   5 {
   6     int i, j, n;
   7 
   8     if (argc > 1) {
   9         n = atoi(argv[1]);
  10         for (i = 0; i < n; ++i) {
  11             for (j = 0; j < n - i - 1; ++j)
  12                 putchar(' ');
  13             for (j = 0; j < i + 1; ++j)
  14                 printf(j == i ? "*\n" : "* ");
  15         }
  16     }
  17 
  18     return 0;
  19 }

(387B)

After (clear):

   1 i,n;main(int a,char**v){for(n=a>1?atoi(v[1]):0;i<n;++i){for(a=n-i-1;a--;)putchar(32);while(a++<i)printf(a==i?"*\n":"* ");}}

(124B)

Revised (renda):

   1 i,n;main(int a,char**v){for(n=a>1?atoi(v[1]):0;i<n;++i){for(a=n-i-1;a--;)putchar(32);while(a++<i)printf(a-i?"* ":"*\n");}}

(123B)

Revised (ytoku):

   1 i,n;main(int a,char**v){for(n=a>1?atoi(v[1]):0;i++<n;){for(a=n-i;a--;)putchar(32);while(++a<i)printf(a-i+1?"* ":"*\n");}}

(122B)

Revised (nomeaning):

   1 i;main(a,b){for(i=a=atoi(*((int*)b+a-1));i--;)for(b=0;b++<a;)printf(i<b?b-a?"* ":"*\n":" ");}

(93B)

2011年度センター試験 数学IIB 第6問 C移植版

Original

100 INPUT M
101 FOR N=1 TO M
110 LET I=N
120 LET C=0
130 IF I=1 THEN GOTO 210
140 IF INT(I/2)*2=I THEN
150     LET I=I/2
160     GOTO 190
170 END IF
180 LET I=3*I+1
190 LET C=C+1
200 GOTO 130
210 IF C<=10 THEN PRINT "F(";N;")=";C
211 NEXT N

Before

   1 #include <stdio.h>
   2 int main(){
   3         int i,m,n,c;
   4 
   5         scanf("%d",&m);
   6         for(i = 1;i <= m;i++){
   7                 c=0;
   8                 n=i;
   9                 while(n > 1){
  10                         if(n / 2 * 2 == n){
  11                                 n /= 2;
  12                         }else{
  13                                 n = n * 3 + 1;
  14                         }
  15                         c++;
  16                 }
  17                 if(c <= 10) printf("F(%d)=%d\n",i,c);
  18         }
  19         return 0;
  20 }

(257B)

After

   1 c,i,n;main(m){scanf("%d",&m);for(;++i<=m;c=0){for(n=i;n>1;c++)n=n&1?n*3+1:n/2;if(c<11)printf("F(%d)=%d\n",i,c);}}

(115B)

Revised (ytoku):

   1 c,i,n;main(m){scanf("%d",&m);for(;i++<m;c=0){for(n=i;n>1;c++)n=n&1?n*3+1:n/2;if(c<11)printf("F(%d)=%d\n",i,c);}}

(114B)

Revised (nomeaning):

   1 c,i;main(n,m){for(scanf("%d",&m);i++<m;c=0){for(n=i;n>1;c++)n=n&1?n*3+1:n/2;c<11&&printf("F(%d)=%d\n",i,c);}}

(109B)

Revised (nomeaning):

   1 c,i;main(n,m){for(scanf("%d",&m);i++<m;c=c<11&&!printf("F(%d)=%d\n",i,c))for(n=i;n>1;c++)n=n&1?n*3+1:n/2;}

(106B)

完全数を求める

問題

N以下の完全数を全て求め、昇順に出力せよ。

入力

N(≦10000)を含む1行で与えられる

出力

完全数を1行ごとに、降順に出力せよ。

入力例1

9999

出力例1

6
28
496
8128

Before

   1 #include <stdio.h>
   2 
   3 int main(void){
   4     int i, j, n, m;
   5 
   6     scanf("%d", &m);
   7 
   8     for(i = 2;i <= m;i++){
   9         n = 0;
  10         for(j = 1;j < i;j++){
  11             if(i%j == 0){
  12                 n += j;
  13             }
  14         }
  15         if(i == n){
  16             printf("%d\n", i);
  17         }
  18     }
  19     return 0;
  20 }

After (ytoku):

   1 j,n,m;main(i){for(scanf("%d",&m);i++<m;i-n||printf("%d\n",i))for(n=j=0;++j<i;i%j||(n+=j));}

(91B)

Revised (ytoku):

   1 j,n,m;main(i){for(scanf("%d",&m);i++<m;n||printf("%d\n",i))for(n=j=i;--j;n-=i%j?0:j);}

(86B)

After (nomeaning):

   1 m,j=6;main(i){for(scanf("%d",&m);j<=m&&printf("%d\n",j);j=(2<<i*2)-(1<<i))i=(i&6)+2;}

(85B)

Revised (ytoku):

   1 m,j=6;main(i){for(scanf("%d",&m);j<=m;printf("%d\n",j),j=2<<i*2,j-=1<<i)i=(i&6)+2;}

(83B)

Revised (nomeaning):

   1 m,j;main(i){for(scanf("%d",&m);j=2<<i*2,j-=1<<i,i;j>m||printf("%d\n",j))i=i+2&6;}

(81B)

Revised(nomeaning):

   1 m,j=6;main(i){for(scanf("%d",&m);j<=m;j-=1<<i++*2)printf("%d\n",j),j=2<<i*4;}

(77B)

Revised(nomeaning):

   1 i,j=7;main(m){for(scanf("%d",&m);j<=m;j=2<<++i*4)printf("%d\n",j-=1<<i*2);}

(75B)

Shortest Palindrome

問題

10文字以下の英大文字のみからなる文字列が与えられる。文字列の後ろになんらかの文字列を足して回文にするとき、最短となる文字列を答えよ。

入出例

入力

出力

ABA

ABA

APB

APBPA

ABAB

ABABA

Before

   1 #include <stdlib.h>
   2 #include <stdio.h>
   3 #include <string.h>
   4 
   5 int is_palindrome(const char *s){
   6   int i, j;
   7   for(i = 0, j = strlen(s) - 1; i < j; i++, j--){
   8     if(s[i] != s[j]){
   9       return 0;
  10     }
  11   }
  12   return 1;
  13 }
  14 
  15 int main(void) {
  16   char str[15];
  17   int i, len, j;
  18   scanf("%10s",str);
  19   
  20   for(len = strlen(str); len < 100; len++){
  21     char s[100];
  22     s[len] = '\0';
  23     for(i = len - 1, j = 0; str[j]; i--, j++){
  24       s[i] = str[j];
  25     }
  26     memcpy(s, str, strlen(str));
  27     if(is_palindrome(s)){
  28       puts(s);
  29       return EXIT_SUCCESS;
  30     }
  31   }
  32   return EXIT_FAILURE;
  33 }

ICPC 国内予選2013 C問題

After (ytoku): (Perl)

perl -MList::Util=sum -pe'INIT{<>}s/^,// while s%\[(,?)([,\d]+)]%@s=sort{$a-$b}split/,/,$2;",".($1?sum@s[0..@s/2]:$2+1>>1)%ge'

126bytes

After (nomeaning): (Ruby)

ruby -lpe'BEGIN{gets};$_.strip!while$_.gsub!(/\[( )?([ \d]+)\]/){" #{a=$2.split.map(&:to_i);$1?a.sort[0..a.size/2].inject(:+):a[0]/2+1}"}'

138bytes

After (nomeaning): (Ruby)

ruby -ne'def f(*a);a[1]?a.sort[0..a.size/2].inject(:+): a[0]/2+1;end;p eval$_.gsub("[","f(").gsub("]","),")[0..-3] if$.>1

122bytes

Revised (nomeaning): (Ruby)

ruby -ne'def f(*a);a[1]?a.sort[0..a.size/2].inject(:+):-~a[0]/2;end;p eval$_.gsub(?[,"f(").gsub(?],"),")[0..-3]if$.>1

118bytes

After (nomeaning): (Goruby)

goruby -ne'def f(*a)a[1]?a.so[0..a.sz/2].ij(:+):-~a[0]/2;end;p v$_.gs(?[,"f(").gs(?],"),")[0..-3]if$.>1'

102bytes

After (ytoku): (Perl)

perl -MList::Util=sum -lpe'INIT{<>}sub f{@_-1?sum@_[0..@_/2]:1+pop>>1}s/\[/f(sort{\$a-\$b}/g,s/]/),/g,$_=eval'

(110B)


CategoryContest

CodeGolf (最終更新日時 2013-07-24 21:33:54 更新者 ytoku)