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¢£AtCoder Beginner Contest 138 A - Red or Not

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AtCoder Beginner Contest 138 A - Red or Not

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  1.   char s[11];
  2.   scanf("%s", s);

 

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  1.   printf("%s¥n", s);

 

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  1.   printf("%c¥n", s[2]);

 

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  1. #include <stdio.h>
  2. int main()
  3. {
  4.   int a;
  5.   char s[11];
  6.   scanf("%d %s", &a, s);
  7.   if (3200<=a) {
  8.     printf("%s¥n", s);
  9.   } else {
  10.     printf("red¥n");
  11.   }
  12.   return 0;
  13. }

 

 

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¢£¶¥µ»¥×¥í¥°¥é¥ß¥ó¥° AtCoder Beginner Contest 133 D ÌäÂê ¹ç³Ê¤Î´¬

AtCoder Beginner Contest 133 D ÌäÂê  Rain Flows into Dams ¤ò²ò¤¤¤Æ¤ß¤Þ¤·¤¿¤¬¡¢TLE (¼Â¹Ô»þ´ÖÀ©¸ÂĶ²á) ¤È¤Ê¤Ã¤ÆÉÔ¹ç³Ê¤Ç¤·¤¿¡£

 

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´×ÏõÙÂê

 

·ë²Ì¤¬ TLE ¤Ë¤Ê¤Ã¤¿¤È¤¤¤¦¤³¤È¤Ï·×»»Î̤¬Â¿¤¹¤®¤ë¤Ã¤Æ¤³¤È¡£²ò·èºö¤òµá¤á¤Æ²òÀâ¤ò¤ß¤Æ¤ß¤ë¤³¤È¤Ë¤·¤Þ¤¹¡£¤ó¡Á¤È… ¤½¤¦¤«¡¢Á²²½¼°¤ò¤½¤Î¤Þ¤Þ½çÈ֤˷׻»¤·¤Æ¤¤¤±¤Ð O(n) ¤Ë¤Ç¤­¤ë¤ó¤À¡£¤Ê¤ë¤Û¤É (^_^;)

 

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Á°²ó¤ÈƱ¤¸¤¯»³¤Ï 5 ¸Ä¤È¤·¤Þ¤¹¡£¹ß±«Î̤ÎÁíÎ̤ò S ¤È¤¹¤ë¤È¡¢

 

S = A1 + A2 + A3 + A4 + A5 = X1 + X2 + X3 + X4 + X5

X1 = S - ( X2 + X3 + X4 + X5 )

 

X2 + X3 = 2 × A2 ¡¢X4 + X5 = 2 × A4¡¡¤Ç¤¹¤«¤é¡¢

 

X1 = S - 2 × ( A2 + A4 )

 

¤³¤ì¤Ç¤Þ¤º X1 ¤¬µá¤Þ¤ê¤Þ¤·¤¿¡£¤¢¤È¤Ï¡¢

 

X2 = 2 × A1 - X1

X3 = 2 × A2 - X2

X4 = 2 × A3 - X3

X5 = 2 × A4 - X4

 

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/*  AtCoder Beginner Contest 133 D - Rain Flows into Dams
    2019.07.28 by meyon */

 

#include <stdio.h>
#include <stdlib.h>

 

void err()
{
  printf("Error!¥n");
  exit(1);
}

 

int main()
{
  // input N
  int N;
  scanf("%d", &N);
  if(3>N || 99999<N || 0==N%2) {
    err();
  }

 

  // input A
  int A[N];
  for(int i=0; i<N; i++) {
    scanf("%d", &A[i]);
    if(0>A[i] || 1000000000<A[i]) {
      err();
    }
  }

 

  // rainfall X
  int X[N];
  for(int i=0; i<N; i++) {
    X[i]=0;
  }

 

  int S=0;
  for(int i=0; i<N; i++) {
    S += A[i];
  }
  int P=0;
  for(int i=1; i<N; i+=2) {
    P += A[i];
  }
  X[0]=S-2*P;   // X1=S-2(A2+A4+...+A[n-1])

 

  for(int i=0; i<N-1; i++) {
    X[i+1]=2*A[i]-X[i];   // X[i+1]=2Ai-Xi

  }

 

  // display answer
  for(int i=0; i<N; i++) {
    printf("%d ", X[i]);
  }
  printf("¥n");

  return 0;
}

 

ÆþÎϤ·¤Æ¡¢Èϰϳ°¤Î¾ì¹ç¤Ï¥¨¥é¡¼½ªÎ»¡£¤³¤³¤ÏËèÅÙƱ¤¸¤Ç¤¹¡£

²þÁ±¤·¤¿ rainfall X ¤ÎÉôʬ¡£

 

  1. ÇÛÎóÊÑ¿ô X[N] ¤òÀë¸À¤·¤Æ½é´ü²½
  2. Áí¹ß±«ÎÌ S = A1 + A2 + ... + A5 ¤ò·×»»
  3. X1 ¤Î»»½Ð¤Î¤¿¤á¤Ë P = (A2 + A4) ¤ò·×»»
  4. X1 = S - 2 × P ¤Ç X1 ¤òµá¤á¤ë
  5. X2 = 2 × A1 - X1 ¤òµá¤á¡¢·«ÊÖ¤· X3¡¢X4¡¢X5 ¤ò»»½Ð

 

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¢£¶¥µ»¥×¥í¥°¥é¥ß¥ó¥° AtCoder Beginner Contest 133 D ÌäÂê ÉÔ¹ç³Ê¤Î´¬

ÌÌÇò¤¬¤Ã¤Æ²áµîÌäÂê¤ò²ò¤¤¤Æ¤¤¤ë¤¦¤Á¤Ë¡¢¤Ê¤ó¤È D ÌäÂê¤Þ¤Ç¤­¤Æ¤·¤Þ¤¤¤Þ¤·¤¿¡£

AtCoder Beginner Contest 133 D ÌäÂê ¤Ï Rain Flows into Dams ¤Ç¤¹¡£

 

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±ß·Á¤Ë  N ¸Ä¤Î»³¤¬Ï¢¤Ê¤Ã¤Æ¤ª¤ê¡¢»þ·×²ó¤ê¤Ë»³  1 , »³  2 …… , »³  N  ¤È¸Æ¤Ð¤ì¤Þ¤¹¡£ N ¤Ï´ñ¿ô¤Ç¤¹¡£

¤³¤ì¤é¤Î»³¤Î´Ö¤Ë  N  ¸Ä¤Î¥À¥à¤¬¤¢¤ê¡¢¥À¥à  1 , ¥À¥à  2 …… , ¥À¥à  N  ¤È¸Æ¤Ð¤ì¤Þ¤¹¡£¥À¥à  i  ( 1≤i≤N ) ¤Ï»³  i  ¤È»³  i+1  ¤Î´Ö¤Ë¤¢¤ê¤Þ¤¹ (»³  N+1  ¤Ï»³  1  ¤Î¤³¤È¤ò»Ø¤·¤Þ¤¹)¡£

»³  i  ( 1≤i≤N ) ¤Ë  2x  ¥ê¥Ã¥È¥ë¤Î±«¤¬¹ß¤ë¤È¡¢¥À¥à  i−1  ¤È¥À¥à  i ¤Ë¤½¤ì¤¾¤ì  x ¥ê¥Ã¥È¥ë¤º¤Ä¿å¤¬Î¯¤Þ¤ê¤Þ¤¹ (¥À¥à  0 ¤Ï¥À¥à  N ¤Î¤³¤È¤ò»Ø¤·¤Þ¤¹)¡£

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¤½¤Î·ë²Ì¡¢¥À¥à  i ( 1≤i≤N ) ¤Ë¤Ï¹ç·×¤Ç  Ai ¥ê¥Ã¥È¥ë¤Î¿å¤¬Î¯¤Þ¤ê¤Þ¤·¤¿¡£

³Æ»³¤Ë¹ß¤Ã¤¿±«¤ÎÎ̤òµá¤á¤Æ¤¯¤À¤µ¤¤¡£¤³¤ÎÌäÂê¤ÎÀ©Ì󲼤Ǥϲò¤¬°ì°Õ¤ËÄê¤Þ¤ë¤³¤È¤¬¾ÚÌÀ¤Ç¤­¤Þ¤¹¡£

 

¤µ¤ÆƬ¤¬Äˤ¯¤Ê¤ê¤½¤¦¤Ç¤¹¤¬ (^_^;) ¤Þ¤º¤ÏÌäÂê¤òÍý²ò¤·¤Ê¤±¤ì¤Ð¤Ê¤ê¤Þ¤»¤ó¡£»æ¤Ë»³¤È¥À¥à (ë) ¤òÉÁ¤¤¤Æ¹Í¤¨¤Æ¤ß¤Þ¤·¤ç¤¦¡£

»³¤¬ 3 ¸Ä¤¢¤ë¤È¤·¤Þ¤¹¡£¥À¥à 1 ¤Îº¸¤¬»³ 1 ¡¢±¦¤¬»³ 2 ¤Ç¤¹¡£¥À¥à 1 ¤Ëί¤Þ¤Ã¤¿¿å A1 ¤Ï¡¢»³ 1 ¤È»³ 2 ¤Ë¹ß¤Ã¤¿±« X1¡¢X2 ¤ÎȾʬ¤Ç¤¹¡£

 

A1 = ( X1 + X2 ) ÷ 2

 

¤·¤¿¤¬¤Ã¤Æ»³ 1 ¤Ë¹ß¤Ã¤¿±« X1 ¤Ï¡¢

 

X1 = 2 × A1 - X2

 

¤È¤Ê¤ê¤Þ¤¹¡£Æ±Íͤˤ½¤ì¤¾¤ì¤Î»³¤Ë¹ß¤Ã¤¿±«¤Ï¡¢

 

X2 = 2 × A2 - X3

X3 = 2 × A3 - X1

 

¤Ç¤¹¡£¤È¤¤¤¦¤³¤È¤Ç¡¢¤½¤ì¤¾¤ì¤Î¼°¤òÂåÆþ¤·¤Æ X ¤òµá¤á¤ë¤È¡¢

 

X1 = A1 - A2 + A3

X2 = A2 - A3 + A1

X3 = A3 - A1 + A2

 

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»³¤¬ 5 ¸Ä¤Î¾ì¹ç¤Ï¤É¤¦¤Ç¤·¤ç¤¦¡£Æ±Íͤ˹ͤ¨¤Æ¤¤¤¯¤È

 

X1 = A1 - A2 + A3 - A4 + A5

X2 = A2 - A3 + A4 - A5 + A1

X3 = A3 - A4 + A5 - A1 + A2

X4 = A4 - A5 + A1 - A2 + A3

X5 = A5 - A1 + A2 - A3 + A4

 

¤È¤Ê¤ê¤Þ¤¹¡£

¤µ¤Æ¥×¥í¥°¥é¥à¤Ç¤¹¤¬¡¢¤³¤Î¼°¤ò¤É¤¦¤·¤Þ¤·¤ç¤¦¤«¡£

 

for ( i = 0 ; i < N ; i++ ) {

  for ( j = i ; j < N+i ; j++ ) { 

    A[ j%N ];

 

¤Ã¤Æ´¶¤¸¤Ç¥ë¡¼¥×¤¹¤ì¤Ð¡¢

 

A[0] A[1] A[2] A[3] A[4]

A[1] A[2] A[3] A[4] A[0]

A[2] A[3] A[4] A[0] A[1]

A[3] A[4] A[0] A[1] A[2]

A[4] A[0] A[1] A[2] A[3]

 

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¤È¤¤¤¦¤³¤È¤Ç¡¢

 

  1. »³¤Î¿ô N ¤òÆþÎÏ
  2. ¥À¥à¤Î¿åÎÌ A ¤òÆþÎÏ
  3. ¹ß±«ÎÌ X ¤ò·×»»
  4. ¤¹¤Ù¤Æ¤Î»³¤ËÂФ·¤Æ·«ÊÖ¤·
  5. ·ë²Ì¤òɽ¼¨

 

¤È¤¤¤¦Î®¤ì¤Ç¥×¥í¥°¥é¥à¤ò½ñ¤¤¤Æ¤ß¤Þ¤·¤¿¡£

 

/*  AtCoder Beginner Contest 133 D - Rain Flows into Dams
    2019.07.26 by meyon */

 

#include <stdio.h>
#include <stdlib.h>

 

void err()
{
  printf("Error!¥n");
  exit(1);
}

 

int main()
{
  // input N
  int N;
  scanf("%d", &N);
  if(3>N || 99999<N || 0==N%2) {
    err();
  }

 

  // input A
  int A[N];
  for(int i=0; i<N; i++) {
    scanf("%d", &A[i]);
    if(0>A[i] || 1000000000<A[i]) {
      err();
    }
  }

 

  // rainfall X
  int X[N];
  for(int i=0; i<N; i++) {
    X[i]=0;
  }
  for(int i=0; i<N; i++) {
    int s=-1;
    for(int j=i; j<N+i; j++) {
      int amou=A[j%N];
      s=-s;
      X[i] += s*amou;
    }
  }

 

  // display answer
  for(int i=0; i<N; i++) {
    printf("%d ", X[i]);
  }
  printf("¥n");

 

  return 0;
}

 

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¢£¶¥µ»¥×¥í¥°¥é¥ß¥ó¥° AtCoder Beginner Contest 133 C ÌäÂê

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¶¥µ»¥×¥í¥°¥é¥ß¥ó¥° AtCoder Beginner Contest 133 C - Remainder Minimization 2019 ¤ò²ò¤¤¤Æ¤ß¤Þ¤¹¡£

 

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remain = i * j % 2019

 

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¹Í¤¨¤ë¤³¤È¾®°ì»þ´Ö (^_^;)¡¡¤¢¤Ã¤¿¡ª¹çƱ¼°¤Ç¤¹¡£

 

¹çƱ¼°¡¡a ≡ b ( mod n )         a ¤ò n ¤Ç³ä¤Ã¤¿Í¾¤ê¤Ï b 

 

¹çƱ¼°¤Î³Ý¤±»»¤Ï

 

a ≡ c ( mod n ) ¡¢b ≡ d ( mod n ) ¤Î¤È¤­¡¢a × b ≡ c × d ( mod n )

 

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remain = ( i % 2019 ) * ( j % 2019 ) % 2019

 

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¤Ê¤ª¡¢¹çƱ¼°¤Ë¤Ä¤¤¤Æ¤Ï¡Ö¹çƱ¼°¤Î¾ÚÌÀ¤äÌäÂê¤Î²ò¤­Êý¤ò²òÀ⡪Âç³Ø¼õ¸³¤Ç»È¤¤¤³¤Ê¤½¤¦¡ª¡×¤ò»²¹Í¤Ë¤µ¤»¤Æ¤¤¤¿¤À¤­¤Þ¤·¤¿¡£

 

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/*  AtCoder Beginner Contest 133 C - Remainder Minimization 2019
    2019.07.25 by meyon */

 

#include <stdio.h>

 

int main()
{
  // input date
  int L, R;
  scanf("%d %d", &L, &R);
  if (0>L || L>=R || 2000000000<R) {
    printf("Error!¥n");
    return 1;
  }

 

  // modulo
  int ans=2018;
  for (int i=L; i<R; i++) {
    for (int j=i+1; j<=R; j++) {
      int remain=(i%2019)*(j%2019)%2019;
      if(ans>remain) {
        ans=remain;
      }
      if (0==ans) {
        i=R;
        j=R+1;    // end of modulo
      }
    }
  }

 

  // answer
  printf("%d¥n", ans);

  return 0;
}

 

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¢£¶¥µ»¥×¥í¥°¥é¥ß¥ó¥° AtCoder Beginner Contest 133 BÌäÂê ¤½¤Î3

AtCoder Beginner Contest 133 ¤Î B ÌäÂê¡ÖGood Distance¡×¤Î¥×¥í¥°¥é¥à¤ò½ñ¤¤¤Æ¤ß¤Þ¤·¤¿¡£

 

/*  AtCoder Beginner Contest 133 B - Good Distance
    2019.07.25 by meyon  */

 

#include <stdio.h>
#include <stdlib.h>

 

void err()
{
  printf("Error!¥n");
  exit(1);
}

 

int main()
{
  // input N:point and D:dimension
  int N, D;
  scanf("%d %d", &N, &D);
  if(N<2 || N>10 || D<1 || D>10) {
    err();
  }

 

  // input X:coordinates
  int X[N][D];
  for(int i=0; i<N; i++) {
    for(int j=0; j<D; j++) {
      scanf("%d", &X[i][j]);
      if(X[i][j] < -20 || X[i][j] > 20) {
        err();
      }
    }
  }

 

  // calculate norm
  int ans=0;
  for(int i=0; i<N-1; i++) {
    for(int j=i+1; j<N; j++) {
      int norm=0;
      for(int k=0; k<D; k++) {
        int diff = X[i][k] - X[j][k];
        norm += diff * diff;
      }

 

      // check norm whether integer
      int flag=0;
      for(int k=0; k*k <= norm; k++) {
        if(k*k == norm) {
          flag=1;
        }
      }
      if(flag) {
        ans++;
      }
    }
  }

 

  // display answer
  printf("%d¥n", ans);

 

  return 0;
}

 

err() ´Ø¿ô¤Ï Error! ¤òɽ¼¨¤·¤Æ¥×¥í¥°¥é¥à¤ò¥¨¥é¡¼½ªÎ»¤·¤Þ¤¹¡£exit() ¤òÍøÍѤ¹¤ë¤¿¤á¤Ë stdlib.h ¤ò¥¤¥ó¥¯¥ë¡¼¥É¤·¤Æ¤¤¤Þ¤¹¡£

ÅÀ¤Î¿ô N ¤È¼¡¸µ¿ô D ¤òÆþÎϤ·¤¿¤é¡¢¤½¤Î¿ô¤ÇºÂɸ¤ÎÇÛÎóÊÑ¿ô¤òÄêµÁ¤·¤Þ¤¹¡£for ʸ¤ÇºÂɸ¥Ç¡¼¥¿¤òÆþÎϤ·¤Þ¤¹¡£

calculate norm ¤È check norm whether integer ¤ÎÉôʬ¤Ï²òÀâ¤Î¼ÂÁõÎã¤ò»²¹Í¤Ë¤·¤Þ¤·¤¿¡£¥ë¡¼¥×¤Î¾ò·ï¼°¤¬¤Á¤ç¤Ã¤È°Û¤Ê¤Ã¤Æ¤¤¤ë¤Î¤È¡¢º¹Ê¬¤ÎÀäÂÐÃͤȤϤ»¤º¤½¤Î¤Þ¤Þº¹¤ò¤À¤·¤Æ¤¤¤ë¤È¤³¤í¤¬Áê°ãÅÀ¤Ç¤¹¡£

¤Á¤Ê¤ß¤Ë diff ¤Ïº¹Ê¬¡¢norm ¤Ï¥Ù¥¯¥È¥ë¤Îµ÷Î¥¤ß¤¿¤¤¤Ê°ÕÌ£¤Ç¤¹¤Í¡£

¼ÂÁõÎã¤Î¤è¤¦¤Ë flag ¤ò boolean ·¿¤Ë¤¹¤ë¤Ë¤ÏÊ̤ΥإåÀ¥Õ¥¡¥¤¥ë¤¬É¬Íפˤʤê¤Þ¤¹¡£¤Þ¤¡ int ·¿¤Ç¤âÌäÂê¤Ê¤¤¤«¤Ê¤È¡£

 

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¢£¶¥µ»¥×¥í¥°¥é¥ß¥ó¥° AtCoder Beginner Contest 133 BÌäÂê ¤½¤Î2

AtCoder Beginner Contest 133 B - Good Distance ¤ò¹Í¤¨¤Æ¤¤¤Þ¤¹¡£ÌäÂê¤ÏÍý²ò¤Ç¤­¤¿¤Î¤Ç¡¢¼¡¤Ï¥×¥í¥°¥é¥à¤ò¹Í¤¨¤Æ¤ß¤è¤¦¤È»×¤¤¤Þ¤¹¡£

 

ή¤ì¤È¤·¤Æ¤Ï¡¢

 

  1.  ÅÀ¤Î¿ô¡¢¼¡¸µ¤Î¿ô¤òÆþÎÏ
  2.  ºÂɸ¥Ç¡¼¥¿¤òÆþÎÏ
  3.  Yij ¤ò·×»»
  4.  k2==Yij ¤«¤É¤¦¤«¤òȽÄê
  5.  3. 4. ¤ò¤¹¤Ù¤Æ¤ÎÅÀ´Ö¤Ç·«¤êÊÖ¤¹
  6.  4. ¤¬¿¿¤À¤Ã¤¿¿ô¤ò½ÐÎϤȤ·¤Æɽ¼¨

 

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for (int i = 0; i < N; ++i) {

for (int j = i+1; j < N; ++j) {

  int norm = 0;

  for (int k = 0; k < D; ++k) {

    int diff = abs(X[i][k] - X[j][k]);

    norm += diff * diff;

  }

 

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ºÂɸ¤Ï¡¢

 

        1 ÈÖÌܤÎÅÀ (X00 X01 X02 X03)

        2 ÈÖÌܤÎÅÀ (X10 X11 X12 X13)

        3 ÈÖÌܤÎÅÀ (X20 X21 X22 X23)

 

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Y01 =(X00-X10)2+(X01-x11)2+(X02-X12)2+(X03-X13)2

 

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Y02 =(X00-X20)2+(X01-x21)2+(X02-X22)2+(X03-X23)2

 

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Y12 =(X10-X20)2+(X11-x21)2+(X12-X22)2+(X13-X23)2

 

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X[i][k]-X[j][k] ¤È¤¹¤ë¤È¡¢

 

for (i=0; i<N-1; i++) {

  for (j=i+1; j<N; j++) {

    for (k=0; k<D; k++) {

      X[i][k] - X[j][k]

 

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¢£¶¥µ»¥×¥í¥°¥é¥ß¥ó¥° AtCoder Beginner Contest 133 BÌäÂê ¤½¤Î1

Ä´»Ò¤Ë¾è¤Ã¤¿ meyon ¤µ¤ó¤Ï¤µ¤é¤ËÄ©À路¤Æ¤ß¤Þ¤·¤¿ (^_^;)

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Y12=(-3+12)2¡Ü(7-1)2¡Ü(10-8)2¡Ü(2-2)2=121

 

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2 ÈÖÌÜ¤È 3 ÈÖÌÜ¤Î´Ö¤Ï ¡¢

 

Y23=(-2+12)2¡Ü(8-1)2¡Ü(10-9)2¡Ü(3-2)2=151

 

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Y31=(-2+3)2¡Ü(8-7)2¡Ü(9-8)2¡Ü(3-2)2=4

 

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    ¤Þ¤ºÅżÖĤòºÇ¾®ÃÍ lowestFare ¤È¤·¤Æ¤ª¤­¡¢¼¡¤Ë¥¿¥¯¥·¡¼Âå¤ÈÈæ³Ó¤·¤Æ¡¢¤â¤·¥¿¥¯¥·¡¼Âå¤ÎÊý¤¬°Â¤±¤ì¤Ð lowestFare ¤ò¹¹¿·¤¹¤ë¤È¤¤¤¦Ã±½ã¤ÊÊýË¡¤Ç¤¹¡£
  4. ºÇ¾®ÃͤÎɽ¼¨
    printf() ¤ÇºÇ¾®ÃÍ lowestFare ¤òɽ¼¨¤·¤Þ¤¹¡£

 

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/*  AtCoder Beginner Contest 133 A - T or T
    2019.07.24 by meyon  */

 

#include <stdio.h>

 

int main()
{
  // data input and judgment
  int N;    // number of people
  int A;    // train fare
  int B;    // taxi fare
  scanf("%d %d %d", &N, &A, &B);
  if(N<1 || N>20 || A<1 || A>50 || B<1 || B>50) {
    printf("Error!¥n");
    return 1;
  }

 

  // extract the lowest fare
  int lowestFare = N * A;
  if(lowestFare > B) {
    lowestFare = B;
  }

 

  // display the lowest fare
  printf("%d¥n", lowestFare);

 

  return 0;
}

 

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