Codeforces Beta Round #76(Div.2) report

It's my top score contest!

After starting, I get accepted Problem A(10min). It's very simple problem.

I start problem B.Wrong answer twice(0:19,0:21). I found my wrong in reading, I get accepted(0:22).

Problem C。It's a problem about select folders. I like this problem, it's interesting! I submitted wrong answer twice, however I found a miss, I get accepted(0:42).

It is happy to be able to solve Problem D. Once I read this problem, I think this is greedy algorithm.  Surprisingly, I accepted this only one submit(1:18).

I can't solve Problem E in programming time.

Problems is here。3816,8/1070. It's excellent! I'm grad and celebrate this. I'm ranked up to blue.

A.Only comparing two strings.

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

int main(void){

  char str[81],num[10][11],sub[11];

  int i,j,k;

  scanf("%s",str);

  for(i=0;i<10;i++) scanf("%s",num[i]);

  for(i=0;i<8;i++){

    for(j=0;j<10;j++) sub[j]=str[j+10*i];

    sub[11]='\0';

    for(j=0;j<10;j++){

      if(strncmp(sub,num[j],10)==0){

printf("%d",j);

break;

      }

    }

  }

  putchar('\n');

  return 0;

}

    

B.If we use array, it's simple.

#include<stdio.h>

int main(void){

  int in1,in2,people[5];

  int m,n,i,flg=0;

  for(i=0;i<5;i++) people[i]=0;

  scanf("%d",&m);

  for(i=0;i<m;i++){

    scanf("%d %d",&in1,&in2);

    people[in1-1]++;

    people[in2-1]++;

  }

  for(i=0;i<5;i++){

    if(people[i]!=2){

      puts("WIN");

      flg=1;

      break;

    }

  }

  if(flg!=1) puts("FAIL");

  return 0;

}

C.Think each case. Answer is only "1" "2", or "3".

#include<stdio.h>

typedef unsigned int u_int;

int main(void){

  u_int m,n,a,b,a_pos,b_pos,a_line,b_line,k;

  scanf("%u %u %u %u",&m,&n,&a,&b);

  a_pos=(a-1)%n+1;

  b_pos=(b-1)%n+1;

  a_line=(a-1)/n;

  b_line=(b-1)/n;

  if(b==m) b_pos=n;

  if(a_line==b_line){

    k=1;

  }else if(a_pos==1 && b_pos==n){

    k=1;

  }else if(a_pos==1 || b_pos==n){

    k=2;

  }else if(b_pos+1==a_pos){

    k=2;

  }else if(a_line+1==b_line){

    k=2;

  }else{

    k=3;

  }

  printf("%d\n",k);

  return 0;

}

D.It's typical greedy algorithm.

#include<stdio.h>

#define MIN(x,y) ((x<y)?(x):(y))

int main(void){

  int n,m,w;

  double milk[50];

  double cup[50];

  double goals,pour;

  int i,j,k,now,flg;

  

  scanf("%d %d %d",&n,&w,&m);

  for(i=0;i<n;i++) milk[i]=w;

  for(i=0;i<m;i++) cup[i]=0;

  goals=(double)w*(double)n/(double)m;

  now=0;

  flg=0;

  for(i=0;i<m;i++){

    pour=MIN(goals-cup[i],milk[now]);

    cup[i]+=pour;

    milk[now]-=pour;

    if(milk[now]<1E-6){

      now++;

      flg=0;

    }else{

      flg++;

    }

    if(flg==2){

      puts("NO");

      return 0;

    }

    if(cup[i]<goals-(1E-6)){

      i--;

    }

  }

  for(i=0;i<n;i++) milk[i]=w;

  for(i=0;i<m;i++) cup[i]=0;

  now=0;

  puts("YES");

  for(i=0;i<m;i++){

    pour=MIN(goals-cup[i],milk[now]);

    cup[i]+=pour;

    milk[now]-=pour;

    printf("%d %.6f",now+1,pour);

    if(milk[now]<1E-6) now++;

    if(cup[i]<goals-(1E-6)){

      putchar(' ');

      i--;

    }else{

      putchar('\n');

    }

  }

  return 0;

}