FOJ有奖月赛-2011年11月 Report

Because I knew this contest for the first time, I participated just a little. It is three to have challenged. I do not understand problems of Chinese well.

1: I guessed from translation with the computer and Chinese which I can understand, a sample case and submitted it. It is an extremely simple problem.

 5: I am in time if I search after having sorted it. However ,it's TLE. I should use binary search. I want to find some good algorithm.

9: I thought this problem was the simplest although it was the very last. You should only think about the binary scale.

Problems are here.
Standings=120,Accept=2,Penalty=71.

[A.Ac]

#include<stdio.h>

int main(void){
  int t,a[3],b[3],awin,i,j;
  scanf("%d",&t);
  while(t--){
    scanf("%d %d %d %d %d %d",&a[0],&a[1],&a[2],&b[0],&b[1],&b[2]);
    for(i=0,j=0;i<3 && j<3;(a[i]>b[j]?j++:i++));
    if(j==3) puts("1");
    else puts("2");
  }
  return 0;
}

[E.TLE]

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

int comp(const void *p1, const void *p2);

int main(void){
  int n,m,*array,i,j,k,count;
  while(fscanf(stdin,"%d %d",&n,&m)!=EOF){
    array=(int *)calloc(n,sizeof(int));
    for(i=0;i<n;i++) scanf("%d",array+i);
    qsort(array,n,sizeof(int),comp);
    count=0;
    for(i=0,j=n-1;i<j;i++){
      k=-1;
      while('_'){
if(i>=j){
  if(k!=-1) j=k;
  break;
}
if(*(array+i)+*(array+j)==m){
  count++;
  if(k==-1) k=j;
}else if(*(array+i)+*(array+j)<m){
  if(k!=-1) j=k;
  break;
}
j--;
      }
    }
    printf("%d\n",count);
    free(array);
  }
  return 0;
}

int comp(const void *p1, const void *p2){
  int n1,n2;
  n1=*((const int *)p1);
  n2=*((const int *)p2);
  return n1-n2;
}


[I.Ac]

#include<stdio.h>

int main(void){
  unsigned int n;
  unsigned int i,j,k;
  while(fscanf(stdin,"%u",&n)!=EOF){
    for(i=0;i<33;i++){
      if((1<<i)>n){
printf("%u\n",i);
break;
      }
    }
  }
  return 0;
}