PTA乙级训练题1001-1012解答（C语言）

1001【答案】

#include 
int main(){
    int n;int i=0;
    scanf("%d",&n);
    while(n!=1){
    if(n%2==0)
        n/=2;
    else 
        n=(n*3+1)/2;
        i++;
    }
    printf("%d",i);
}

1002【答案】

#include 
void print(int a){
  switch (a){
   case 0:
        printf ("ling");
        break;
   case 1:
        printf ("yi");
        break;
   case 2:
        printf ("er");
        break;
   case 3:
        printf ("san");
        break;
   case 4:
        printf ("si");
        break;
   case 5:
        printf ("wu");
        break;
   case 6:
        printf ("liu");
        break;
   case 7:
        printf ("qi");
        break;
   case 8:
        printf ("ba");
        break;
   case 9:
        printf ("jiu");
        break;
   default:
        break;

  }
}
int main(){
 char num[105];
 int sum=0;int i = 0;
 int hund = 0,dec = 0,unit = 0;//百、十、个
 scanf("%s",num);
  for(;num[i]!='\0';i++){
    sum += (num[i]-'0'); 
  }
  //printf("%d",sum);
  if(sum>=100){
        unit = sum % 10;
        //printf("%d ",unit);
        dec = (sum/10)%10;
        //printf("%d ",dec);
        hund = sum/100;
        //printf("%d ",hund);
        print(hund);
        printf(" ");
        print(dec);
        printf(" ");
        print(unit);

  } else if(sum >= 10){
    unit = sum%10;
        dec = sum / 10;
        print(dec);
        printf(" ");
        print(unit);
  }else {
    unit = sum;
        print(unit);
  }
  return 0;
}

1003【答案】

#include 
int Judge(char ch[]){
  int i=0;int tag = 1;int T_time=0,P_time=0;
  int A_time_front=0,A_time_last=0;
  int A_time_mid =0;
  int A_time=0;

  while(ch[i]!='\0'){
    if(ch[i]!='A'&&ch[i]!='P'&&ch[i]!='T'){
      tag=0;
      break;
    }//字符串中必须仅有P,A,T三种字符 
    else if(ch[i]=='P'){
      P_time++;
      if(P_time!=1){
        tag=0;
        break;
      }//记录P的个数 
    }else if(ch[i]=='T'){
        T_time++;
        if(P_time==0||T_time!=1){
            tag = 0;
            break;
        }
    } //T的个数 
    else if(ch[i]=='A'){
        A_time++;
        if(P_time==0){
            A_time_front++;
        } else if(P_time==1&&T_time==0){
            A_time_mid++;
        }else if(P_time==1&&T_time==1){
            A_time_last++;
        }
    }
    i++;
}
    if(P_time==0||T_time==0||A_time==0)
        tag=0;
    else if(A_time_front!=0&&A_time_last!=0){
    if((A_time_last/A_time_front)!=(A_time_mid))
        tag=0;
    }
    return tag;
}
int main(){
  int n,i=0;
  char s[105];
  int result[15];
 scanf("%d",&n);
  for(;iscanf("%s",s);
  result[i]=Judge(s);
  }
  for(i=0;iif(result[i])
      printf("YES\n");
    else
      printf("NO\n");
  }   
  return 0;
}

1004 【答案】

#include 
#include 
typedef struct student ST;
struct student{
char name[15];
char stunumber[15];
int score;
};
int main(){
ST T;
int n;int i=0;
char name_max[15];
char name_min[15];
char stunumber_max[15];
char stunumber_min[15];
int score_max=-1,score_min=101;

scanf("%d",&n);
for(;iscanf("%s%s%d",T.name,T.stunumber,&T.score);
if(score_maxmemcpy(name_max,T.name,sizeof(T.name)); 
memcpy(stunumber_max,T.stunumber,sizeof(T.stunumber));
score_max=T.score;
}if(score_min>T.score){
memcpy(name_min,T.name,sizeof(T.name));
memcpy(stunumber_min,T.stunumber,sizeof(T.stunumber));
score_min=T.score;
}

}
printf("%s %s\n",name_max,stunumber_max);
printf("%s %s\n",name_min,stunumber_min);
return 0;
}

1005【答案】

//将数组全部置为零，（1-100）中需要判断的数作为下标，将其置为1
//被覆盖的数作为下标，改其值为0，这样子剩下的值为1的下标就是关键数
//再逆序输出，就是从大到小输出了
#include 
int main(){
    int cover[105]={0};
    int position[105]={0};
    int n,tag;
    int first=1;
    scanf ("%d",&n);
    int i=0;
    for(;i"%d",&position[i]);//这里由于是数，所以前面要用&
        cover[position[i]]=1;   
    }
    for(i=0;iwhile(tag!=0&&tag!=1){
            if(tag%2==0)
                tag = tag/2;
            else
                tag = (3*tag+1)/2;
            if(tag<=100)
            //  printf("%d ",tag);
                cover[tag]=0;
        }
        }

    for(i=104;i>=0;i--){
        if(cover[i]==1)
            if(first){
        printf("%d",i);
        first = 0;
        }
        else{
            printf(" %d",i);
        }

    }

    return 0;
}

1006 【答案】

#include 
int main(){
int B;int S;int G;
int n;
int i=0;
scanf("%d",&n);
B = n/100;
S = (n-100*B)/10;
G = (n-100*B-10*S);
for(;iprintf("B");
for(i=0;iprintf("S");
for(i=1;i<=G;i++)
printf("%d",i);
return 0;
}

1007 【答案】

#include 
int judge(int a){
  int i=2;int tag=1;
  for(;i*i<=a;i++){
    if(a%i==0){
      tag=0;
      break;
    }
  }
  return tag;
}
int main(){
  int n;int time=0;
  scanf("%d",&n);
  int i=3;
  for(;i<=n-2;i++){
  if(judge(i)&&judge(i+2))
    time++;
  }   
  printf("%d",time);
}

1008 【答案】

#include 
int main(){
  int N,M;
  int i=0;int first=1;
  int num[105];
  scanf("%d%d",&N,&M);
  for(;i"%d",&num[i]);
  }
 int time = --i;      //关键
  while(M--){
    int tag=num[time];//关键
    i = time;         //关键
    for(;i>0;i--)
    num[i]=num[i-1];

    num[0]=tag;
  }

  for(i=0;iif(first){
    printf("%d",num[i]);
      first=0;
    }else
      printf(" %d",num[i]);

  return 0;
}

1009 【答案】

//关于字符串指针数组应该好好研究一下
#include 
#include 
#include 
char ch[90];
int main(){
int position[40];
int i=0;int j,k=1;
while((ch[i++]=getchar())!='\n');
i--;
ch[i]='\0';

position[0]=0;
for(j=0;jif(ch[j]==' '){ 
ch[j]='\0';
position[k++]=j+1;
} 
int first=1;
for(i=k-1;i>=0;i--){
    if(first){ 
printf("%s",&ch[position[i]]);
first=0;
}else{
printf(" %s",&ch[position[i]]);
} 
}
  return 0;

}

1010 【答案】

#include 
int main(){
  int position;int number,max=0;
  int a[1005]={0};//记着初始化
  int i=0;
  scanf("%d%d",&number,&position);


  if(position>0){
    max = position-1;
    a[position-1]=number*position;
  }else{
    a[0]=0;
  }
  int tag=position-1;
  while(tag>0){
    scanf("%d%d",&number,&position);
  if(position>0){
    a[position-1]=number*position;
  }
  tag--;
  }
  int first=1;
  for(i=max;i>=0;i--){
   if(a[i]!=0){ 
     if(first){
    printf("%d %d",a[i],i);
    first=0;
     }else 
    printf(" %d %d",a[i],i);
   }
  }
  if(first)
  printf("0 0");
  return 0;
}

1011 【答案】

//注意数据大小，如果超过long的-2^31 ~ 2^31 - 1
//就要使用字符数组来计算数字加减乘除了
#include
int main(){

long a,b,c;
int p[11]={0};
int n,j,i=0;
scanf("%d",&n);
while(n--){
  scanf("%ld%ld%ld",&a,&b,&c);
 if((a+b)>c) 
   p[i++]=0;
  else
    p[i++]=1;
}

for(j=0;jif(p[j]==0)
printf("Case #%d: true\n",j+1);
  else
printf("Case #%d: false\n",j+1);
}

return 0;
}

1012 【答案】

#include 
int num[1005]={0};
int N;
void A1(int n);
void A2(int n);
void A3(int n);
void A4(int n);
void A5(int n);


int main(){
scanf("%d",&N);
int i=0;
for(;i"%d",&num[i]);
A1(N);
A2(N);
A3(N);
A4(N);
A5(N);

return 0;

}

void A1(int n){
int i=0;int sum=0;
for(;iif(num[i]%2==0&&num[i]%5==0)
sum+=num[i];

if(sum!=0)
printf("%d",sum);
else
printf("N");
}

void A2(int n){
int i=0;int sum=0;int time=0;int tag=1;
for(;iif(num[i]%5==1){
sum+=(num[i]*tag);
time++;
tag*=-1;
}
if(time!=0)
printf(" %d",sum);
else
printf(" N");

}

void A3(int n){
int tag=0;
int i =0;
for(;iif(num[i]%5==2)
tag++;
if(tag)
printf(" %d",tag);
else
printf(" N");

}
void A4(int n){
int sum=0;
float average =0;
int tag=0,i=0;
for(;iif(num[i]%5==3){
sum+=num[i];
tag++;
}
if(tag==0)
printf(" N");
else {
average = (float)sum/tag;/*要先将sum转换为float，不然答案直接去掉小数部分*/
printf(" %.1f",average);
}
}
void A5(int n){
int tag=0,max=0,i=0;
for(;iif(num[i]%5==4){
tag++;
if(maxif(tag)
printf(" %d",max);
else
printf(" N");
}

1013 【答案】

#include 
int prime[10005]={0};
void judge(int a){
  int k=3;
  int i;
  int tag=1;
  int temp;
  int j=2;
  while(tag1;
    i = 2;
    for(;i*i<=k;i++)
    if(k%i==0)
      temp=0;
    if(temp){
      tag++;
      prime[j++]=k;
    }
    k++;
  }
}
int main(){
int N,M;int i=1;int posi=1;int first=1;
prime[0]=1;
  prime[1]=2;
  scanf("%d%d",&N,&M);
  judge(M);
  for(;N<=M;N++){
    if(posi%10!=0&&first==1){
    printf("%d",prime[N]);
    first=0;
    posi++;
  }else if(posi%10==0){
    printf(" %d\n",prime[N]);
    first=1;
    posi++;
  }else if(posi%10!=0&&first!=1){
    printf(" %d",prime[N]);
    posi++;
  }
  }
  return 0;
}