1001【答案】
#include <stdio.h>
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 <stdio.h>
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 <stdio.h>
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]==