Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 34610 Accepted
Submission(s): 14259
解题报告:
水题,遍历每个字符串的字母,用switch case解决。开一个一维数组保存元音的个数。
1 #include <stdio.h> 2 int main() 3 { 4 int i,n,a[5]; 5 char s[100]; 6 scanf("%d",&n); 7 while(n--) 8 { 9 for(i=0;i<5;i++) 10 a[i]=0; 11 scanf("%s",s); 12 for(i=0;s[i];i++) 13 { 14 switch(s[i]) 15 { 16 case ‘a‘:a[0]++;break; 17 case ‘e‘:a[1]++;break; 18 case ‘i‘:a[2]++;break; 19 case ‘o‘:a[3]++;break; 20 case ‘u‘:a[4]++;break; 21 } 22 } 23 printf("a:%d\ne:%d\ni:%d\no:%d\nu:%d\n",a[0],a[1],a[2],a[3],a[4]); 24 if(n) 25 printf("\n"); 26 } 27 return 0; 28 }
原文:http://www.cnblogs.com/la0bei/p/3652942.html