第二次循环 p=s+1 输出为改亩876
然后p=s+2 不满足条件p<s+2 退出循环
12 a=p1==&m 因为p1=&k 所以判断不成立,a=0
b=(*p1)/(*p2)+7=k/m+7=4/6+7=7
于是型歼液执行后
a值为卜物0 b值为7
#include<stdio.h>#include<string.h>
int main(){
int i = 0
char ch[3][100]
char max[100]
for( 并源或i <裂纳 3 ++i)
绝伍 scanf("%s", ch[i])
strcpy(max, ch[0])
for(i = 1 i < 3 ++i)
if(strcmp(max, ch[i]) < 0)
strcpy(max, ch[i])
printf("the max string is : %s", max)
return 0
}
#include<stdio.h>
#include<string.h>
#include<ctype.h>
//有一篇文章,共有三行字符,每行80个字符,分别统计其中英文大写字母,小写字母,数字,空格以及其他字符的个数
void calc(char ch[3][80], int col){
int i = 0, j = 9, upp_cnt, low_cnt, num_cnt, space_cnt, other_cnt
upp_cnt = low_cnt = num_cnt = space_cnt = other_cnt = 0
for( i < col ++i)
for(j = 0 j < 80 ++j){
if(isupper(ch[i][j]))
upp_cnt++
else if(islower(ch[i][j]))
low_cnt++
else if(' ' == ch[i][j])
space_cnt++
else if (isdigit(ch[i][j]))
num_cnt++
else
other_cnt++
}
printf("upper count : %d\n",upp_cnt )
printf("lower count : %d\n",low_cnt )
printf("space count : %d\n",space_cnt )
printf("nun count : %d\n",num_cnt )
printf("other count : %d\n",other_cnt )
}
int main(){
char ch[3][80] = {
{"Hello, my name is Ann. I'm taking an American accent training."},
{"And I have paid $ 100. I think I can do it. So insist it."},
{"Now I am making a programming and I want to mannage it."}
}
calc(ch, 3)
return 0
}
这些都是很基础的,希望LZ好好学习,不要谈恋爱了
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