(1)输入一个年份,判断该年份是否是闰年
int main(int argc, char *argv[])
{
int y;
printf("please input year: ");
scanf("%d", &y);
if(y % 4 == 0 && y % 100 != 0){
puts("闰年");
}else if(y % 100 == 0 && y % 400 == 0){
puts("闰年");
}else{
puts("不是闰年");
}
return 0;
}
(2)输出0~100(或者0~任意整数)以内的偶数
int main(int argc, char *argv[])
{
for(int i = 0; i <= 100; i++){
if(i % 2 == 0)
printf("%d ", i);
}
puts("");
return 0;
}
(3)输入整数n,输出含有n层的 *号金字塔
此题只需要考虑每一层左边的空格和买一行的*号输出就行,右边的空格可用\n或者puts("")代替
核心代码如下:
int main(int argc, char *argv[])
{
int N, i, j, z;
printf("please input num: ");
scanf("%d", &N);
for(i = 1; i <= N; i++){
for(z = 1; z <= N-i; z++)
printf(" ");
for(j = 1; j <= 2*i-1; j++)
printf("*");
puts("");
}
return 0;
}
(4)输入n个整数,求输入数据之和:(通过for循环输入数据)
int main(int argc, char *argv[])
{
int n;
printf("please input num: ");
scanf("%d", &n);
int a[n],i,j;
for(i = 0; i < n; i++)
{
printf("please input arr[%d]: ", i);
scanf("%d", &a[i]);
}
int sum = 0;
for(j = 0; j < n; j++)
{
sum += a[j];
}
printf("%d\n", sum);
return 0;
}
(5)输入字符串,将小写字母转换为大写字母,大写字母转换为小写字母
int main(int argc, char *argv[])
{
char buf[20] = {0};
puts("input string:");
gets(buf);
for(int i = 0; buf[i] != '\0'; i++)
{
if('A' <= buf[i] && buf[i] <= 'Z'){
buf[i] += 32;
}
else if('a' <= buf[i] && buf[i] <= 'z'){
buf[i] -= 32;
}
}
puts(buf);
return 0;
}