#829. 循环及奇偶判断
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ID: 829
Type: Objective
Tried: 71
Accepted: 26
Difficulty: 5
Uploaded By:
循环及奇偶判断
以下程序段的输出结果是（{{ input(1) }} ）
int i,x;
for(i=1;i<=5;i++)
{
if (i%2) printf("*");
else continue;
printf("#");
}
printf("$\n");