#828. 循环及求余
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ID: 828
Type: Objective
Tried: 64
Accepted: 30
Difficulty: 5
Uploaded By:
循环及求余
以下程序段的输出结果是（{{ input(1) }} ）
int y=10;
for ( ;y>0;y--)
if (y%3==0)
{
printf("%d",--y);
continue;
}