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#include<iostream>
using namespace std;
int main(){
int n;
cin >> n;
cout << n - 3;
return 0;
}
-n
0
9
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当你输⼊ 3 时,程序会执⾏以下步骤:
cin >> n; | n ??!? 3 b
cout << n - 3; 9? n - 3' 3 - 3 = 0i{?T 0 b
yN???{?T^ :0
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下列C++选项中,输出结果为 0的是?( )
cout << "5 - 5";
cout << 5 - 5;
cout << 2 * 3;
cout << 7 / 2;
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A.cout << "5 - 5";
{=?"5 - 5"3?1?^9?VrT
L={5 - 5?^ 0
Bcout << 5 - 5;
9?5 - 5 = 0
L={0
Ccout << 2 * 3;
9?2 * 3 = 6
L={6?^ 0
Dcout << 7 / 2;
9?7 / 2 C++ ?^??"E?T? 3Ŕ?
L={3?^ 0
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cout << 10 % 3;
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这道题考察的是 取模运算( %),即计算两个数相除后的 余数。????6547???ڲ?ѧϰʹ??
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计算 10 % 3
10 ÷ 3 = 3^ 3
3 × 3 = 9
Ŕ 10 - 9 = 1 ?[10 % 3 ??T^ 1 b
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