在下面的代码中,仅更改/添加一个字符并精确打印20次“ *”。
int main()
{
int i, n = 20;
for (i = 0; i < n; i--)
printf("*");
getchar();
return 0;
}
解决方案:
1.在for循环的第三个表达式中将i替换为n
C
#include
int main()
{
int i, n = 20;
for (i = 0; i < n; n--)
printf("*");
getchar();
return 0;
}
C++
#include
using namespace std;
int main()
{
int i, n = 20;
for (i = 0; i < n; n--)
cout << "*";
getchar();
return 0;
}
Java
// Java code
class GfG {
public static void main(String[] args)
{
int i, n = 20;
for (i = 0; i < n; n--)
System.out.print("*");
}
}
Python3
# Python3 programe to implement
# the above approach
if __name__ == '__main__':
n = 20;
for i in range(0, n):
print("*"); n -= 1;
# This code is contributed by gauravrajput1
C#
// C# code
using System;
class GfG
{
public static void Main()
{
int i, n = 20;
for (i = 0; i < n; n--)
Console.Write("*");
}
}
// This code is contributed by SoumikMondal
C
#include
int main()
{
int i, n = 20;
for (i = 0; -i < n; i--)
printf("*");
getchar();
return 0;
}
C++
#include
using namespace std;
int main()
{
int i, n = 20;
for (i = 0; -i < n; i--)
cout<<"*";
return 0;
}
// This code is contributed by rutvik_56.
Java
// Java code
import java.util.*;
public class GFG
{
public static void main(String[] args)
{
int i, n = 20;
for (i = 0; -i < n; i--)
System.out.print("*");
}
}
// This code is contributed by divyesh072019.
C#
// C# code
using System;
class GfG
{
public static void Main()
{
int i, n = 20;
for (i = 0; -i < n; i--)
Console.Write("*");
}
}
// This code is contributed by divyeshrabadiya07.
c
#include
int main()
{
int i, n = 20;
for (i = 0; i + n; i--)
printf("*");
getchar();
return 0;
}
c
#include
int main()
{
int i, n = 20;
for (i = 0; ~i < n; i--)
printf("*");
getchar();
return 0;
}
2.在for循环的第二个表达式中,在我之前加上“-”
C
#include
int main()
{
int i, n = 20;
for (i = 0; -i < n; i--)
printf("*");
getchar();
return 0;
}
C++
#include
using namespace std;
int main()
{
int i, n = 20;
for (i = 0; -i < n; i--)
cout<<"*";
return 0;
}
// This code is contributed by rutvik_56.
Java
// Java code
import java.util.*;
public class GFG
{
public static void main(String[] args)
{
int i, n = 20;
for (i = 0; -i < n; i--)
System.out.print("*");
}
}
// This code is contributed by divyesh072019.
C#
// C# code
using System;
class GfG
{
public static void Main()
{
int i, n = 20;
for (i = 0; -i < n; i--)
Console.Write("*");
}
}
// This code is contributed by divyeshrabadiya07.
3.在for循环的第二个表达式中用<代替+
C
#include
int main()
{
int i, n = 20;
for (i = 0; i + n; i--)
printf("*");
getchar();
return 0;
}
让我们扩展一下这个问题。
仅更改/添加一个字符,并精确打印21次“ *”。
解决方案:将否定运算符放在for循环的第二个表达式中的i之前。
说明:负运算符将数字转换为其补码。
No. One's complement
0 (00000..00) -1 (1111..11)
-1 (11..1111) 0 (00..0000)
-2 (11..1110) 1 (00..0001)
-3 (11..1101) 2 (00..0010)
...............................................
-20 (11..01100) 19 (00..10011)
C
#include
int main()
{
int i, n = 20;
for (i = 0; ~i < n; i--)
printf("*");
getchar();
return 0;
}
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