打印n 0和m 1,以使C程序中没有两个0和三个1在一起
应该有N0和M1的序列,这样形成的序列不应包含两个连续的0和三个连续的1。
输入 -N=5M=9
输出 −11011011010101
注意-为了使上面的序列,语句(m<n-1)||m>=2*(n+1)如果为true,则应为false,这是我们无法建立上述序列的原因。
建议先通过问题逻辑来尝试一下,而不要跳到下面直接给出的解决方案。
算法
START Step 1 -> take values in ‘n’ and ‘m’ Step 2 -> Loop IF m=n-1 Loop While m>0 and n>0 Print 01 Decrement m and n by 1 End Loop While Loop IF n!=0 Print 0 End IF Loop IF m!=0 Print 1 End IF Step 3-> Else (m < n-1) || m >= 2 * (n + 1) Print cn’t have sequence for this Step 4 -> Else Loop While m-n > 1 && n > 0 Print 1 1 0 Decrement m by 2 and n by 1 End While Loop While n>0 Print 1 0 Decrement m and n by 1 End While Loop While m>0 Print 1 Decrement m by 1 End While Step 5-> End Else STOP
示例
#include <stdio.h> #include <math.h> int main() { int n =5, m=9; if( m == n-1 ) { //If m is 1 greater than n then consecutive 0's and 1's while( m > 0 && n > 0 ) { //Loop until all m's and n's printf("01"); m--; n--; } if ( n!=0 ) //Print the remaining 0 printf("0"); if( m!=0 ) //Print the remaining 1 printf("1"); } else if ( (m < n-1) || m >= 2 * (n + 1) ) { //If this is true the sequence can't be made printf("Can't have sequence for this\n"); } else { while( m-n > 1 && n > 0 ) { printf("1 1 0 "); m -= 2; n--; } while ( n > 0 ) { printf("1 0 "); n--; m--; } while ( m > 0 ) { printf("1 "); m--; } } return 0; }
输出结果
如果我们运行上面的程序,那么它将生成以下输出。
1 1 0 1 1 0 1 1 0 1 0 1 0 1