Description
Usually we use number in the decimal system, for it is so convenient for us to remember and calculate.
But it is not the same in the computer world where numbers are always stored in the binary system. For example, the number 21 in decimal can be presented as (21)10
= (10101)2 = 24+22+20. It is the sum of the power of 2. Note that in the first item, the power is 4, then the number 4 can be presented as (4)10
= (100)2=22, so , and it is much more convenient for computer to display as 21=2(2(2))+2(2)+2(0). Every positive integer can be written
in this form following these principles:
= (10101)2 = 24+22+20. It is the sum of the power of 2. Note that in the first item, the power is 4, then the number 4 can be presented as (4)10
= (100)2=22, so , and it is much more convenient for computer to display as 21=2(2(2))+2(2)+2(0). Every positive integer can be written
in this form following these principles:
- Number 1 is presented as 2(0), while number 2 is presented as 2. Then other numbers must be combined by these two basic numbers;
- The powers of 2 are always sorted in descending order .
Input
Each line of the Input is the number n (0 < n < 1000000) in the binary system. Input file is ended with -1.
Output
For each case, you should only export the equation as the sample output. Be careful of the space before and after the equal sign. And there mustn’t be any more space in your output.
Sample Input
8
21
1315
-1
Sample Output
8 = 2(2+2(0))
21 = 2(2(2))+2(2)+2(0)
1315 = 2(2(2+2(0))+2)+2(2(2+2(0)))+2(2(2)+2(0))+2+2(0)
题意:把一个十进制的数转化成以2为底的若干个整数和。
思路:递归。n=2(a)+2(b)+...+2(x)。而a,b,...,x分别又是相当于n。注意边界。
#include<stdio.h>
void work(int n)
{
int a[30],i=0;
while(n>0) {
a[i++]=n%2;
n/=2;
}
for(int j=i-1;j>=0;j--)
if(a[j]) {
if(j<i-1) printf("+");
if(j!=1 && j!=0) {
printf("2(");
work(j);
printf(")");
}
else if(j==1) printf("2");
else printf("2(0)");
}
}
int main()
{
int n;
while(~scanf("%d",&n) && n!=-1) {
printf("%d = ",n);
work(n);
printf("\n");
}
return 0;
}
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