A. Raising Bacteria
2015-09-19 10:02
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time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly x bacteria
in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input
The only line containing one integer x (1 ≤ x ≤ 109).
Output
The only line containing one integer: the answer.
Sample test(s)
input
output
input
output
Note
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the
box. Now we put one more resulting 5 in the box. We added 2 bacteria
in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
解题说明:此题其实是一道数学题,由于每个数放入后会扩大一倍,为了得到最少的数字,迭代除以2,直到满足最终的条件即可。
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
int main()
{
int i,j,x,count=0;;
scanf("%d",&x);
while(x>1)
{
if(x%2==1)
{
count++;
}
x=x/2;
}
printf("%d\n",count+1);
return 0;
}
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly x bacteria
in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input
The only line containing one integer x (1 ≤ x ≤ 109).
Output
The only line containing one integer: the answer.
Sample test(s)
input
5
output
2
input
8
output
1
Note
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the
box. Now we put one more resulting 5 in the box. We added 2 bacteria
in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
解题说明:此题其实是一道数学题,由于每个数放入后会扩大一倍,为了得到最少的数字,迭代除以2,直到满足最终的条件即可。
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
int main()
{
int i,j,x,count=0;;
scanf("%d",&x);
while(x>1)
{
if(x%2==1)
{
count++;
}
x=x/2;
}
printf("%d\n",count+1);
return 0;
}
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