A Simple Task
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Problem Description
Given a positive integer n and the odd integer o and the nonnegative integer p such that n = o2^p.
Example
For n = 24, o = 3 and p = 3.
Task
Write a program which for each data set:
reads a positive integer n,
computes the odd integer o and the nonnegative integer p such that n = o2^p,
writes the result.
Input
The first line of the input contains exactly one positive integer d equal to the number of data sets, 1 <= d <= 10. The data sets follow.
Each data set consists of exactly one line containing exactly one integer n, 1 <= n <= 10^6.
Output
The output should consists of exactly d lines, one line for each data set.
Line i, 1 <= i <= d, corresponds to the i-th input and should contain two integers o and p separated by a single space such that n = o2^p.
Sample Input
1
24
Sample Output
3 3
題意：輸入一個n，將n表示成一個奇數乘以2的p次方的積，輸出這個奇數和p。
解題思路：n對2取整，用一個計數變量記錄次數，直到余數為1時輸出n和次數。
[cpp] #include<stdio.h>
int main()
{
int n,t,p;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
p=0;
while(n%2==0) //奇數除以2余數為1
{
p++; //記錄次數
n/=2;
}
printf("%d %d\n",n,p);
}
return 0;
}
#include<stdio.h>
int main()
{
int n,t,p;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
p=0;
while(n%2==0) //奇數除以2余數為1
{
p++; //記錄次數
n/=2;
}
printf("%d %d\n",n,p);
}
return 0;
}
