算术基本定理+容斥-17级暑假集训

G - Ekka Dokka

https://vjudge.net/contest/240973#problem/G

基础题水题：

Ekka and his friend Dokka decided to buy a cake. They both love cakes and that's why they want to share the cake after buying it. As the name suggested that Ekka is very fond of odd numbers and Dokka is very fond of even numbers, they want to divide the cake such that Ekka gets a share of N square centimeters and Dokka gets a share of M square centimeters where N is odd and M is even. Both N and M are positive integers.

They want to divide the cake such that N * M = W, where W is the dashing factor set by them. Now you know their dashing factor, you have to find whether they can buy the desired cake or not.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer W (2 ≤ W < 263). And W will not be a power of 2.

Output

For each case, print the case number first. After that print "Impossible" if they can't buy their desired cake. If they can buy such a cake, you have to print N and M. If there are multiple solutions, then print the result where M is as small as possible.

Sample Input

Sample Output

Case 1: 5 2

Case 2: Impossible

Case 3: 3 4

题意：
t组
对于每一组，给出w(2<=w<=2^63),
问是否存在正数n,m，n奇m偶，满足n*m=w,没有输出...
若有输出n,m,若有多组则输出让m尽可能小的那一组

有一点没有注意：奇数*偶数=偶数，所以奇数一定不可能

偶数一定可以(可分解成1乘本身）

代码：

#include <stdio.h>
#include <math.h>
typedef long long ll;
int main()
{ll t,w,k=0,i;scanf("%lld",&t);while(t--){k++;scanf("%lld",&w);if(w%2!=0)//奇数时一定不存在 printf("Case %d: Impossible\n",k);else{for(i=2;i<=w;i=i+2){if(w%i==0){if((w/i)%2!=0){printf("Case %d: %lld %ld\n",k,w/i,i);break;}}}}}}

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算术基本定理+容斥-17级暑假集训

G - Ekka Dokka

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