lightoj-1014【枚举】

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1014 - Ifter Party

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Time Limit: 2 second(s)Memory Limit: 32 MB

I have an Ifter party at the 5^th day of Ramadan for the contestants. For this reason I have invited C contestants and arranged P piaju's (some kind of food, specially made for Ifter). Each contestant ate Q piaju's and L piaju's were left (L < Q).

Now you have to find the number of piaju's each contestant ate.

Input

Input starts with an integer T (≤ 325), denoting the number of test cases.

Each case contains two non-negative integers P and L (0 ≤ L < P < 2³¹).

Output

For each case, print the case number and the number of possible integers in ascending order. If no such integer is found print 'impossible'.

Sample Input

Output for Sample Input

10 0

13 2

300 98

1000 997

Case 1: 1 2 5 10

Case 2: 11

Case 3: 101 202

Case 4: impossible

大意：

有C个人，安排了P个吃的，每个人会吃Q个吃的，最后留下L个吃的；求所有可能的Q，从小到大输出，要保证Q>L;

思路：
其实就是求出 P-L 的所有数的约数，然后这个约数 >L 的话就满足；感觉开数组会爆，但是没想到也能过

#include<cstdio>#include<algorithm>#include<cstring>#include<set> using namespace std;int p,l;int main(){int t,text=0;scanf("%d",&t);while(t--){scanf("%d%d",&p,&l);set<int> a;p-=l;for(long long i=1;i*i<=p;i++) // 这里换 long long 就过了，之前 int 一直 tle {if(p%i==0){if(i>l) a.insert(i);if(p/i>l)a.insert(p/i);}}printf("Case %d:",++text);if(a.size()==0){puts(" impossible");continue;}set<int>::iterator it;//bool flag=0;for(it=a.begin();it!=a.end();it++){//if(flag)//printf(" ");printf(" %d",*it);//flag=1;}puts("");}return 0;}

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