1014 - Ifter Party

IfterParty食物分配问题

最新推荐文章于 2018-05-07 17:55:17 发布

glu_cky

最新推荐文章于 2018-05-07 17:55:17 发布

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本文介绍了一个关于食物分配的问题，即如何计算每位参赛者在IfterParty上吃了多少个piaju。通过输入派对准备的食物总数和剩余数量，程序能够找出符合条件的分配方案，并按升序输出可能的数量。

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

SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET)

求p-l大于l的因子

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn=1<<30;
const int SIZE=1e5+10;
set<int>st;
void get_fac(int x,int l){
    st.clear();
    for(int i=1;i<=(int)sqrt((double)x);i++){
        if(x%i==0){
            if(i>l)st.insert(i);
            if(x/i>l)st.insert(x/i);
        }
    }
}
int main()
{
    int T,x;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++){
        int p,l;
        scanf("%d%d",&p,&l);
        x=p-l;
        get_fac(x,l);
        printf("Case %d: ",cas);
        if(st.empty()){
            printf("impossible\n");
            continue;
        }
        set<int>::iterator it=st.begin();
        printf("%d",*it);
        for(it++;it!=st.end();it++){
            printf(" %d",*it);
        }
        printf("\n");
    }
    return 0;
}