light oj 1014 - Ifter Party分解因子

1014 - Ifter Party

 

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 Lpiaju'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
4 10 0 13 2 300 98 1000 997	Case 1: 1 2 5 10 Case 2: 11 Case 3: 101 202 Case 4: impossible

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>

using namespace std;

int main(void)
{
int T, cas;
int p, l;
vector<int>vec;

scanf("%d", &T);

cas = 0;

while(T--)
{
cas++;
vec.clear();

scanf("%d%d", &p, &l);

int n = p - l;

int m =(int)sqrt(n);

for(int i = 1; i <= m; i++)
{
if(n % i == 0)
{
if(i > l)
vec.push_back(i);
if(i * i != n)
{
if(n / i > l)
vec.push_back(n / i);
}
}
}

sort(vec.begin(), vec.end());
printf("Case %d:", cas);

int mm = vec.size();
if(mm == 0)
printf(" impossible\n");

for(int i = 0; i < mm; i++)
printf(" %d", vec[i]);

printf("\n");
}
return 0;
}

c++容器<vector>文件名#include<vector>   定义vector<int>c;

　　　　　　　 c.clear()         移除容器中所有数据。

c.empty()         判断容器是否为空。

c.erase(pos)        删除pos位置的数据

c.erase(beg,end) 删除[beg,end)区间的数据

c.front()         传回第一个数据。

c.insert(pos,elem)  在pos位置插入一个elem拷贝

c.pop_back()     删除最后一个数据。

c.push_back(elem) 在尾部加入一个数据。

c.resize(num)     重新设置该容器的大小

c.size()         回容器中实际数据的个数。

c.begin()           返回指向容器第一个元素的迭代器

c.end()             返回指向容器最后一个元素的迭代器