I have an Ifter party at the 5th 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
");
for(int i = 0; i < mm; i++)
printf(" %d", vec[i]);
printf("
");
}
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