[2016CCPC东北地区大学生程序设计竞赛]Mr. Frog’s Problem 规律+打表

Mr. Frog’s Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0


Problem Description

One day, you, a clever boy, feel bored in your math class, and then fall asleep without your control. In your dream, you meet Mr. Frog, an elder man. He has a problem for you.

He gives you two positive integers A and B, and your task is to find all pairs of integers (C, D), such that A≤C≤B,A≤D≤B and AB+BA≤CD+DC

 

Input

first line contains only one integer T (T≤125),
which indicates the number of test cases. Each test case contains two integers A and B (1≤A≤B≤1018).

 

Output

For each test case, first output one line "Case #x:", where x is the case number (starting from 1). 

Then in a new line, print an integer s indicating the number of pairs you find.

In each of the following s lines, print a pair of integers C and D. pairs should be sorted by C, and then by D in ascending order.

 

Sample Input

2
10 10
9 27

 

Sample Output

Case #1:
1
10 10
Case #2:
2
9 27
27 9

打表可以发现，其就有两种结果，相等时为1   不相等时为2

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define PI acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};

ll a,b;
int main()
{
ll i,j;
int T;
scanf("%d",&T);
for(int k=1;k<=T;k++)
{
scanf("%lld %lld",&a,&b);
printf("Case #%d:\n",k);
if(a==b)
{
printf("1\n");
printf("%lld %lld\n",a,b);
}
else
{
printf("2\n");
printf("%lld %lld\n",a,b);
printf("%lld %lld\n",b,a);
}
}
return 0;
}