Mr. Frog’s Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 312    Accepted Submission(s): 219

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

 

Source

2016CCPC东北地区大学生程序设计竞赛 - 重现赛

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5932 5931 5930 5929 5928 

 

Statistic | Submit | Discuss | Note

题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5924

题目大意：

　　题目给定A和B(<=10¹⁸)，求满足的C和D个个数，并全部输出。

题目思路：

　　【模拟】

　　+1s 这题一看样例就猜大概只有AB，BA两种情况。不妨假设C<=D

　　假设A/B+B/A>A/(B-1)+(B-1)/A，解得条件为B²-B>A²,所以如果要满足，则需要A²>=B²-B，而只有当A=B时满足(B>=A,B=A+1时，B²-B=A(A+1)>A²)

　　所以除了A=B的情况，其余都是减的，(C,D)可视为(A,B)->(A,D)->(C,D)，每次只改一维，递减，所以最终也是递减的。

　　所以最终答案就是AB,BA 注意A=B的情况。

 //HDU 5924

 //by coolxxx

 //#include<bits/stdc++.h>

 #include<iostream>

 #include<algorithm>

 #include<string>

 #include<iomanip>

 #include<map>

 #include<stack>

 #include<queue>

 #include<set>

 #include<bitset>

 #include<memory.h>

 #include<time.h>

 #include<stdio.h>

 #include<stdlib.h>

 #include<string.h>

 //#include<stdbool.h>

 #include<math.h>

 #pragma comment(linker,"/STACK:1024000000,1024000000")

 #define min(a,b) ((a)<(b)?(a):(b))

 #define max(a,b) ((a)>(b)?(a):(b))

 #define abs(a) ((a)>0?(a):(-(a)))

 #define lowbit(a) (a&(-a))

 #define sqr(a) ((a)*(a))

 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))

 #define mem(a,b) memset(a,b,sizeof(a))

 #define eps (1e-10)

 #define J 10000

 #define mod 1000000007

 #define MAX 0x7f7f7f7f

 #define PI 3.14159265358979323

 #define N 100004

 using namespace std;

 typedef long long LL;

 double anss;

 LL aans;

 int cas,cass;

 LL n,m,lll,ans;

 int main()

 {

     #ifndef ONLINE_JUDGEW

 //    freopen("1.txt","r",stdin);

 //    freopen("2.txt","w",stdout);

     #endif

     int i,j,k;

     int x,y,z;

 //    init();

 //    for(scanf("%d",&cass);cass;cass--)

     for(scanf("%d",&cas),cass=;cass<=cas;cass++)

 //    while(~scanf("%s",s))

 //    while(~scanf("%d%d",&n,&m))

     {

         printf("Case #%d:\n",cass);

         scanf("%lld%lld",&n,&m);

         if(n==m)printf("%d\n%lld %lld\n",,n,m);

         else printf("%d\n%lld %lld\n%lld %lld\n",,n,m,m,n);

     }

     return ;

 }

 /*

 //

 //

 */