HDU 5924 Mr. Frog’s Problem 模拟 (2016CCPC东北地区大学生程序设计竞赛)

Posted 2020-08-15

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

 

 

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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。

 

 1 //HDU 5924
 2 //by coolxxx
 3 //#include<bits/stdc++.h>
 4 #include<iostream>
 5 #include<algorithm>
 6 #include<string>
 7 #include<iomanip>
 8 #include<map>
 9 #include<stack>
10 #include<queue>
11 #include<set>
12 #include<bitset>
13 #include<memory.h>
14 #include<time.h>
15 #include<stdio.h>
16 #include<stdlib.h>
17 #include<string.h>
18 //#include<stdbool.h>
19 #include<math.h>
20 #pragma comment(linker,"/STACK:1024000000,1024000000")
21 #define min(a,b) ((a)<(b)?(a):(b))
22 #define max(a,b) ((a)>(b)?(a):(b))
23 #define abs(a) ((a)>0?(a):(-(a)))
24 #define lowbit(a) (a&(-a))
25 #define sqr(a) ((a)*(a))
26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
27 #define mem(a,b) memset(a,b,sizeof(a))
28 #define eps (1e-10)
29 #define J 10000
30 #define mod 1000000007
31 #define MAX 0x7f7f7f7f
32 #define PI 3.14159265358979323
33 #define N 100004
34 using namespace std;
35 typedef long long LL;
36 double anss;
37 LL aans;
38 int cas,cass;
39 LL n,m,lll,ans;
40 
41 int main()
42 {
43     #ifndef ONLINE_JUDGEW
44 //    freopen("1.txt","r",stdin);
45 //    freopen("2.txt","w",stdout);
46     #endif
47     int i,j,k;
48     int x,y,z;
49 //    init();
50 //    for(scanf("%d",&cass);cass;cass--)
51     for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
52 //    while(~scanf("%s",s))
53 //    while(~scanf("%d%d",&n,&m))
54     {
55         printf("Case #%d:\\n",cass);
56         scanf("%lld%lld",&n,&m);
57         if(n==m)printf("%d\\n%lld %lld\\n",1,n,m);
58         else printf("%d\\n%lld %lld\\n%lld %lld\\n",2,n,m,m,n);
59     }
60     return 0;
61 }
62 /*
63 //
64 
65 //
66 */

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