HDU 5924 Mr. Frog’s Problem(想看證明的來)——2016CCPC東北地區大學生程序設計競賽 - 重現賽

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 HDU 5924 Mr. Frog’s Problem

Accept: 0    Submit: 0
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

 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 A/B+B/A≤C/D+D/C.

 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≤10^18).

 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

 Problem Idea

解題思路:

【題意】
給你兩個正整數A和B

要求找出所有的整數對(C,D)

滿足A≤C≤B,A≤D≤B且A/B+B/A≤C/D+D/C

【類型】
數學證明
【分析】
網上的很多題解貌似都直接說是規律就完事了

作爲一個合格的Acmer,我們應該要糾結一下爲什麼

∵C/D+D/C是對稱的

∴我們不妨假設D≥C

爲了簡化運算,我們令D=C+k(k≥0)

由上式可知,當k越大,C越小時,D/C+C/D越大

∵A≤C≤B

∴C(min)=A

此時,當k達到最大時,即爲D(max)=B

而該情況下,C/D+D/C恰好等於A/B+B/A

故滿足A/B+B/A≤C/D+D/C的解僅有A==C&&B==D||A==D&&B==C

而當A==B時,解唯一,即A==B==C==D

【時間複雜度&&優化】
O(1)

題目鏈接→HDU 5924 Mr. Frog’s Problem

 Source Code

/*Sherlock and Watson and Adler*/
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<bitset>
#include<cmath>
#include<complex>
#include<string>
#include<algorithm>
#include<iostream>
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define bitnum(a) __builtin_popcount(a)
using namespace std;
const int N = 100005;
const int M = 100005;
const int inf = 1000000007;
const int mod = 1000000007;
int main()
{
    int t,p=1;
    __int64 A,B;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d",&A,&B);
        printf("Case #%d:\n",p++);
        if(A!=B)
            printf("2\n%I64d %I64d\n%I64d %I64d\n",A,B,B,A);
        else
            printf("1\n%I64d %I64d\n",A,B);
    }
    return 0;
}

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