UVA - 10673
| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
Problem A
Play with Floor and Ceil
Input: standard input
Output: standard output
Time Limit: 1 second
For any two integers x and k there exists two more integersp and q such that:
It’s a fairly easy task to prove this theorem, so we’d not ask you to do that. We’d ask for something even easier! Given the values ofx and k, you’d only need to find integers p and q that satisfies the given equation.
Input
The first line of the input contains an integer, T (1≤T≤1000) that gives you the number of test cases. In each of the following T lines you’d be given two positive integersx and k. You can safely assume that x andk will always be less than 108.
For each of the test cases print two integers: p and
q in one line. These two integers are to be separated by a single space. If there are multiple pairs ofp and
q that satisfy the equation, any one would do. But to help us keep our task simple, please make sure that the values,
and
fit
in a64 bit signed integer.
|
3 5 2 40 2 24444 6 |
1 1 1 1 0 6
|
Problem setter: Monirul Hasan, Member of Elite Problemsetters‘ Panel
Special Thanks: Shahriar Manzoor, Member of Elite Problemsetters‘ Panel
Source
Root :: Prominent Problemsetters :: Monirul Hasan
数学类简单题。。
题意：就是找有没有符合题中那个式子的p和q，有就输出p和q
思路：先从p入手，0到k扫一边，如果存在有q可以使式子满足就break，再输出p和q
简单说下floor和ceil，他们都是math头文件中的库函数，floor表示向下取整，ceil表示向上取整
AC代码：
/*************************************************************************
> File Name: b.cpp
> Author: zzuspy
> Mail: zzuspy@qq.com
> Created Time: 2014年12月01日 星期一 21时41分53秒
************************************************************************/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#define LL long long
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
using namespace std;
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int x, k;
scanf("%d %d", &x, &k);
int p, q, fl = (int)floor((double)x/k), ce = (int)ceil((double)x/k);
for(p = 0; p <= k; p++)
{
q = (x-p*fl)/ce;
if((LL)p*fl+(LL)q*ce == (LL)x) //判断q是否成立，这里p*fl要加个（LL），防止int溢出
break;
}
printf("%d %d\n", p, q);
}
return 0;
}UVA - 10673 - Play with Floor and Ceil （简单数学！）
原文地址：http://blog.csdn.net/u014355480/article/details/41654159
