zoukankan      html  css  js  c++  java
  • bzoj 3834 [Poi2014]Solar Panels 数论分块

    3834: [Poi2014]Solar Panels

    Time Limit: 20 Sec  Memory Limit: 128 MB
    Submit: 367  Solved: 285
    [Submit][Status][Discuss]

    Description

    Having decided to invest in renewable energy, Byteasar started a solar panels factory. It appears that he has hit the gold as within a few days  clients walked through his door. Each client has ordered a single rectangular panel with specified width and height ranges.
    The panels consist of square photovoltaic cells. The cells are available in all integer sizes, i.e., with the side length integer, but all cells in one panel have to be of the same size. The production process exhibits economies of scale in that the larger the cells that form it, the more efficient the panel. Thus, for each of the ordered panels, Byteasar would like to know the maximum side length of the cells it can be made of.
    n组询问,每次问smin<=x<=smax, wmin<=y<=wmax时gcd(x, y)的最大值。

    Input

    The first line of the standard input contains a single integer N(1<=N<=1000): the number of panels that were ordered. The following   lines describe each of those panels: the i-th line contains four integers Smin,Smax,Wmin,Wmax(1<=Smin<=Smax<=10^9,1<=Wmin<=Wmax<=10^9), separated by single spaces; these specify the minimum width, the maximum width, the minimum height, and the maximum height of the i-th panel respectively.

    Output

    Your program should print exactly n lines to the standard output. The i-th line is to give the maximum side length of the cells that the i-th panel can be made of.

    Sample Input

    4
    3 9 8 8
    1 10 11 15
    4 7 22 23
    2 5 19 24

    Sample Output

    8
    7
    2
    5

    HINT

    Explanation: Byteasar will produce four solar panels of the following sizes: 8*8 (a single cell), 7*14 (two cells), 4*22 or 6*22 (22 or 33 cells respectively), and 5*20 (four cells).

     

    Source

    鸣谢zhonghaoxi

    发现可以数论分块 

     1 #include<cstring>
     2 #include<cmath>
     3 #include<cstdio>
     4 #include<algorithm>
     5 #include<iostream>
     6 
     7 #define N 1007
     8 
     9 #define Wb putchar(' ')
    10 #define We putchar('
    ')
    11 #define rg register int
    12 using namespace std;
    13 inline int read()
    14 {
    15     int x=0,f=1;char ch=getchar();
    16     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    17     while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    18     return x*f;
    19 }
    20 inline void write(int x)
    21 {
    22     if(x<0) putchar('-'),x=-x;
    23     if (x==0) putchar(48);
    24     int num=0;char c[15];
    25     while(x) c[++num]=(x%10)+48,x/=10;
    26     while(num) putchar(c[num--]);
    27 }
    28 
    29 int ans;
    30 int mx1,mn1,mx2,mn2;
    31 
    32 
    33 int main()
    34 {
    35     int T=read();
    36     while(T--)
    37     {
    38         mn1=read(),mx1=read();
    39         mn2=read(),mx2=read();
    40         if (mx1>mx2) swap(mx1,mx2),swap(mn1,mn2);
    41         ans=1;
    42         if (mx1>=mn2) ans=mx1;
    43         else
    44         {
    45             mn1--,mn2--;
    46             for (rg i=mx1,last;i>=1;i=last)
    47             {
    48                 last=max(mx1/(mx1/i+1),mx2/(mx2/i+1));
    49                 if (mn1>=i) last=max(last,mn1/(mn1/i+1));
    50                 if (mn2>=i) last=max(last,mn2/(mn2/i+1));
    51                 if (mx1/i-mn1/i>0&&mx2/i-mn2/i>0)
    52                 {
    53                     ans=i;
    54                     break;
    55                 }
    56             }
    57         }
    58         write(ans),We;
    59     }
    60 }
  • 相关阅读:
    HDU1542矩形面积并
    HDU5869树状数组+gcd预处理
    HDU5845 Best Division
    整体二分
    Lattice 的 Framebuffer IP核使用调试笔记之datasheet笔记
    DDR3详解(以Micron MT41J128M8 1Gb DDR3 SDRAM为例)
    ISD1700系列多段语音录放系列
    Lattice Diamond 学习之编译、检查和设置约束
    欧拉函数的几个性质及证明
    CF776B Sherlock and his girlfriend
  • 原文地址:https://www.cnblogs.com/fengzhiyuan/p/8983619.html
Copyright © 2011-2022 走看看