zoukankan      html  css  js  c++  java
  • POJ_2100_Graveyard_Design_(尺取法)

    描述


    http://poj.org/problem?id=2100

    求连续平方和=n的序列个数,并输出序列.

    Graveyard Design
    Time Limit: 10000MS   Memory Limit: 64000K
    Total Submissions: 5987   Accepted: 1416
    Case Time Limit: 2000MS

    Description

    King George has recently decided that he would like to have a new design for the royal graveyard. The graveyard must consist of several sections, each of which must be a square of graves. All sections must have different number of graves.
    After a consultation with his astrologer, King George decided that the lengths of section sides must be a sequence of successive positive integer numbers. A section with side length s contains s2 graves. George has estimated the total number of graves that will be located on the graveyard and now wants to know all possible graveyard designs satisfying the condition. You were asked to find them.

    Input

    Input file contains n --- the number of graves to be located in the graveyard (1 <= n <= 1014 ).

    Output

    On the first line of the output file print k --- the number of possible graveyard designs. Next k lines must contain the descriptions of the graveyards. Each line must start with l --- the number of sections in the corresponding graveyard, followed by l integers --- the lengths of section sides (successive positive integer numbers). Output line's in descending order of l.

    Sample Input

    2030

    Sample Output

    2
    4 21 22 23 24
    3 25 26 27

    Source

    Northeastern Europe 2004, Northern Subregion

    分析


    直接尺取.

    注意:

    1.如果要用开根计算的话要写成 " ll ub=(ll)sqrt(n*1.0); "写成 " ll ub=sqrt(n); "会CE.

    所以干脆写成 " r*r<=n "

     1 #include<cstdio>
     2 #include<queue>
     3 #define ll long long
     4 using std :: queue;
     5 
     6 struct node 
     7 {
     8     ll len,fst;
     9     node() {}
    10     node(ll a,ll b) : len(a),fst(b) {}
    11 };
    12 queue <node> q;
    13 ll n;
    14 
    15 inline ll val(ll x) { return x*x; }    
    16 
    17 int main()
    18 {
    19 #ifndef ONLINE_JUDGE
    20     freopen("grave.in","r",stdin);
    21     freopen("grave.out","w",stdout);
    22 #endif
    23     scanf("%lld",&n);
    24     ll l=1,r=0,k=0,len=0;
    25     ll sum=0;
    26     while(val(r)<=n)
    27     {
    28         if(sum<n)
    29         {
    30             sum+=val(++r);
    31             len++;
    32         }
    33         else if(sum>n)
    34         {
    35             sum-=val(l++);
    36             len--;
    37         }
    38         else
    39         {
    40             q.push(node(len,l));
    41             k++;
    42             sum-=val(l++);
    43             len--;
    44         }
    45     }
    46     printf("%lld",k);
    47     while(!q.empty())
    48     {
    49         node t=q.front(); q.pop();
    50         ll len=t.len,fst=t.fst;
    51         printf("
    %lld ",len);
    52         for(ll i=0;i<len;i++) printf("%lld ",fst+i);
    53     }
    54 #ifndef ONLINE_JUDGE
    55     fclose(stdin);
    56     fclose(stdout);
    57 #endif
    58     return 0;
    59 }
    View Code

     

  • 相关阅读:
    如何判断touch到子视图或离开视图
    NSString属性声明中的copy和retain区别
    iOS创建PDF文件
    NSString的内存分配及管理
    清除新微博Cookie
    Object System (对象系统)
    基于组件的游戏编程
    继电器srd05vdcslc
    JavaScript专题(二):深入理解iframe
    Eclipse插件安装maven svn ibatis openExplorer PropertiesEditor
  • 原文地址:https://www.cnblogs.com/Sunnie69/p/5428034.html
Copyright © 2011-2022 走看看