usaco——ariprog

最新推荐文章于 2018-11-01 00:16:14 发布

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本文介绍了一种寻找特定长度算术级数的方法，该级数由双平方数构成。通过有效的算法设计，实现了对给定范围内所有可能序列的有效搜索，并确保结果按差值和起始值有序排列。

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这道题我一看蛮简单，写了50来行，结果time limit execeeded.后来搜了搜，知道了，是自己把等差写在外面而且是从1开始增。。。

要是遇到p》100的话，最大数字是100*100*2=20000；从1开始看，再在1下找是否存在，差不多有20000*100*100/2就是一亿了。。。，这个的确是太多了。当初这么做也是因为看答案是要求按d来升序排的。要是在生成的双平方数中搜，双平方大约有100*100/2个，5000个外层循环，里面可以精简不少，因为d可以从相邻的两个双平方数中得到，即d的初值和递增再也不是1和++，这样就非常高效。看了许多代码，下面一个我最喜欢，我也学习到了algorithm一个排序函数sort（），很好可以通过自己巧妙地定义比较函数，来实现加权排序。了不起。

Arithmetic Progressions

An arithmetic progression is a sequence of the form a, a+b, a+2b, ..., a+nb where n=0,1,2,3,... . For this problem, a is a non-negative integer and b is a positive integer.

Write a program that finds all arithmetic progressions of length n in the set S of bisquares. The set of bisquares is defined as the set of all integers of the form p² + q² (where p and q are non-negative integers).

TIME LIMIT: 5 secs

PROGRAM NAME: ariprog

INPUT FORMAT

Line 1:	N (3 <= N <= 25), the length of progressions for which to search
Line 2:	M (1 <= M <= 250), an upper bound to limit the search to the bisquares with 0 <= p,q <= M.

SAMPLE INPUT (file ariprog.in)

5
7

OUTPUT FORMAT

If no sequence is found, a singe line reading `NONE'. Otherwise, output one or more lines, each with two integers: the first element in a found sequence and the difference between consecutive elements in the same sequence. The lines should be ordered with smallest-difference sequences first and smallest starting number within those sequences first.

There will be no more than 10,000 sequences.

SAMPLE OUTPUT (file ariprog.out)

为了复习一遍，这次代码我自己打一遍；

#include <iostream>

#include <algorithm>

using namespace std;

const int MAX=250*250*2+10; //+10,I don't understand clearly;

bool flag[MAX]={};

int leg[MAX],size=0,res_cnt=0;

struct re

{

int a,b;

bool operator < (const re& x) const

{

return b<x.b||b==x.b&&a<x.a;

}

}res[10000];

int main()

{

freopen("ariprog.in","r",stdin);

freopen("ariprog.out","w",stdout);

int n,m;

cin>>n>>m;

int max=m*m*2;

for(int i=0; i<=m; ++i)

for(int j=i; j<=m; ++j)

flag[i*i+j*j]=1;

for(int i=0; i<=max;++i) if(flag[i]) leg[size++]=i;

for(int i=0; i<size; ++i)

for(int j=i+1; j<size; ++j)

{

int d=leg[j]-leg[i];

if(leg[i]+(n-1)*d>max) break;

for(int k=2; k<n; ++k)

if(!flag[leg[i]+k*d]) goto L1;

res[res_cnt].a=leg[i];

res[res_cnt++].b=d;

L1: ;

}

sort(res,res+res_cnt);

if(!res_cnt) cout<<"NONE"<<endl;

else

for(int i=0; i<res_cnt; ++i) cout<<res[i].a<<' '<<res[i].b<<endl;

return 0;

}

代码非常清楚，精简。测试数据中最大的0.41s，很不错。

学习！

#include <iostream> #include <algorithm> using namespace std; const int MAX=250*250*2+10; //+10,I don't understand clearly; bool flag[MAX]={}; int leg[MAX],size=0,res_cnt=0; struct re { int a,b; bool operator < (const re& x) const { return b<x.b||b==x.b&&a<x.a; } }res[10000]; int main() { freopen("ariprog.in","r",stdin); freopen("ariprog.out","w",stdout); int n,m; cin>>n>>m; int max=m*m*2; for(int i=0; i<=m; ++i) for(int j=i; j<=m; ++j) flag[i*i+j*j]=1; for(int i=0; i<=max;++i) if(flag[i]) leg[size++]=i; for(int i=0; i<size; ++i) for(int j=i+1; j<size; ++j) { int d=leg[j]-leg[i]; if(leg[i]+(n-1)*d>max) break; for(int k=2; k<n; ++k) if(!flag[leg[i]+k*d]) goto L1; res[res_cnt].a=leg[i]; res[res_cnt++].b=d; L1: ; } sort(res,res+res_cnt); if(!res_cnt) cout<<"NONE"<<endl; else for(int i=0; i<res_cnt; ++i) cout<<res[i].a<<' '<<res[i].b<<endl; return 0; }