USACO 1.4 Arithmetic Progressions

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 single 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)

1 4
37 4
2 8
29 8
1 12
5 12
13 12
17 12
5 20
2 24

题目大意：给你n和m，n表示目标等差数列的长度（等差数列由一个非负的首项和一个正整数公差描述），m表示p,q的范围，目标等差数列的长度必须严格等于n且其中每个元素都得属于集合{x|x=p^2+q^2}(0<=p<=m,0<=q<=m)，按顺序输出所有的目标数列。
思路：其实很简单，就是枚举，枚举起点和公差，一开始有点担心会超时，弄的自己神烦意乱的，但是实际上并没有。。。。。下面附上代码

 /*
 ID:fffgrdcc1
 PROB:ariprog
 LANG:C++
 */
 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 int a[*],cnt=,n,m;
 int bo[];
 struct str
 {
     int a;
     int b;
 }ans[];
 int tot=;
 bool check(int a,int b)
 {
     int temp=a+b+b,tt=m-;
     while(tt--)
     {
         if(temp>n*n*||!bo[temp])return ;
         temp+=b;
     }
     return ;
 }
 bool kong(str xx,str yy)
 {
     return xx.b<yy.b||(xx.b==yy.b&&xx.a<yy.a);
 }
 int main()
 {
     freopen("ariprog.in","r",stdin);
     freopen("ariprog.out","w",stdout);
     scanf("%d%d",&m,&n);
     for(int i=;i<=n;i++)
     {
         for(int j=i;j<=n;j++)
         {
                 bo[i*i+j*j]=;
         }
     }
     for(int i=;i<=n*n*;i++)
         if(bo[i])
             a[cnt++]=i;
     for(int i=;i<cnt;i++)
     {
         for(int j=i+;j<cnt;j++)
         {
             if(check(a[i],a[j]-a[i]))
             {
                 ans[tot].a=a[i];
                 ans[tot++].b=a[j]-a[i];
             }
         }
     }
     sort(ans,ans+tot,kong);
     if(!tot)printf("NONE\n");
     for(int i=;i<tot;i++)
     {
         printf("%d %d\n",ans[i].a,ans[i].b);
     }
     return ;
 }