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 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)

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



期末啦，事情多啊，作业一大堆。。。  有很久没有A题啦~
发现不A题好像脑子都变笨了。。。
看样子还是要有经常做题的好习惯~

这是一道搜索+简单剪枝的题目：

题意：
    我研究了好半天。。。真难懂。。。
    大意是求长度为n的等差数列a+n*b；变为公差，使数列中的每一个数可以是在0~m范围内任意两个数的平方和。

暴搜的话，研究了一下时间复杂度，发现如果有5s说不定可以过。。。
直接枚举a，b的值；结果果然会超时。。。
想了一下，剪枝：
重点是：如果a+x*b>m*m*2，那么就肯定不存在这种等差数列。。。可以直接跳出对这种形式的a，b的研究，枚举下一个a
       还有，对于每一个a，先判断它是否可以被表示成平方和，可以的话再往下，不可以就直接判断下一个a。


代码：

/*
ID: jings_h1
PROG: ariprog
LANG: C++
*/

#include<iostream>
#include<fstream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[250*250*2+10];
struct node{
       int start;
       int end;
       };
bool cmp (const node & c,const node & d){
     if(c.end!=d.end)
         return c.end<d.end;
     else
         return c.start<d.start;
}
bool istrue(int v,int temp){
     if(a[v]==1){
                  return true;
                  }
     return false;
}
int main(){
    FILE *fin = fopen("ariprog.in", "r");
    FILE *fout = fopen("ariprog.out", "w");
   // ofstream fout ("ariprog.out");
   // ifstream fin ("ariprog.in");
    int n,m;
    memset(a,0,sizeof(a));
    fscanf(fin,"%d%d",&n,&m);
    //   cin>>n>>m;
       int times=0;
       node res[10005];
       int sum=0;
      // int a[250*250];
       for(int i=0;i<=m;i++){
               for(int j=i;j<=m;j++){
                       a[i*i+j*j]=1;
                     // a[times++]=i*i+j*j;
              //        cout<<a[times-1]<<endl;
                      }
                      }
       for(int k=0;(k+n-1)<=m*m*2;k++){
               if(!istrue(k,times))
                    continue;
               for(int g=1;;g++){
                       if((k+g*(n-1))>m*m*2)
                            break;
                       int j;
                       for(j=0;j<n;j++){
                               int temp=k+g*j;
                               if(!istrue(temp,times))
                                   break;
                                   }
                       if(j>=n){
                           res[sum].start=k;
                           res[sum].end=g;
                           sum++;
                           }
                           }
                           }
       if(sum==0){
          fprintf(fout,"NONE\n");
        // cout<<"NONE"<<endl;
        //  continue;
          }
       else{
          sort(res,res+sum,cmp);
          for(int k=0;k<sum;k++){
                 // cout<<res[k].start<<" "<<res[k].end<<endl;
                                   fprintf(fout,"%d %d\n",res[k].start,res[k].end);
          }
          }
      cin>>n;
    return 0;
}