1.4.3---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 searchLine 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)

57

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)

1 437 42 829 81 125 1213 1217 125 202 24

#include <iostream>#include <vector>#include <cstdio>#include <fstream>#include <cstring>#include <algorithm>using namespace std;struct node{//定义结构体存储a,b结果int a;int b;};int exist[126002];//标识2*250*250内各数是否出现int st[126002];//用于记录可用的双平方数集合int main(){ofstream fout ("ariprog.out");ifstream fin ("ariprog.in");int n,m,temp,length,count;vector <node> no1;no1.clear();//count记录结果规模count = 0;fin>>n>>m;//length用于记录可用双平方数集合的规模length = 0;memset (exist,0,sizeof(exist));for(int p = 0;p <= m;p ++)for(int q = p;q <= m; q ++){temp = p * p + q *q;if(!exist[temp]){exist[temp] = 1;st[length] = temp;length ++;}}sort(st,st+length-1);//若可用的双平方数个数小于等差数列长度，则直接失败if((length-1)<n){fout<<"NONE"<<endl;return 0;}for(int i = 0; i <= length - n;i ++)for(int j = i+1; j <= length - n +1;j ++){//由于b>0 所以排除b=0的情况if(st[i]==st[j])continue; //检测假设等差数列最大值能否满足if((st[i] + (n-1)*(st[j]-st[i]))>st[length -1])break;int tempn;tempn = n-2;//检测除当前两点外其余各点能否满足while(tempn){if(exist[st[i]+(n-tempn)*(st[j]-st[i])])tempn--;elsebreak;}//通过检测则记录当前a,bif(!tempn){node nodetemp;nodetemp.a = st[i];nodetemp.b = st[j]-st[i];no1.push_back(nodetemp);count++;}}if(!count)fout<<"NONE"<<endl;else{//对结果集按要求进行排序，肯定有比选择排序更适合的方法，目前还没想如何实现，姑且先用选择吧。int k;for(int i=0;i<count-1;i++){k=i;node notemp;for(int j=i+1;j<count;j++)//先按b进行升序排序，若b相等则按a进行升序排序if(no1[j].b<no1[k].b||((no1[j].b==no1[k].b)&&no1[j].a<no1[k].a))k=j;if(k!=i){notemp = no1[i];no1[i] = no1[k];no1[k] = notemp;}}for(int i=0;i<count;i++)fout<<no1[i].a<<" "<<no1[i].b<<endl;}return 0;}