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 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

时间不超过5s,之前考虑到的只有双平方对称、从小到大排序、以及a+(n-1)b〉maxx的情况，a和b还分别从0，1穷举弱爆了。。然后发现要快就直接抽table里面的元素，然后生成table要做个标记。。。还是超时，所以有关搜索的就只有判断是否存在table的函数了，既然用了sort,那就再用一下<algorithm>的二分搜索函数binary_search(a,a+b,ele)嗯

/*ID: des_jas1PROG: ariprogLANG: C++*/#include <iostream>#include <fstream>#include <string>#include <string.h>#include <algorithm>#include <vector>#define fin cin#define fout coutusing namespace std;typedef struct seq{int a;int b;}seq;const int MAXN=10000,MAX=62500;int table[MAX+5],n,M,maxx,r=0;seq ans[MAXN+5];bool flag[MAX]={0};bool cmp(seq t1,seq t2){if(t1.b!=t2.b)return t1.b<t2.b;elsereturn t1.a<t2.a;}void createtable(){int i,j,temp;for(i=0;i<=M;i++)for(j=i;j<=M;j++) //矩阵的一半{temp=i*i+j*j;if(!flag[temp]){table[r]=temp;r++;flag[temp]=true;}}}/*bool IS_IN(int p){int i;for(i=0;i<r;i++){if(p==table[i])return true;if(p<table[i])return false;}return false;} */int main() {ofstream fout ("ariprog.out");    ifstream fin ("ariprog.in");int i,j,d,tp=1,cn=0,q=0;fin>>n>>M;    maxx=M*M*2;createtable();sort(table,table+r);for(i=0;i<r;i++)for(j=i+1;j<r;j++){d=table[j]-table[i];if(table[i]+(n-1)*d>maxx)break;for(cn=2;cn<n;cn++){tp=table[i]+d*cn;if(!binary_search(table,table+r,tp))break;}if(cn==n) {ans[q].a=table[i];ans[q].b=d;q++;}}sort(ans,ans+q,cmp);if(!q)fout<<"NONE"<<endl;else{for(i=0;i<q;i++)fout<<ans[i].a<<" "<<ans[i].b<<endl;}fout.close();fin.close();    return 0;}