USACO-cha1-sec1.4 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 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

————————————————————口渴的分割线————————————————————

前言：这题还是蛮简单的。数据范围小可以直接暴搜。但是如果不注意剪枝是不行的。

思路：既然全部都是用p^2+q^2得到的结果构成的等差数列，那么开一个vis数组就行了。经过计算剪枝是这样的：

d（公差） <= 2*m*m/(len-1)

a1（首项） <= 2*m*m - d*(len-1)

代码如下：

/*ID: j.sure.1PROG: ariprogLANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <iostream>using namespace std;/****************************************/bool vis[1000000];int len, m;struct Node{int d;int a1;}ans[1000000];int main(){freopen("ariprog.in", "r", stdin);freopen("ariprog.out", "w", stdout);scanf("%d%d", &len, &m);for(int i = 0; i <= m; i++)for(int j = i; j <= m; j++) {vis[i*i + j*j] = true;}int cas = 0;for(int d = 1; d <= 2*m*m/(len-1); d++) {for(int a1 = 0; a1 <= 2*m*m - d*(len-1); a1++) if(vis[a1]) {for(int l = 1; l < len; l++) {int u = a1 + l*d;if(vis[u]) {if(l == len-1) {cas++;ans[cas].d = d;ans[cas].a1 = a1;}}else {break;}}}}if(!cas)puts("NONE");elsefor(int i = 1; i <= cas; i++) {printf("%d %d\n", ans[i].a1, ans[i].d);}fclose(stdin);fclose(stdout);return 0;}