USACO Section 1.4 Arithmetic Progressions（枚举）

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

题意：求出一个等差数列的首项 a 和公差 d 使得前 n 项都是双平方数即可分解成 p^2+q^2 且 p、q <= m 。
分析：打表所有的双平方数，生成两个数组，一个数组用做标记 flag 另一个存储所有的双平方数 bis。然后开始枚举存储数组 bis。
心得：起初我直接打表 250*250 的所以造成 bis 数组中的值是否满足 p、q <= m 这个不好直接判断，于是我又用一个vector把所以的 p、q 打表时都存起来，于是又多了一层循环。如果输入一个 m ，再打表的话可就巧了，不仅不需要判断而且速度也快多了。

View Code

 1 /*
 2 ID: dizzy_l1
 3 LANG: C++
 4 TASK: ariprog
 5 */
 6 #include<iostream>
 7 #include<algorithm>
 8 #include<cstring>
 9 #include<cstdio>
10 #include<vector>
11 #define MAXN 10001
12 
13 using namespace std;
14 
15 struct ANS
16 {
17     int a,d;
18 } ans[MAXN];
19 bool flag[250*250*2+10];
20 int bis[250*250*2+10],cnt,cntn,N,M;
21 
22 void Init()
23 {
24     int i,j;
25     cnt=0;
26     memset(flag,false,sizeof(flag));
27     for(i=0; i<=M; i++)
28     {
29         for(j=i; j<=M; j++)
30         {
31             if(!flag[i*i+j*j])
32                 bis[cnt++]=i*i+j*j;
33             flag[i*i+j*j]=true;
34         }
35     }
36     sort(bis,bis+cnt);
37 }
38 
39 bool cmp(ANS A,ANS B)
40 {
41     if(A.d<B.d) return true;
42     if(A.d==B.d&&A.a<B.a) return true;
43     return false;
44 }
45 
46 int main()
47 {
48     //freopen("ariprog.in","r",stdin);
49     //freopen("ariprog.out","w",stdout);
50     int i,j,k,d;
51     scanf("%d%d",&N,&M);N--;
52     Init();
53     cntn=0;
54     for(i=0; i<cnt; i++)
55     {
56         for(j=i+1; j<cnt; j++)
57         {
58             d=bis[j]-bis[i];
59             if(bis[i]+d*N>M*M*2) break;
60             for(k=1; k<=N; k++)
61             {
62                 if(!flag[bis[i]+k*d])
63                     break;
64             }
65             if(k==N+1)
66             {
67                 ans[cntn].a=bis[i];
68                 ans[cntn++].d=d;
69             }
70         }
71     }
72     sort(ans,ans+cntn,cmp);
73     if(cntn==0)
74     {
75         printf("NONE\n");
76     }
77     else
78     {
79         for(i=0; i<cntn; i++)
80             printf("%d %d\n",ans[i].a,ans[i].d);
81     }
82     return 0;
83 }