【模拟枚举】Arithmetic Progressions等差数列(Usaco_Training 1.4)

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

 

依次枚举a和b即可，最后由于按照b排序，所以可以把b的循环放在外层，也可以用一个优先队列存储

C++ Code

/*ID: jiangzh15TASK: ariprogLANG: C++http://blog.csdn.net/jiangzh7*/#include<cstdio>#include<utility>#include<queue>using namespace std;bool h[250*250*2+10];int n,m,maxv;bool check(int a,int b){    for(int i=0;i<n;i++)        if(!h[a+i*b]) return false;    return true;}int main(){    freopen("ariprog.in","r",stdin);    freopen("ariprog.out","w",stdout);    scanf("%d%d",&n,&m);    int i,j;    for(i=0;i<=m;i++)        for(j=0;j<=m;j++)            h[i*i+j*j]=true;    maxv=m*m*2;    bool flag=false;    for(j=1;j<=maxv;j++)//由于按照b排序  所以b放在外层        for(i=0;i<=maxv;i++)        {            if(i+(n-1)*j>maxv)break;            if(check(i,j))            {                printf("%d %d\n",i,j);                flag=true;            }        }    if(!flag)printf("NONE\n");    return 0;}