Arithmetic Progressions USACO1.4
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).
INPUT: N (3 <= N <= 25), the length of progressions for which to search M (1 <= M <= 250), an upper bound to limit the search to the bisquares with 0 <= p,q <= M.
OUTPUT: If no sequence is found, a single 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.
此题目需要一些小的剪枝，详见注释。 1
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using namespace std;
const int MAXN=100000+5;
int a[MAXN];
int aa[10005],bb[10005];
bool tab[MAXN];
int n,m,cnt,tot,mx;
int main()
{
freopen("ariprog.in","r",stdin);
freopen("ariprog.out","w",stdout);
scanf("%d%d",&n,&m);
for(rint i=0;i<=m;++i)
for(rint j=0;j<=m;++j)
a[++cnt]=i*i+j*j,tab[a[cnt]]=1;//预处理双平方数表，快速查表
sort(a+1,a+cnt);
cnt=unique(a+1,a+cnt+1)-a-1;
mx=m*m<<1;
int r=mx/(n-1);//公差上界，最大的数除以要求的长度
for(rint i=1;i<=r;++i)
{
for(rint j=1;j<=cnt;++j)
{
rint c=0;
for(rint k=n-1;k>0&&a[j]+i*k<=mx;--k)//若超过max退出循环
//从大到小枚举，不符合情况易退出
if(!tab[a[j]+i*k]) //若有一个不符合条件即break
break;
else ++c;
if(c==n-1)
{
aa[++tot]=a[j];
bb[tot]=i;
}
}
}
if(tot==0)
puts("NONE");
else
for(int i=1;i<=tot;++i)
printf("%d %d\n",aa[i],bb[i]);
return 0;
}