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
解题思路：时限5S，果断爆搜。用一个数组has[i]记录数字i是否是bisquares，然后再用一个数组f记录所有的bisquares，这个预处理能够减少很多时间，因为在极限数据下has数组长度将达到125000，而实际bisquares总数才20000多。接下来的任务就是枚举a和b了，如果a+k*b>2*m*m或者a+k*b不是bisquares,那么可以break 循环，这个剪枝也能够减少时间。
程序写得比较丑陋o(╯□╰)o。
View Code
1 /* 2 ID:spcjv51 3 PROG:ariprog 4 LANG:C 5 */ 6 #include<stdio.h> 7 void qsort(long aa[],long bb[],long l,long r) 8 { 9 long i,j,mid1,mid2,temp; 10 i=l; 11 j=r; 12 mid1=aa[(l+r)/2]; 13 mid2=bb[(l+r)/2]; 14 while(i<=j) 15 { 16 while(aa[i]<mid1||(aa[i]==mid1&&bb[i]<mid2)) i++; 17 while(aa[j]>mid1||(aa[j]==mid1&&bb[j]>mid2)) j--; 18 if(i<=j) 19 { 20 temp=aa[i]; 21 aa[i]=aa[j]; 22 aa[j]=temp; 23 temp=bb[i]; 24 bb[i]=bb[j]; 25 bb[j]=temp; 26 i++; 27 j--; 28 } 29 } 30 if(i<r) qsort(aa,bb,i,r); 31 if(j>l) qsort(aa,bb,l,j); 32 33 } 34 35 int main(void) 36 { 37 freopen("ariprog.in","r",stdin); 38 freopen("ariprog.out","w",stdout); 39 long has[150000],f[30000],aa[10005],bb[10005]; 40 memset(has,0,sizeof(has)); 41 long i,j,ans,m,n,total,a,b,k; 42 scanf("%ld",&n); 43 scanf("%ld",&m); 44 ans=0; 45 total=0; 46 for(i=0; i<=m; i++) 47 for(j=0; j<=m; j++) 48 if(has[i*i+j*j]==0) 49 has[i*i+j*j]=1; 50 for(i=0;i<=2*m*m;i++) 51 if(has[i]) 52 { 53 f[ans]=i; 54 ans++; 55 56 } 57 for(i=0; i<ans-1; i++) 58 for(j=i+1; j<ans; j++) 59 { 60 a=f[i]; 61 b=f[j]-f[i]; 62 for(k=1; k<n; k++) 63 { 64 if((a+b*k)>f[ans-1]) break; 65 if(has[a+b*k]==0) break; 66 } 67 if(k==n) 68 { 69 aa[total]=a; 70 bb[total]=b; 71 total++; 72 73 } 74 } 75 qsort(bb,aa,0,total-1); 76 if(total==0) printf("NONE\n"); 77 for(i=0; i<total; i++) 78 printf("%ld %ld\n",aa[i],bb[i]); 79 return 0; 80 }