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：求由p*p+q*q形成的方和等差数列。a+n*b 输出按b从小到大a从小到大输出 a,b;
思路：我使用枚举险过4.871ms。用bool数组存符合的方数判断。
注意点三个：1：第一个数a需要是方数，别忘判断
2：方数是由1—n的方和形成的，若用数组一次存1—250的方和寻找时会有可能出错哦，
如样例5 7中65 8 是2 是98内的方和数组，但不和题意 最后一个 97=81+16 但81明显不是7内的方数
3：剪枝：但a+n*b>2*n*n时后面的可以都不用搜了，直接break;
/*
ID:nealgav1
PROG:ariprog
LANG:C++
*/
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#define N 255
using namespace std;
bool vis[2*N*N];
int l,n;
int f[N];
class node
{
public:
int a,b;
}q[11000];
int num;
void search(int l,int n)
{
num=0;
int m=2*n*n;
int flag=0;
int z;
for(int i=0;i<m;i++)//a
{
for(int j=1;j<m;j++)//b
{
if(!vis[i])
break;
z=i;
flag=0;
for(int k=1;k<l;k++)
{
z+=j;
if(z>m)
{flag=2;break;}
if(!vis[z])
{flag=1;break;}
}
if(flag==2)break;
else if(flag==0)
{
q[num].a=i;q[num].b=j;num++;
}
}
}
}
bool cmp(node a,node b)
{
if(a.b==b.b)return a.a<b.a;
else return a.b<b.b;
}
int main()
{
freopen("ariprog.in","r",stdin);
freopen("ariprog.out","w",stdout);
for(int i=0;i<251;i++)
f[i]=i*i;
while(scanf("%d%d",&l,&n)!=EOF)
{memset(vis,0,sizeof(vis));
for(int i=0;i<=n;i++)
for(int j=0;j<=n;j++)
vis[f[i]+f[j]]=1;
search(l,n);
sort(q,q+num,cmp);
if(num==0)
printf("NONE\n");
else
for(int i=0;i<num;i++)
printf("%d %d\n",q[i].a,q[i].b);
}
}
USER: Neal Gavin Gavin [nealgav1] TASK: ariprog LANG: C++ Compiling... Compile: OK Executing... Test 1: TEST OK [0.011 secs, 3560 KB] Test 2: TEST OK [0.011 secs, 3560 KB] Test 3: TEST OK [0.011 secs, 3560 KB] Test 4: TEST OK [0.011 secs, 3560 KB] Test 5: TEST OK [0.043 secs, 3560 KB] Test 6: TEST OK [0.194 secs, 3560 KB] Test 7: TEST OK [2.106 secs, 3560 KB] Test 8: TEST OK [4.871 secs, 3560 KB] Test 9: TEST OK [4.806 secs, 3560 KB] All tests OK.
Your program ('ariprog') produced all correct answers! This is your submission #2 for this problem. Congratulations!
Here are the test data inputs:
------- test 1 ---- 3 2 ------- test 2 ---- 5 7 ------- test 3 ---- 14 10 ------- test 4 ---- 10 13 ------- test 5 ---- 12 50 ------- test 6 ---- 18 100 ------- test 7 ---- 21 200 ------- test 8 ---- 22 250 ------- test 9 ---- 25 250Keep up the good work!
Thanks for your submission!
This can be done by brute force, enumerating over all possible sequences. It has to be done a little carefully in order to get it to run in time, however.
Precalculate the bisquares, first of all. Calculate both a sorted list of the bisquares, along with a boolean array saying whether each number between 1 and 125000 (the maximum bisquare possible, for p,q < 250) is a bisquare.
Go through the skips in increasing order, starting at 1, and continuing along as the sequence starting at 1 with that skip doesn't exceed the maximum bisquare. For each bisquare, determine if the sequence starting at that location and with the current skip value consists of all of bisquares. If so, output it.
Here is the solution of Felix Arends from Germany (modified by Iran's Saber Fadaee):
#include <stdio.h>
#include <assert.h>
#include <string>
using namespace std;
// open files
FILE *fin = fopen ("ariprog.in", "r");
FILE *fout = fopen ("ariprog.out", "w");
// global variables
unsigned int N, M, maxMM;
unsigned int numbers [65000];
unsigned int number_size = 0;
unsigned char num_available [125001];
unsigned char dist_available [125001];
int have_res = 0;
int skipstep = 1;
// read the input
int read_input () {
fscanf (fin, "%d %d", &N, &M);
return 0;
}
int cmp_int (const void *a, const void *b) {
return (*(int *)a - *(int *)b);
}
void asm_num (int a, int b) {
for (unsigned int n = 1; n < N; n++)
if (num_available [a + n * b] == 0) return;
fprintf (fout, "%d %d\n", a, b);
have_res ++;
if (have_res==1)
skipstep = b;
}
void asm_num () {
for (unsigned int b = 1; b < maxMM; b+=skipstep) {
if (dist_available [b]) {
for (unsigned int p = 0; p < number_size && numbers [p] + (N -
1) * b <= maxMM; p++)
asm_num (numbers [p], b);
}
}
}
int process () {
memset (num_available, 0, sizeof (unsigned char) * 125001);
memset (dist_available, 0, sizeof (unsigned char) * 125001);
for (unsigned int m1 = 0; m1 <= M; m1++) {
for (unsigned int m2 = m1; m2 <= M; m2++) {
int n = m1 * m1 + m2 * m2;
if (!num_available [n]) {
num_available [n] = 1;
numbers [number_size++] = n;
}
}
}
qsort (numbers, number_size, sizeof (unsigned int), cmp_int);
maxMM = M * M + M * M;
for (unsigned int n1 = 0; n1 < number_size; n1++) {
unsigned int _n1 = numbers [n1];
for (unsigned int n2 = n1 + 1; n2 < number_size && _n1 + (numbers
[n2] - _n1) * (N - 1) <= maxMM; n2++) {
assert (numbers [n2] - _n1 >= 0 && numbers [n2] - _n1 < 125001);
if (num_available [_n1 + (numbers [n2] - _n1) * (N - 1)] &&
num_available [_n1 + (numbers [n2] - _n1) * (N - 2)])
dist_available [numbers [n2] - _n1] = true;
}
}
asm_num ();
if (!have_res) fprintf (fout, "NONE\n");
return 0;
}
int main () {
read_input ();
process ();
fclose (fin);
fclose (fout);
return 0;
}
Here is an even faster solution of "UaE ProGrammeR":
#include <fstream>
#include <iostream>
using namespace std;
void quicksort (int[], int ,int);
int pivotlist (int[], int,int);
ofstream out ("ariprog.out");
int n;
int main () {
ifstream in ("ariprog.in");
bool array[125001] = {false}, noneF;
int m, upper, upperdef, def, p;
int places[300000], pl = 0;
noneF = true;
in>>n>>m;
for (int i = 0; i <= m; i++)
for (int j = 0; j <= m; j++) {
if (!array[i*i+j*j]) {
places[pl] = i*i+j*j; //Saving generated numbers
pl++;
}
array[i*i+j*j] = true;
}
upper = 2*m*m;
upperdef = upper/(n-1);
quicksort (places, 0, pl-1);
for ( def = 1; def<=upperdef; def++) // Loop to check for solutions
// It looks for solutions in
// correct order so you
// print the solution directly
// without sorting first, thnx to who said:
// Trade Memory for Speed !!
{
for ( p = 0; places[p]<=(upper-((n-2)*def)); p++) {
bool is;
is = true;
int where;
for (int c = (n-1); c>=0 ; c--)
if (!array[places[p]+c*def]) {
is = false;
where = (p+c*def);
break;
}
if (is) {
noneF = false;
out<<places[p]<<" "<<def<<endl;
}
}
}
if (noneF)
out<<"NONE"<<endl;
return 0;
}
void quicksort (int array[], int start, int last) {
int pivot;
if (start < last) {
pivot = pivotlist(array, start,last);
quicksort (array, start,pivot-1);
quicksort (array, pivot+1,last);
}
}
int pivotlist(int array[], int f, int l) {
int pivotpoint;
int pivotvalue, temp;
pivotvalue = array[f];
pivotpoint = f;
for (int i = f+1;i<=l; i++) {
if (array[i]<pivotvalue) {
pivotpoint++;
temp = array[i];
array[i] = array[pivotpoint];
array[pivotpoint] = temp;
}
}
temp = array[f];
array[f] = array[pivotpoint];
array[pivotpoint] = temp;
return pivotpoint;
}
