数论
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Magic Pairs
Problem's Link
Mean:
已知N、A0、B0,对于给定X、Y,若A0X+B0Y能被N整除,则AX+BY也能被N整除,求所有的A、B.(0<=A、B<N)
analyse:
这道题目就是先把A0和B0和N都除以他们的最大公约数,然后就是枚举A0和B0的0到N-1倍.
最后快排,注意一下有0的情况.
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-01-08-10.51
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
int main()
{
pair<int, int> ans[10010];
int n,a0,b0;
scanf("%d%d%d",&n,&a0,&b0);
int i = a0 %= n;
int j = b0 %= n;
int num = 0;
do
{
ans[num++] = make_pair(i, j);
i = (i+a0)%n;
j = (j+b0)%n;
}
while(i != a0 || j != b0);
sort(ans, ans+num);
printf("%d\n",num);
for(i = 0; i < num; i++)
printf("%d %d\n",ans[i].first,ans[i].second);
return 0;
}
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-01-08-10.51
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
int main()
{
pair<int, int> ans[10010];
int n,a0,b0;
scanf("%d%d%d",&n,&a0,&b0);
int i = a0 %= n;
int j = b0 %= n;
int num = 0;
do
{
ans[num++] = make_pair(i, j);
i = (i+a0)%n;
j = (j+b0)%n;
}
while(i != a0 || j != b0);
sort(ans, ans+num);
printf("%d\n",num);
for(i = 0; i < num; i++)
printf("%d %d\n",ans[i].first,ans[i].second);
return 0;
}
0 0