题目链接:codeforces 492e vanya and field 留个扩展gcd求逆元的板子. 设i,j为每颗苹果树的位置,因为gcd(n,dx) = 1,gcd(n,dy) = 1,所以当走了n步后,x从0~n-1,y从0~n-1都访问过,但x,y不相同. 所以,x肯定要经过0点,所以我只需要求y点就可以了. i,j为每颗苹果树的位置,设在经过了a步后,i到达了0,j到达了M. 则有 1----------------------(i + b * dx) % n = 0 2------…

E. Vanya and Field time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vanya decided to walk in the field of size n × n cells. The field contains m apple trees, the i-th apple tree is at the c…

E. Vanya and Field time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vanya decided to walk in the field of size n × n cells. The field contains m apple trees, the i-th apple tree is at the c…

E. Vanya and Field Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/492/problem/E Description Vanya decided to walk in the field of size n × n cells. The field contains m apple trees, the i-th apple tree is at the cell with…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud 本场题目都比较简单,故只写了E题. E. Vanya and Field Vanya decided to walk in the field of size n × n cells. The field contains m apple trees, the i-th apple tree is at the cell with coordinates(xi, yi). Van…

E. Vanya and Field time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vanya decided to walk in the field of size n × n cells. The field contains m apple trees, the i-th apple tree is at the c…

Vanya and Field 题目链接:http://www.codeforces.com/problemset/problem/492/E 逆元 刚看到这题的时候一脸懵逼不知道从哪下手好,于是打表找规律.但是打出来的东西完全不能看啊,有个鬼规律(╯‵□′)╯︵┻━┻,是我数据处理不当?按x排序后发现从一个点出发可以到达任意一个x坐标,回过去看题,发现有这么一句话:The following condition is satisfied for the vector: ,恩,好像有点思路了.(…

E. Vanya and Field time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vanya decided to walk in the field of size n × n cells. The field contains m apple trees, the i-th apple tree is at the c…

链接 Codeforces 677D Vanya and Treasure 题意 n*m中有p个type,经过了任意一个 type=i 的各自才能打开 type=i+1 的钥匙,最初有type=1的钥匙, 问拿到type=p的钥匙最少需要走多少步 思路 第一想法就是按type来递推, 将type相同的存到一起,dp[i][j]=min(dp[i][j], dp[k][l]+distance([i][j], [k][l])),其中 a[i][j] = a[k][l]+1. 但这样type相同的个数…

http://codeforces.com/contest/492/problem/E 一开始没时间想,,诶真是.. 挺水的一道题.. 将每个点的横坐标都转换成0,然后找纵坐标有多少即可..即解方程 $$a \times dx \equiv x(mod n)$$ 然后注意开long long #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <…