Codeforces Round #392 (Div. 2)C Unfair Poll

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题目大意：

一个班有n行m列座位，老师要点k个学生，老师点名的顺序是从第1行。。。。第n-1行，第n行，第n-1行。。。。。第2行，这样重复，同一行中一定是从左到右的顺序点名，问最后被点到次数最多的人被点了几次，被点到次数最少的人被点了几次，还有坐在x行y列的人被点了几次。

题目解法：

本质上就是找规律题，除了n=1和n=2的情况要分开考虑，剩下的都可以一起考虑。然后就是边界的判断了。

代码：

#include "iostream"#include "cstdio"#include "math.h"#include "algorithm"#include "string"#include "string.h"#include "vector"#include "map"#include "queue"using namespace std;long long n, m, k, x, y;int main() {while (scanf("%lld%lld%lld%lld%lld", &n, &m, &k, &x, &y) != EOF){if (n == 1){long long num = k / (n*m);k -= num*n*m;long long ans1 = num, ans2 = num, ans3 = num;if (k != 0)ans1++;if (y <= k)ans3++;printf("%lld %lld %lld\n", ans1, ans2, ans3);}else if (n == 2){long long num = k / (n*m);k -= num*n*m;long long ans1 = num, ans2 = num, ans3 = num;if (k)ans1++;if ((x - 1)*m + y <= k)ans3++;printf("%lld %lld %lld\n", ans1, ans2, ans3);}else{long long num = k / (n*m + (n - 2)*m);long long mxnum = num << 1;long long mnnum = num;k -= num*(n*m + (n - 2)*m);long long ans1 = mnnum, ans2 = mxnum, ans3 = mxnum;if (k)ans1++;if (k>m)ans2++;if (k>n*m)ans2++;ans1 = max(ans2, ans1);ans2 = mnnum;if (k >= n*m)ans2++;if (x == 1 || x == n)ans3 = mnnum;if ((x - 1)*m + y <= k)ans3++;if (n*m<k){if (x != 1 && x != n){if (n*m + (n - x - 1)*m + y <= k)ans3++;}}printf("%lld %lld %lld\n", ans1, ans2, ans3);}}return 0;}

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