C. Pashmak and Buses
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which hasn students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for alld days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n,?k,?d (1?≤?n,?d?≤?1000; 1?≤?k?≤?109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
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Sample test(s)
input
3 2 2
output
1 1 2 1 2 1
input
3 2 1
output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.
题目大意：
n个人，k个公交，出去游玩d天，每天每个人可以选择任意一辆公交乘坐，最后要你求每天每个人选择的公交并输出，要求所有的d天中，不能至少有两个人一直在同一辆公交上。
解法：
把这题抽象一下，每个人，在1~K个数中选数字，组成d长度的序列，要求每个序列不能完全相同，数字可以重复选。
进一步抽象一下，每个人的乘坐公交的序列，就是d位k进制数，最多可以分配的序列个数为 k^d。如若 n
举个例子进一步说说明：
n = 8, k = 3, d = 4。 这里的0表示标记为1的公交，1为标记为2的公交，以此类推
第一个人的序列： 0, 0, 0, 0
第二个人的序列： 0, 0, 0, 1
第三个人的序列： 0, 0, 0, 2
第四个人的序列： 0, 0, 1, 0
第五个人的序列： 0, 0, 1, 1
第六个人的序列： 0, 0, 1, 2
第七个人的序列： 0, 0, 2, 0
第八个人的序列： 0, 0, 2, 1
代码：
#include#include #define LL long long#define N_max 1234using namespace std;int d, n, k;int f[N_max][N_max];void init() { cin >> n >> k >> d;}bool check() { LL cnt=1; for (int i = 1; i <= d; i++) { cnt *= k; if (n <= cnt) return true; } return false;}void solve() { if (check()) { for (int i = 2; i <= n; i++) { int tmp=0; f[i][1] = 1; for (int j = 1; j <= d; j++) { f[i][j] = f[i-1][j] + f[i][j] + tmp; if (f[i][j] >= k) { f[i][j] -= k; tmp = 1; } else tmp = 0; } } for (int j = 1; j <= d; j++) { for (int i = 1; i <= n; i++) printf("%d ", f[i][j]+1); printf("\n"); } } else printf("-1\n");}int main() { init(); solve();}
