题目链接：http://codeforces.com/problemset/problem/459/C
题意：n个人， k辆车， d天，每天将所有 任意人安排到k辆车， 问怎样安排， 可时不存在 2人或2人以上 d天都在一起。
题解：可以构造一个数组A， n行d列，0<a[i][j] <= k. 不存在两行完全相等，由此可得 k^d < n 可以构造以个k进制数， 从0 加到n
不得不说智商是硬伤T T，看了赛后题解才恍然大悟。
1 /***Good Luck***/ 2 #define _CRT_SECURE_NO_WARNINGS 3 #include <iostream> 4 #include <cstdio> 5 #include <cstdlib> 6 #include <cstring> 7 #include <string> 8 #include <algorithm> 9 #include <stack> 10 #include <map> 11 #include <queue> 12 #include <vector> 13 #include <set> 14 #include <functional> 15 #include <cmath> 16 17 #define Zero(a) memset(a, 0, sizeof(a)) 18 #define Neg(a) memset(a, -1, sizeof(a)) 19 #define All(a) a.begin(), a.end() 20 #define PB push_back 21 #define inf 0x3f3f3f3f 22 #define inf2 0x7fffffffffffffff 23 #define ll long long 24 using namespace std; 25 //#pragma comment(linker, "/STACK:102400000,102400000") 26 void get_val(int &a) { 27 int value = 0, s = 1; 28 char c; 29 while ((c = getchar()) == ' ' || c == ' '); 30 if (c == '-') s = -s; else value = c - 48; 31 while ((c = getchar()) >= '0' && c <= '9') 32 value = value * 10 + c - 48; 33 a = s * value; 34 } 35 const int maxn = 1005; 36 int ret[maxn][maxn]; 37 int n, k, d; 38 39 40 41 int main() { 42 //freopen("data.out", "w", stdout); 43 //freopen("data.in", "r", stdin); 44 //cin.sync_with_stdio(false); 45 cin >> n >> k >> d; 46 bool flag = false; 47 int tmp = 1; 48 for (int i = 0; i <= d; ++i) 49 if (tmp < n) tmp *= k; 50 else { 51 flag = true; 52 break; 53 } 54 if (!flag) { 55 cout << -1 << endl; 56 return 0; 57 } 58 int jz = 0; 59 for (int i = 2; i <= n; ++i) { 60 jz = 1; 61 for (int j = 1; j <= d; ++j) { 62 ret[i][j] = (ret[i - 1][j] + jz) % k; 63 if (ret[i][j] < ret[i - 1][j]) jz = 1; 64 else jz = 0; 65 } 66 } 67 for (int i = 1; i <= d; ++i) { 68 for (int j = 1; j < n; ++j) { 69 printf("%d ", ret[j][i] + 1); 70 } 71 printf("%d ", ret[n][i] + 1); 72 } 73 return 0; 74 }