Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse.

First the playing order is randomly decided for N P programmers. Then every N G programmers are grouped in a match. The fattest mouse in a group wins and enters the next turn. All the losers in this turn are ranked the same. Every N G winners are then grouped in the next match until a final winner is determined.

For the sake of simplicity, assume that the weight of each mouse is fixed once the programmer submits his/her code. Given the weights of all the mice and the initial playing order, you are supposed to output the ranks for the programmers.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N P and N G (≤1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than N G mice at the end of the player’s list, then all the mice left will be put into the last group. The second line contains N P distinct non-negative numbers W i (i=0,⋯,N P − 1) where each W i is the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,⋯,N P − 1 (assume that the programmers are numbered from 0 to N P - 1). All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the final ranks in a line. The i-th number is the rank of the i-th programmer, and all the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

11 3
25 18 0 46 37 3 19 22 57 56 10
6 0 8 7 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

注意点

在计算当前循环所要确定的排名时，需要分为是否能整除两种情况：

1. 如果剩下还未确定排名的数量（len）不能整除每组的人数（ng），则该轮循环可以确定的排名是 len / ng + 2
2. 否则，该轮循环可以确定的排名是 len / ng + 1

#include <cstdio>
#include <queue>
using namespace std;

const int maxn = 1000;

int main() {
    int np, ng, index, w[maxn], rank[maxn];
    scanf("%d %d", &np, &ng);
    queue<int> que;
    for (int i = 0; i < np; i++) {
        scanf("%d", &w[i]);
    }
    for (int i = 0; i < np; i++) {
        scanf("%d", &index);
        que.push(index);
    }
    int len = que.size(), nowRank;
    while (len > 1) {
        if (len % ng == 0) {
            nowRank = len / ng + 1;
        } else {
            nowRank = len / ng + 2;
        }
        for (int i = 0; i < len; i += ng) {
            int maxW = -1, maxI = -1;
            for (int j = i; j < i + ng && j < len; j++) {
                int tempI = que.front();
                if (w[tempI] > maxW) {
                    maxW = w[tempI];
                    maxI = tempI;
                }
                rank[tempI] = nowRank;
                que.pop();
            }
            que.push(maxI);
        }
        len = que.size();
    }
    rank[que.front()] = 1;
    for (int i = 0; i < np; i++) {
        printf("%d", rank[i]);
        if (i != np - 1) {
            printf(" ");
        }
    }
    return 0;
}