国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

A1056. Mice and Rice (25)

2019-11-08 03:16:01
字體:
來源:轉載
供稿:網友

1056. Mice and Rice (25)

時間限制100 ms內存限制65536 kB代碼長度限制16000 B判題程序Standard作者CHEN, Yue

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 NP programmers. Then every NG 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 NGwinners 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: NP and NG (<= 1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than NG 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 NP distinct non-negative numbers Wi (i=0,...NP-1) where each Wiis the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,...NP-1 (assume that the programmers are numbered from 0 to NP-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:
11 325 18 0 46 37 3 19 22 57 56 106 0 8 7 10 5 9 1 4 2 3Sample Output:
5 5 5 2 5 5 5 3 1 3 5
模擬排序
#include<cstdio>const int maxn = 1e3 + 10;int order[maxn], Rank[maxn], wgh[maxn], contest[maxn];  int main(){	int np, ng;	scanf("%d %d", &np, &ng);	for(int i = 0; i < np; ++i)		scanf("%d", &wgh[i]);	for(int i = 0; i < np; ++i)  //編號和競爭順序往后挪一位		scanf("%d", &order[i]);	while(1){		int total = 0;		for(int i = 0; i < np; ++i){			if(!Rank[order[i]]){   //位序i還沒有排序  ,則加入比較數組contest 				contest[total++] = order[i];   //total始終指向元素的后一位置			}		}		if(total == 1){         //剩余最后一位勝者 			Rank[contest[total - 1]] = 1;			break;		}		int low = 0, high = 0;   //處理競爭區間 		while(low < total){			if(low % ng == 0){				if(low + ng < total){					high = low + ng;				}else{					high = total;				}			}else{				++low;				continue;			}			int now = low, MAX = -1;			for(;now < high; ++now){				if(wgh[contest[now]] > MAX){					MAX = wgh[contest[now]];				}			}			for(now = low; now < high; ++now){				if(wgh[contest[now]] != MAX){					Rank[contest[now]] = total / ng + 1 + (total % ng ? 1 : 0);				}			}			++low;		}	}	for(int i = 0; i < np; ++i)		printf("%s%d", i ? " " : "", Rank[i]);	return 0;}
發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 措勤县| 赣州市| 晋城| 门源| 佳木斯市| 鄂托克旗| 三穗县| 防城港市| 康平县| 夹江县| 江都市| 汝南县| 阿拉尔市| 定西市| 屏山县| 类乌齐县| 博白县| 巩义市| 平顺县| 剑川县| 黄浦区| 浏阳市| 德清县| 大竹县| 安乡县| 张家口市| 岑溪市| 台湾省| 永州市| 安达市| 临邑县| 沙河市| 三明市| 当涂县| 芜湖县| 文登市| 大邑县| 城固县| 古交市| 深水埗区| 修水县|