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最新推荐文章于 2019-02-08 18:45:53 发布

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最新推荐文章于 2019-02-08 18:45:53 发布

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分类专栏：排序

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本文介绍了一种基于桶排序的游戏算法实现。通过设定数组记录各数值出现次数，并使用累积数组确定每个数值的位置，实现了快速查找特定数值的第一出现位置。适用于数值范围固定且较小的情况。

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Where is the Marble?

Raju and Meena love to play with Marbles. They have got a lotof marbles with numbers written on them. At the beginning, Rajuwould place the marbles one after another in ascending order ofthe numbers written on them. Then Meena would ask Raju tofind the first marble with a certain number. She would count1...2...3. Raju gets one point for correct answer, and Meena getsthe point if Raju fails. After some fixed number of trials thegame ends and the player with maximum points wins. Today it'syour chance to play as Raju. Being the smart kid, you'd be takingthe favor of a computer. But don't underestimate Meena, she hadwritten a program to keep track how much time you're taking togive all the answers. So now you have to write a program, whichwill help you in your role as Raju.

Input

There can be multiple test cases. Total no of test cases is less than 65. Each test case consistsbegins with 2 integers:N the number of marbles andQ the number of queries Mina wouldmake. The nextN lines would contain the numbers written on theN marbles. These marblenumbers will not come in any particular order. FollowingQ lines will haveQ queries. Beassured, none of the input numbers are greater than 10000 and none of them are negative.

Input is terminated by a test case where N = 0 andQ = 0.

Output

For each test case output the serial number of the case.

For each of the queries, print one line of output. The format of this line will depend uponwhether or not the query number is written upon any of the marbles. The two different formatsare described below:

`x found at y', if the first marble with numberx was found at positiony.Positions are numbered1, 2,...,N.
`x not found', if the marble with numberx is not present.

Look at the output for sample input for details.

Sample Input

Sample Output

CASE# 1:
5 found at 4
CASE# 2:
2 not found
3 found at 3

通过此题又见识到一种新排序: 桶排序
题意容易理解, 不讲
桶排序在此题的应用:
设定一个num数组, 下标为i, 表示标号为i的石块有几块;
再设定一个sum数组, 下标为j, 表示累计到标号为j的石块时, 一共有多少块石头
这样一来, sum[i-1]+1 即是 标号为i的石块在第几个, 即位置:

AC代码:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>

#define MAXN 10000+10

int N, Q;
int marble[MAXN];
int sum[MAXN];
int cas = 0;

int main() {
	int t;
	while(scanf("%d%d", &N, &Q) != EOF) {
		if(N == 0 && Q == 0)
			break;
		printf("CASE# %d:\n", ++cas);
		memset(marble, 0, sizeof(marble));
		memset(sum, 0, sizeof(sum));
		for(int i = 0; i < N; i++) {
			scanf("%d", &t);
			marble[t]++;	
		}
		for(int i = 1; i <= 10000; i++) 
			sum[i] = sum[i-1] + marble[i];
		for(int i = 0; i < Q; i++) {
			scanf("%d", &t);	
			if(marble[t] == 0)
				printf("%d not found\n", t);
			else
				printf("%d found at %d\n", t, sum[t-1] + 1);
		}
	}
}