UVA 10474 (13.08.04)

 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 and Q the number of queries Mina wouldmake. The next N lines would contain the numbers written on the N marbles. These marblenumbers will not come in any particular order. FollowingQ lines will have Q 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 position y.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

4 1
2
3
5
1
5
5 2
1
3
3
3
1
2
3
0 0

Sample Output

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

题意:

每一组数据的第一行包含两个数, 第一个数N是石头数, 第二个数Q是要找的石头编号

接下来N个石头, 数值是它们的编号

然后排序, 看看我们要找的石头是排在第几个.

做法:

我第一反应是桶排序, 不过自己还有一种方法, 所以先用自己想的方法写了一遍, 结果WA了

我的想法是: 不用排序, 一个for循环过去, 统计那些 编号值 比 我们要找的石头的编号 要小的石头个数, 并且, 标志一下数组里是否有我们要找的石头.

这个想法是简单直接的, 但是我也不知道错在哪里, 最后乖乖桶排序AC了

下面两份代码, 第一份是AC代码, 第二份是WA代码, 个人还是在深究第二份为何错, 我喜欢自己的想法~

AC:

#include<stdio.h>
#include<string.h>

int main() {
	int N, Q;
	int cas = 0;
	while(scanf("%d %d", &N, &Q) != EOF) {
		if(N == 0 && Q == 0)
			break;

		int num[10005], sum[10005];
		int tmp, aim;

		memset(sum, 0, sizeof(sum));
		memset(num, 0, sizeof(num));

		for(int i = 1; i <= N; i++) {
			scanf("%d", &tmp);
			num[tmp]++;
		}

		printf("CASE# %d:\n", ++cas);

		for(int i = 1; i <= 10000; i++)
			sum[i] = num[i] + sum[i-1];

		for(int i = 1; i <= Q; i++) {
			scanf("%d", &aim);
			if(num[aim])
				printf("%d found at %d\n", aim, sum[aim-1] + 1);
			else
				printf("%d not found\n", aim);
		}
	}
	return 0;
}

WA:

#include<stdio.h>

int N, Q;
int cas = 0;

int main() {
	while(scanf("%d %d", &N, &Q) != EOF) {
		if(N == 0 && Q == 0)
			break;

		int aim, pos, mark;
		int num[10001];

		for(int i = 0; i < N; i++)
			scanf("%d", &num[i]);

		printf("CASE# %d:\n", ++cas);

		for(int i = 0; i < Q; i++) {
			scanf("%d", &aim);
			pos = 1;
			mark = 0;
			for(int j = 0; j < N; j++) {
				if(num[j] < aim)
					pos++;
				if(num[j] == aim)
					mark = 1;
			}
			if(mark == 1)
				printf("%d found at %d\n", aim, pos);
			else
				printf("%d not found\n", aim);
		}
	}
	return 0;
}