Moist-google

Problem

Moist has a hobby -- collecting figure skating trading cards. His card collection has been growing, and it is now too large to keep in one disorganized pile. Moist needs to sort the cards in alphabetical order, so that he can find the cards that he wants on short notice whenever it is necessary.

The problem is -- Moist can't actually pick up the cards because they keep sliding out his hands, and the sweat causes permanent damage. Some of the cards are rather expensive, mind you. To facilitate the sorting, Moist has convinced Dr. Horrible to build him a sorting robot. However, in his rather horrible style, Dr. Horrible has decided to make the sorting robot charge Moist a fee of $1 whenever it has to move a trading card during the sorting process.

Moist has figured out that the robot's sorting mechanism is very primitive. It scans the deck of cards from top to bottom. Whenever it finds a card that is lexicographically smaller than the previous card, it moves that card to its correct place in the stack above. This operation costs $1, and the robot resumes scanning down towards the bottom of the deck, moving cards one by one until the entire deck is sorted in lexicographical order from top to bottom.

As wet luck would have it, Moist is almost broke, but keeping his trading cards in order is the only remaining joy in his miserable life. He needs to know how much it would cost him to use the robot to sort his deck of cards.

Input

The first line of the input gives the number of test cases, TT test cases follow. Each one starts with a line containing a single integer, N. The next N lines each contain the name of a figure skater, in order from the top of the deck to the bottom.

Output

For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the number of dollars it would cost Moist to use the robot to sort his deck of trading cards.

Limits

1 ≤ T ≤ 100.
Each name will consist of only letters and the space character.
Each name will contain at most 100 characters.
No name with start or end with a space.
No name will appear more than once in the same test case.
Lexicographically, the space character comes first, then come the upper case letters, then the lower case letters.

Small dataset

1 ≤ N ≤ 10.

Large dataset

1 ≤ N ≤ 100.

Sample


Input 
 

Output 
 
2
2
Oksana Baiul
Michelle Kwan
3
Elvis Stojko
Evgeni Plushenko
Kristi Yamaguchi
Case #1: 1
Case #2: 0

推荐指数:※※

来源:google

这道题就是一个插入(排序)的次数。

注意读入数据的格式。

#include <iostream>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
const int N=110;
int compare_char(char *a,char *b){
	int i;
	for(i=0;a[i]!='\0'||b[i]!='\0';i++){
		if(a[i]!=b[i])
			return a[i]-b[i];
	}
}
void name_copy(char *a,char *b){
	int i=0;
	while(b[i]!='\0'){
		a[i]=b[i];
		i++;
	}
	a[i]='\0';
}
int main()
{
    freopen("C-small-1-attempt0.in", "r", stdin);
    freopen("a2.out", "w", stdout);

	int t_num,i,j,loop;
	scanf("%d",&t_num);
	for(loop=1;loop<=t_num;loop++){
		int pnum;
		scanf("%d",&pnum);
		getchar();
		char curr[N];
		int cost=0;
		if(pnum>0){
			gets(curr);
			for(i=1;i<pnum;i++){
				char tmp[N];
				gets(tmp);
				int co=compare_char(tmp,curr);
				if(co<0){
					cost++;
				}else if(co>0){
					name_copy(curr,tmp);
				}
			}
		}
		printf("Case #%d: %d\n",loop,cost);
	}
    return 0;
}


内容概要:论文提出了一种基于空间调制的能量高效分子通信方案(SM-MC),将传输符号分为空间符号和浓度符号。空间符号通过激活单个发射纳米机器人的索引来传输信息,浓度符号则采用传统的浓度移位键控(CSK)调制。相比现有的MIMO分子通信方案,SM-MC避免了链路间干扰,降低了检测复杂度并提高了性能。论文分析了SM-MC及其特例SSK-MC的符号错误率(SER),并通过仿真验证了其性能优于传统的MIMO-MC和SISO-MC方案。此外,论文还探讨了分子通信领域的挑战、优势及相关研究工作,强调了空间维度作为新的信息自由度的重要性,并提出了未来的研究方向和技术挑战。 适合人群:具备一定通信理论基础,特别是对纳米通信和分子通信感兴趣的科研人员、研究生和工程师。 使用场景及目标:①理解分子通信中空间调制的工作原理及其优势;②掌握SM-MC系统的具体实现细节,包括发射、接收、检测算法及性能分析;③对比不同分子通信方案(如MIMO-MC、SISO-MC、SSK-MC)的性能差异;④探索分子通信在纳米网络中的应用前景。 其他说明:论文不仅提供了详细的理论分析和仿真验证,还给出了具体的代码实现,帮助读者更好地理解和复现实验结果。此外,论文还讨论了分子通信领域的标准化进展,以及未来可能的研究方向,如混合调制方案、自适应调制技术和纳米机器协作协议等。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值