1041 Be Unique (20 分)   Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1]. The first one who bets on a unique number wins. For examp…

PAT (Advanced Level) Practice 1041 Be Unique (20 分) 凌宸1642 题目描述: Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10 4 ]. The first…

Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1]. The first one who bets on a unique number wins. For example, if there are 7 peopl…

Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1]. The first one who bets on a unique number wins. For example, if there are 7 peopl…

题目 Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10​4 ]. The first one who bets on a unique number wins. For example, if there ar…

题意: 输入一个正整数N(<=1e5),接下来输入N个正整数.输出第一个独特的数(N个数中没有第二个和他相等的),如果没有这样的数就输出"None". AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; multiset<int>st; ]; int main(){ ios::sync_with_stdio(false);…

1041. Be Unique (20) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen fr…

题目链接:http://pat.zju.edu.cn/contests/pat-a-practise/1041 题目描述: Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1, 104]. The first one…

Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1, 104]. The first one who bets on a unique number wins. For example, if there are 7…

题目链接 https://www.patest.cn/contests/pat-a-practise/1041 思路 可以用 map 标记 每个数字的出现次数 然后最后再 遍历一遍 找到那个 第一个 第一次出现的数字 AC代码 #include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climit…

题目 Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1, 10^4]. The first one who bets on a unique number wins. For example, if there ar…

简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<string> #include<algorithm> using namespace std; int n; +]; +]; int main(…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6789189.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 水题,找出哪个数只出现过一次,输出那个数如果没有的话,输出None #include <iostream> #include <cstdio> #include <algorithm> #include <string.h&g…

一.技术总结 这题在思考的时候遇见了,不知道怎么处理输入顺序问题,虽然有记录每个的次数,事后再反过来需要出现一次的且在第一次出现, 这时我们其实可以使用另一个数组用来存储输入顺序的字符,然后再用另一个数组记录出现的次数,这样就可以解决这个问题了. 如果使用cin出现运行超时的情况可以改用scanf,但是这个我运行的时候没有出现超时. 我把下列代码 int number[N]; 改成 int number[10010]; 然后提交出现段错误.不知道什么原因..... 二.参考代码: #includ…

1041. Be Unique (20) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen fr…

1041 Be Unique (20 分) Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10​4​​]. The first one who bets on a unique number wins. For…

1041 Be Unique (20 分) Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10​4​​]. The first one who bets on a unique number wins. For…

1041 Be Unique(20 分) Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10​4​​]. The first one who bets on a unique number wins. For e…

1041 Be Unique (20 分) Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10​4​​]. The first one who bets on a unique number wins. For…

1068 万绿丛中一点红(20 分) 对于计算机而言,颜色不过是像素点对应的一个 24 位的数值.现给定一幅分辨率为 M×N 的画,要求你找出万绿丛中的一点红,即有独一无二颜色的那个像素点,并且该点的颜色与其周围 8 个相邻像素的颜色差充分大. 输入格式: 输入第一行给出三个正整数,分别是 M 和 N(≤ 1000),即图像的分辨率:以及 TOL,是所求像素点与相邻点的颜色差阈值,色差超过 TOL 的点才被考虑.随后 N 行,每行给出 M 个像素的颜色值,范围在 [0,2​24​​) 内.所有同…

1061 Dating (20 分) Sherlock Holmes received a note with some strange strings: Let's date! 3485djDkxh4hhGE 2984akDfkkkkggEdsb s&hgsfdk d&Hyscvnm. It took him only a minute to figure out that those strange strings are actually referring to the coded…

1061 Dating (20 分)   Sherlock Holmes received a note with some strange strings: Let's date! 3485djDkxh4hhGE 2984akDfkkkkggEdsb s&hgsfdk d&Hyscvnm. It took him only a minute to figure out that those strange strings are actually referring to the cod…

Poiuyt_cyc 博客园首页新随笔联系订阅管理随笔 - 11  文章 - 0  评论 - 111 抛弃EF,20分构建一个属于自己的ORM框架 相信EF大家都不陌生了,因为数据库表跟程序实体是一一对应的原因,我们能够通过lambda这种函数式的编程方式进行操作数据库,感觉非常清晰明了.与我们直接写SQL相比,lambda是强类型,拥有更好的扩展性,伸缩性,而且编程更加的方便,快捷..下面我们就基于Expression和lambda来与大家构建一个属于自己的ORM框架. 思路的话很简单,就是将…

6-2 邻接表存储图的广度优先遍历(20 分) 试实现邻接表存储图的广度优先遍历. 函数接口定义: void BFS ( LGraph Graph, Vertex S, void (*Visit)(Vertex) ); 其中LGraph是邻接表存储的图,定义如下: /* 邻接点的定义 */ typedef struct AdjVNode *PtrToAdjVNode; struct AdjVNode{ Vertex AdjV; /* 邻接点下标 */ PtrToAdjVNode Next; /*…

后面的测试点过不去,两个错误一个超时. 目前未解决   L1-009 N个数求和 (20 分)   本题的要求很简单,就是求N个数字的和.麻烦的是,这些数字是以有理数分子/分母的形式给出的,你输出的和也必须是有理数的形式. 输入格式: 输入第一行给出一个正整数N(≤100).随后一行按格式a1/b1 a2/b2 ...给出N个有理数.题目保证所有分子和分母都在长整型范围内.另外,负数的符号一定出现在分子前面. 输出格式: 输出上述数字和的最简形式 —— 即将结果写成整数部分 分数部分,其中分数部…

L1-023 输出GPLT (20 分) 给定一个长度不超过10000的.仅由英文字母构成的字符串.请将字符重新调整顺序,按GPLTGPLT....这样的顺序输出,并忽略其它字符.当然,四种字符(不区分大小写)的个数不一定是一样多的,若某种字符已经输出完,则余下的字符仍按GPLT的顺序打印,直到所有字符都被输出. 输入格式: 输入在一行中给出一个长度不超过10000的.仅由英文字母构成的非空字符串. 输出格式: 在一行中按题目要求输出排序后的字符串.题目保证输出非空. 输入样例: pcTclnG…

1074 宇宙无敌加法器 (20 分) 地球人习惯使用十进制数,并且默认一个数字的每一位都是十进制的.而在 PAT 星人开挂的世界里,每个数字的每一位都是不同进制的,这种神奇的数字称为“PAT数”.每个 PAT 星人都必须熟记各位数字的进制表,例如“……0527”就表示最低位是 7 进制数.第 2 位是 2 进制数.第 3 位是 5 进制数.第 4 位是 10 进制数,等等.每一位的进制 d 或者是 0(表示十进制).或者是 [2,9] 区间内的整数.理论上这个进制表应该包含无穷多位数字,但从实…

1044 火星数字 (20 分) 火星人是以 13 进制计数的: 地球人的 0 被火星人称为 tret. 地球人数字 1 到 12 的火星文分别为:jan, feb, mar, apr, may, jun, jly, aug, sep, oct, nov, dec. 火星人将进位以后的 12 个高位数字分别称为:tam, hel, maa, huh, tou, kes, hei, elo, syy, lok, mer, jou. 例如地球人的数字 29 翻译成火星文就是 hel mar:而火星文…

1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (one) from l (L in lowercase), or 0 (zero) fr…

package com.Summer_0420.cn; /** * @author Summer * 获取数值型数组中大于60的元素个数 * 给数值型数组中不足60分的加20分 */ public class TestMethod02 { public static void main(String[] args) { int [] a = {1,35,60,80,75,123,156,32,1}; show(a); } private static void show(int[] a) { i…