From 253becc0eaead9744fc748c1a026cf7212f30df9 Mon Sep 17 00:00:00 2001 From: Oliver Lew Date: Thu, 23 Apr 2020 09:54:17 +0800 Subject: [PATCH] add advanced 1035, 1036, 1040, 1041 --- _articles/a1035.md | 67 ++++++++++++++++++ _articles/a1036.md | 64 +++++++++++++++++ _articles/a1040.md | 55 +++++++++++++++ _articles/a1041.md | 46 ++++++++++++ _articles/analysis/a1035.md | 8 +++ _articles/analysis/a1036.md | 6 ++ _articles/analysis/a1040.md | 8 +++ _articles/analysis/a1041.md | 8 +++ _articles/others/a1035.md | 137 ++++++++++++++++++++++++++++++++++++ _articles/others/a1036.md | 113 +++++++++++++++++++++++++++++ _articles/others/a1040.md | 77 ++++++++++++++++++++ _articles/others/a1041.md | 86 ++++++++++++++++++++++ 12 files changed, 675 insertions(+) create mode 100644 _articles/a1035.md create mode 100644 _articles/a1036.md create mode 100644 _articles/a1040.md create mode 100644 _articles/a1041.md create mode 100644 _articles/analysis/a1035.md create mode 100644 _articles/analysis/a1036.md create mode 100644 _articles/analysis/a1040.md create mode 100644 _articles/analysis/a1041.md create mode 100644 _articles/others/a1035.md create mode 100644 _articles/others/a1036.md create mode 100644 _articles/others/a1040.md create mode 100644 _articles/others/a1041.md diff --git a/_articles/a1035.md b/_articles/a1035.md new file mode 100644 index 0000000..547c34b --- /dev/null +++ b/_articles/a1035.md @@ -0,0 +1,67 @@ +--- +layout: post +date: 2020-04-22 14:08:00 +0800 +title: "PAT Advanced 1035. Password (20) (C语言实现)" +categories: Advanced +tags: [string processing] +permalink: Advanced/1035.html +--- + +## 题目 + +{% include_relative html/a1035.md %} + +## 思路 + +{% include_relative analysis/a1035.md %} + +## 代码 + +[Github最新代码](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1035.c)，欢迎交流 + +```c +{% raw %}#include + +int main() +{ + int N, count = 0, modified; + char username[1000][11], password[1000][11]; + + scanf("%d", &N); + for(int i = 0; i < N; i++) + { + modified = 0; + scanf("%s %s", username[i], password[i]); + for(char *p = password[i]; *p; p++) + { + switch(*p) + { + case '1': *p = '@'; modified = 1; break; + case '0': *p = '%'; modified = 1; break; + case 'l': *p = 'L'; modified = 1; break; + case 'O': *p = 'o'; modified = 1; break; + default: break; + } + } + if(modified) + count++; + else /* mark unmodified password */ + password[i][0] = '\0'; + } + + if(count) + { + printf("%d\n", count); + for(int i = 0; i < N; i++) + if(password[i][0] != '\0') + printf("%s %s\n", username[i], password[i]); + } + else + { + printf("There %s %d %s and no account is modified", + N > 1 ? "are" : "is", N, N > 1 ? "accounts" : "account"); + } + + return 0; +} +{% endraw %}``` \ No newline at end of file diff --git a/_articles/a1036.md b/_articles/a1036.md new file mode 100644 index 0000000..58c7f0c --- /dev/null +++ b/_articles/a1036.md @@ -0,0 +1,64 @@ +--- +layout: post +date: 2020-04-22 14:34:20 +0800 +title: "PAT Advanced 1036. Boys vs Girls (25) (C语言实现)" +categories: Advanced +tags: [maxima and minima] +permalink: Advanced/1036.html +--- + +## 题目 + +{% include_relative html/a1036.md %} + +## 思路 + +{% include_relative analysis/a1036.md %} + +## 代码 + +[Github最新代码](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1036.c)，欢迎交流 + +```c +{% raw %}#include + +typedef struct student { + char name[11]; + char gender; + char ID[11]; +} Student; + +int main() +{ + int N, grade, gradeF = -1, gradeM = 101; + Student students[101] = {0}, s; + + scanf("%d", &N); + for(int i = 0; i < N; i++) + { + scanf("%s %c %s %d", s.name, &s.gender, s.ID, &grade); + students[grade] = s; + if(s.gender == 'F') + gradeF = grade > gradeF ? grade : gradeF; + else + gradeM = grade < gradeM ? grade : gradeM; + } + + if(gradeF != -1) + printf("%s %s\n", students[gradeF].name, students[gradeF].ID); + else + printf("Absent\n"); + + if(gradeM != 101) + printf("%s %s\n", students[gradeM].name, students[gradeM].ID); + else + printf("Absent\n"); + + if(gradeM == 101 || gradeF == -1) + printf("NA"); + else + printf("%d", gradeF - gradeM); + + return 0; +} +{% endraw %}``` \ No newline at end of file diff --git a/_articles/a1040.md b/_articles/a1040.md new file mode 100644 index 0000000..808089d --- /dev/null +++ b/_articles/a1040.md @@ -0,0 +1,55 @@ +--- +layout: post +date: 2020-04-23 01:27:45 +0800 +title: "PAT Advanced 1040. Longest Symmetric String (25) (C语言实现)" +categories: Advanced +tags: [string processing] +permalink: Advanced/1040.html +--- + +## 题目 + +{% include_relative html/a1040.md %} + +## 思路 + +{% include_relative analysis/a1040.md %} + +## 代码 + +[Github最新代码](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1040.c)，欢迎交流 + +```c +{% raw %}#include +#include + +int main() +{ + int maxlen = 0, l, L; + char string[1002]; + + fgets(string, 1001, stdin); + + L = strlen(string); + for(int i = 0; i < L; i++) + { + /* odd symmetric */ + l = 0; + while(i - l >= 0 + && i + l < L + && string[i - l] == string[i + l]) + l++; + maxlen = (l * 2 - 1) > maxlen ? (l * 2 - 1) : maxlen; + /* even symmetric */ + l = 0; + while(i - l >= 0 + && i + l + 1 < L + && string[i - l] == string[i + l + 1]) + l++; + maxlen = (l * 2) > maxlen ? (l * 2) : maxlen; + } + + printf("%d", maxlen); + return 0; +} +{% endraw %}``` \ No newline at end of file diff --git a/_articles/a1041.md b/_articles/a1041.md new file mode 100644 index 0000000..ff9566c --- /dev/null +++ b/_articles/a1041.md @@ -0,0 +1,46 @@ +--- +layout: post +date: 2020-04-23 01:50:40 +0800 +title: "PAT Advanced 1041. Be Unique (20) (C语言实现)" +categories: Advanced +tags: [] +permalink: Advanced/1041.html +--- + +## 题目 + +{% include_relative html/a1041.md %} + +## 思路 + +{% include_relative analysis/a1041.md %} + +## 代码 + +[Github最新代码](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1041.c)，欢迎交流 + +```c +{% raw %}#include + +int main() +{ + int N, counts[10001] = {0}, bets[100000] = {0}; + + scanf("%d", &N); + for(int i = 0; i < N; i++) + { + scanf("%d", bets + i); + counts[bets[i]]++; + } + + for(int i = 0; i < N; i++) + if(counts[bets[i]] == 1) + { + printf("%d", bets[i]); + return 0; + } + + printf("None"); + return 0; +} +{% endraw %}``` \ No newline at end of file diff --git a/_articles/analysis/a1035.md b/_articles/analysis/a1035.md new file mode 100644 index 0000000..7afaab1 --- /dev/null +++ b/_articles/analysis/a1035.md @@ -0,0 +1,8 @@ + + +将密码中的几个字符转化为另一个字符，很简单的题目。 + +我的具体做法是边读边改，并且将无需更改的密码首字符置为`'\0'`，就是当做标记以便区分。后面输出就容易了。 diff --git a/_articles/analysis/a1036.md b/_articles/analysis/a1036.md new file mode 100644 index 0000000..c733bd8 --- /dev/null +++ b/_articles/analysis/a1036.md @@ -0,0 +1,6 @@ + + +记录最大和最小，应该无需讲解了。 diff --git a/_articles/analysis/a1040.md b/_articles/analysis/a1040.md new file mode 100644 index 0000000..5c97f10 --- /dev/null +++ b/_articles/analysis/a1040.md @@ -0,0 +1,8 @@ + + +寻找字符串的对称部分，类似于回文数，但字符串的优势在于可以随时访问任意位置的字符，因此实现起来更简单。 + +注意：对称部分有可能是奇数个，也可能是偶数个，要考虑全面。 diff --git a/_articles/analysis/a1041.md b/_articles/analysis/a1041.md new file mode 100644 index 0000000..749f9de --- /dev/null +++ b/_articles/analysis/a1041.md @@ -0,0 +1,8 @@ + + +输出第一个独特的数字。 + +很简单，唯一要注意的是这里的*第一个*是指出现顺序。因此我实现中用了两个大数组。 diff --git a/_articles/others/a1035.md b/_articles/others/a1035.md new file mode 100644 index 0000000..2d9cf3e --- /dev/null +++ b/_articles/others/a1035.md @@ -0,0 +1,137 @@ +### 我的PAT系列文章更新重心已移至Github，欢迎来看PAT题解的小伙伴请到[Github Pages](https://oliverlew.github.io/PAT)浏览最新内容([本篇文章链接](https://oliverlew.github.io/PAT/Advanced/1035.html))。此处文章目前已更新至与Github Pages同步。欢迎star我的[repo](https://github.com/OliverLew/PAT)。 + +## 题目 + +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) +from `O` (`o` in uppercase). One solution is to replace `1` (one) by `@`, `0` +(zero) by `%`, `l` by `L`, and `O` by `o`. Now it is your job to write a +program to check the accounts generated by the judge, and to help the juge +modify the confusing passwords. + +### Input Specification: + +Each input file contains one test case. Each case contains a positive integer +$N$ ( $\le 1000$ ), followed by $N$ lines of accounts. Each account consists +of a user name and a password, both are strings of no more than 10 characters +with no space. + +### Output Specification: + +For each test case, first print the number $M$ of accounts that have been +modified, then print in the following $M$ lines the modified accounts info, +that is, the user names and the corresponding modified passwords. The accounts +must be printed in the same order as they are read in. If no account is +modified, print in one line `There are N accounts and no account is modified` +where `N` is the total number of accounts. However, if `N` is one, you must +print `There is 1 account and no account is modified` instead. + +### Sample Input 1: + + + + 3 + Team000002 Rlsp0dfa + Team000003 perfectpwd + Team000001 R1spOdfa + + +### Sample Output 1: + + + + 2 + Team000002 RLsp%dfa + Team000001 R@spodfa + + +### Sample Input 2: + + + + 1 + team110 abcdefg332 + + +### Sample Output 2: + + + + There is 1 account and no account is modified + + +### Sample Input 3: + + + + 2 + team110 abcdefg222 + team220 abcdefg333 + + +### Sample Output 3: + + + + There are 2 accounts and no account is modified + + + + +## 思路 + + +将密码中的几个字符转化为另一个字符，很简单的题目。 + +我的具体做法是边读边改，并且将无需更改的密码首字符置为`'\0'`，就是当做标记以便区分。后面输出就容易了。 + +## 代码 + +[最新代码@github](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1035.c)，欢迎交流 +```c +#include + +int main() +{ + int N, count = 0, modified; + char username[1000][11], password[1000][11]; + + scanf("%d", &N); + for(int i = 0; i < N; i++) + { + modified = 0; + scanf("%s %s", username[i], password[i]); + for(char *p = password[i]; *p; p++) + { + switch(*p) + { + case '1': *p = '@'; modified = 1; break; + case '0': *p = '%'; modified = 1; break; + case 'l': *p = 'L'; modified = 1; break; + case 'O': *p = 'o'; modified = 1; break; + default: break; + } + } + if(modified) + count++; + else /* mark unmodified password */ + password[i][0] = '\0'; + } + + if(count) + { + printf("%d\n", count); + for(int i = 0; i < N; i++) + if(password[i][0] != '\0') + printf("%s %s\n", username[i], password[i]); + } + else + { + printf("There %s %d %s and no account is modified", + N > 1 ? "are" : "is", N, N > 1 ? "accounts" : "account"); + } + + return 0; +} +``` \ No newline at end of file diff --git a/_articles/others/a1036.md b/_articles/others/a1036.md new file mode 100644 index 0000000..559bf82 --- /dev/null +++ b/_articles/others/a1036.md @@ -0,0 +1,113 @@ +### 我的PAT系列文章更新重心已移至Github，欢迎来看PAT题解的小伙伴请到[Github Pages](https://oliverlew.github.io/PAT)浏览最新内容([本篇文章链接](https://oliverlew.github.io/PAT/Advanced/1036.html))。此处文章目前已更新至与Github Pages同步。欢迎star我的[repo](https://github.com/OliverLew/PAT)。 + +## 题目 + +This time you are asked to tell the difference between the lowest grade of all +the male students and the highest grade of all the female students. + +### Input Specification: + +Each input file contains one test case. Each case contains a positive integer +$N$ , followed by $N$ lines of student information. Each line contains a +student's `name`, `gender`, `ID` and `grade`, separated by a space, where +`name` and `ID` are strings of no more than 10 characters with no space, +`gender` is either `F` (female) or `M` (male), and `grade` is an integer +between 0 and 100. It is guaranteed that all the grades are distinct. + +### Output Specification: + +For each test case, output in 3 lines. The first line gives the name and ID of +the female student with the highest grade, and the second line gives that of +the male student with the lowest grade. The third line gives the difference +$grade_F-grade_M$ . If one such kind of student is missing, output `Absent` in +the corresponding line, and output `NA` in the third line instead. + +### Sample Input 1: + + + + 3 + Joe M Math990112 89 + Mike M CS991301 100 + Mary F EE990830 95 + + +### Sample Output 1: + + + + Mary EE990830 + Joe Math990112 + 6 + + +### Sample Input 2: + + + + 1 + Jean M AA980920 60 + + +### Sample Output 2: + + + + Absent + Jean AA980920 + NA + + + + +## 思路 + + +记录最大和最小，应该无需讲解了。 + +## 代码 + +[最新代码@github](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1036.c)，欢迎交流 +```c +#include + +typedef struct student { + char name[11]; + char gender; + char ID[11]; +} Student; + +int main() +{ + int N, grade, gradeF = -1, gradeM = 101; + Student students[101] = {0}, s; + + scanf("%d", &N); + for(int i = 0; i < N; i++) + { + scanf("%s %c %s %d", s.name, &s.gender, s.ID, &grade); + students[grade] = s; + if(s.gender == 'F') + gradeF = grade > gradeF ? grade : gradeF; + else + gradeM = grade < gradeM ? grade : gradeM; + } + + if(gradeF != -1) + printf("%s %s\n", students[gradeF].name, students[gradeF].ID); + else + printf("Absent\n"); + + if(gradeM != 101) + printf("%s %s\n", students[gradeM].name, students[gradeM].ID); + else + printf("Absent\n"); + + if(gradeM == 101 || gradeF == -1) + printf("NA"); + else + printf("%d", gradeF - gradeM); + + return 0; +} +``` \ No newline at end of file diff --git a/_articles/others/a1040.md b/_articles/others/a1040.md new file mode 100644 index 0000000..b48c407 --- /dev/null +++ b/_articles/others/a1040.md @@ -0,0 +1,77 @@ +### 我的PAT系列文章更新重心已移至Github，欢迎来看PAT题解的小伙伴请到[Github Pages](https://oliverlew.github.io/PAT)浏览最新内容([本篇文章链接](https://oliverlew.github.io/PAT/Advanced/1040.html))。此处文章目前已更新至与Github Pages同步。欢迎star我的[repo](https://github.com/OliverLew/PAT)。 + +## 题目 + +Given a string, you are supposed to output the length of the longest symmetric +sub-string. For example, given `Is PAT&TAP symmetric?`, the longest symmetric +sub-string is `s PAT&TAP s`, hence you must output `11`. + +### Input Specification: + +Each input file contains one test case which gives a non-empty string of +length no more than 1000. + +### Output Specification: + +For each test case, simply print the maximum length in a line. + +### Sample Input: + + + + Is PAT&TAP symmetric? + + +### Sample Output: + + + + 11 + + + + +## 思路 + + +寻找字符串的对称部分，类似于回文数，但字符串的优势在于可以随时访问任意位置的字符，因此实现起来更简单。 + +注意：对称部分有可能是奇数个，也可能是偶数个，要考虑全面。 + +## 代码 + +[最新代码@github](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1040.c)，欢迎交流 +```c +#include +#include + +int main() +{ + int maxlen = 0, l, L; + char string[1002]; + + fgets(string, 1001, stdin); + + L = strlen(string); + for(int i = 0; i < L; i++) + { + /* odd symmetric */ + l = 0; + while(i - l >= 0 + && i + l < L + && string[i - l] == string[i + l]) + l++; + maxlen = (l * 2 - 1) > maxlen ? (l * 2 - 1) : maxlen; + /* even symmetric */ + l = 0; + while(i - l >= 0 + && i + l + 1 < L + && string[i - l] == string[i + l + 1]) + l++; + maxlen = (l * 2) > maxlen ? (l * 2) : maxlen; + } + + printf("%d", maxlen); + return 0; +} +``` \ No newline at end of file diff --git a/_articles/others/a1041.md b/_articles/others/a1041.md new file mode 100644 index 0000000..03cfbf8 --- /dev/null +++ b/_articles/others/a1041.md @@ -0,0 +1,86 @@ +### 我的PAT系列文章更新重心已移至Github，欢迎来看PAT题解的小伙伴请到[Github Pages](https://oliverlew.github.io/PAT)浏览最新内容([本篇文章链接](https://oliverlew.github.io/PAT/Advanced/1041.html))。此处文章目前已更新至与Github Pages同步。欢迎star我的[repo](https://github.com/OliverLew/PAT)。 + +## 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 are 7 people betting on { 5 31 5 88 67 88 17 }, then the +second one who bets on 31 wins. + +### Input Specification: + +Each input file contains one test case. Each case contains a line which begins +with a positive integer $N$ ( $\le 10^5$ ) and then followed by $N$ bets. The +numbers are separated by a space. + +### Output Specification: + +For each test case, print the winning number in a line. If there is no winner, +print `None` instead. + +### Sample Input 1: + + + + 7 5 31 5 88 67 88 17 + + +### Sample Output 1: + + + + 31 + + +### Sample Input 2: + + + + 5 888 666 666 888 888 + + +### Sample Output 2: + + + + None + + + + +## 思路 + + +输出第一个独特的数字。 + +很简单，唯一要注意的是这里的*第一个*是指出现顺序。因此我实现中用了两个大数组。 + +## 代码 + +[最新代码@github](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1041.c)，欢迎交流 +```c +#include + +int main() +{ + int N, counts[10001] = {0}, bets[100000] = {0}; + + scanf("%d", &N); + for(int i = 0; i < N; i++) + { + scanf("%d", bets + i); + counts[bets[i]]++; + } + + for(int i = 0; i < N; i++) + if(counts[bets[i]] == 1) + { + printf("%d", bets[i]); + return 0; + } + + printf("None"); + return 0; +} +``` \ No newline at end of file