liuchuo – 第212页

1035. Password (20)-PAT甲级真题

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 (<= 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
题目大意：给定n个用户的姓名和密码，把密码中的1改为@，0改为%，l改为L，O改为o
如果不存在需要修改的密码，则输出There are n accounts and no account is modified。注意单复数，如果只有一个账户，就输出There is 1 account and no account is modified

分析：把需要改变的字符串改变后存储在字符串数组vector里面，根据数组里面元素的个数是否为0输出相应的结果

#include <iostream>
#include <vector>
using namespace std;
int main() {
    int n;
    scanf("%d", &n);
    vector<string> v;
    for(int i = 0; i < n; i++) {
        string name, s;
        cin >> name >> s;
        int len = s.length(), flag = 0;
        for(int j = 0; j < len; j++) {
            switch(s[j]) {
                case '1' : s[j] = '@'; flag = 1; break;
                case '0' : s[j] = '%'; flag = 1; break;
                case 'l' : s[j] = 'L'; flag = 1; break;
                case 'O' : s[j] = 'o'; flag = 1; break;
            }
        }
        if(flag) {
            string temp = name + " " + s;
            v.push_back(temp);
        }
    }
    int cnt = v.size();
    if(cnt != 0) {
        printf("%d\n", cnt);
        for(int i = 0; i < cnt; i++)
            cout << v[i] << endl;
    } else if(n == 1) {
        printf("There is 1 account and no account is modified");
    } else {
        printf("There are %d accounts and no account is modified", n);
    }
    return 0;
}

#include <iostream>

#include <vector>

using namespace std;

int main() {

int n;

scanf("%d", &n);

vector<string> v;

for(int i = 0; i < n; i++) {

string name, s;

cin >> name >> s;

int len = s.length(), flag = 0;

for(int j = 0; j < len; j++) {

switch(s[j]) {

case '1' : s[j] = '@'; flag = 1; break;

case '0' : s[j] = '%'; flag = 1; break;

case 'l' : s[j] = 'L'; flag = 1; break;

case 'O' : s[j] = 'o'; flag = 1; break;

}

if(flag) {

string temp = name + " " + s;

v.push_back(temp);

}

int cnt = v.size();

if(cnt != 0) {

printf("%d\n", cnt);

for(int i = 0; i < cnt; i++)

cout << v[i] << endl;

} else if(n == 1) {

printf("There is 1 account and no account is modified");

} else {

printf("There are %d accounts and no account is modified", n);

}

return 0;

}

1073. Scientific Notation (20)-PAT甲级真题

Scientific notation is the way that scientists easily handle very large numbers or very small numbers. The notation matches the regular expression [+-][1-9]”.”[0-9]+E[+-][0-9]+ which means that the integer portion has exactly one digit, there is at least one digit in the fractional portion, and the number and its exponent’s signs are always provided even when they are positive.

Now given a real number A in scientific notation, you are supposed to print A in the conventional notation while keeping all the significant figures.

Input Specification:

Each input file contains one test case. For each case, there is one line containing the real number A in scientific notation. The number is no more than 9999 bytes in length and the exponent’s absolute value is no more than 9999.

Output Specification:

For each test case, print in one line the input number A in the conventional notation, with all the significant figures kept, including trailing zeros,

Sample Input 1:
+1.23400E-03
Sample Output 1:
0.00123400
Sample Input 2:
-1.2E+10
Sample Output 2:
-12000000000

题目大意：题目给出科学计数法的格式的数字A，要求输出普通数字表示法的A，并保证所有有效位都被保留，包括末尾的0

分析：n保存E后面的字符串所对应的数字，t保存E前面的字符串，不包括符号位。当n<0时表示向前移动，那么先输出0. 然后输出abs(n)-1个0，然后继续输出t中的所有数字；当n>0时候表示向后移动，那么先输出第一个字符，然后将t中尽可能输出n个字符，如果t已经输出到最后一个字符(j == t.length())那么就在后面补n-cnt个0，否则就补充一个小数点。然后继续输出t剩余的没有输出的字符～

#include <iostream>
using namespace std;
int main() {
    string s;
    cin >> s;
    int i = 0;
    while (s[i] != 'E') i++;
    string t = s.substr(1, i-1);
    int n = stoi(s.substr(i+1));
    if (s[0] == '-') cout << "-";
    if (n < 0) {
        cout << "0.";
        for (int j = 0; j < abs(n) - 1; j++) cout << '0';
        for (int j = 0; j < t.length(); j++)
            if (t[j] != '.') cout << t[j];
    } else {
        cout << t[0];
        int cnt, j;
        for (j = 2, cnt = 0; j < t.length() && cnt < n; j++, cnt++) cout << t[j];
        if (j == t.length()) {
            for (int k = 0; k < n - cnt; k++) cout << '0';
        } else {
            cout << '.';
            for (int k = j; k < t.length(); k++) cout << t[k];
        }
    }
    return 0;
}

#include <iostream>

using namespace std;

int main() {

string s;

cin >> s;

int i = 0;

while (s[i] != 'E') i++;

string t = s.substr(1, i-1);

int n = stoi(s.substr(i+1));

if (s[0] == '-') cout << "-";

if (n < 0) {

cout << "0.";

for (int j = 0; j < abs(n) - 1; j++) cout << '0';

for (int j = 0; j < t.length(); j++)

if (t[j] != '.') cout << t[j];

} else {

cout << t[0];

int cnt, j;

for (j = 2, cnt = 0; j < t.length() && cnt < n; j++, cnt++) cout << t[j];

if (j == t.length()) {

for (int k = 0; k < n - cnt; k++) cout << '0';

} else {

cout << '.';

for (int k = j; k < t.length(); k++) cout << t[k];

}

return 0;

}

1058. A+B in Hogwarts (20)-PAT甲级真题

If you are a fan of Harry Potter, you would know the world of magic has its own currency system — as Hagrid explained it to Harry, “Seventeen silver Sickles to a Galleon and twenty-nine Knuts to a Sickle, it’s easy enough.” Your job is to write a program to compute A+B where A and B are given in the standard form of “Galleon.Sickle.Knut” (Galleon is an integer in [0, 107], Sickle is an integer in [0, 17), and Knut is an integer in [0, 29)).

Input Specification:

Each input file contains one test case which occupies a line with A and B in the standard form, separated by one space.

Output Specification:

For each test case you should output the sum of A and B in one line, with the same format as the input.

Sample Input:
3.2.1 10.16.27
Sample Output:
14.1.28

题目大意：17个Sickle对换一个Galleon，29个Knut对换一个Sickle。根据Galleon.Sickle.Knut的方式相加A和B
分析：像相加算术一样从后往前按位相加，处理好进位～

#include <iostream>
using namespace std;
int main() {
    int a1, b1, c1, a2, b2, c2, a, b, c;
    scanf("%d.%d.%d %d.%d.%d", &a1, &b1, &c1, &a2, &b2, &c2);
    c = c1 + c2;
    b = b1 + b2 + c / 29; c = c % 29;
    a = a1 + a2 + b / 17; b = b % 17;
    printf("%d.%d.%d\n", a, b, c);
    return 0;
}

#include <iostream>

using namespace std;

int main() {

int a1, b1, c1, a2, b2, c2, a, b, c;

scanf("%d.%d.%d %d.%d.%d", &a1, &b1, &c1, &a2, &b2, &c2);

c = c1 + c2;

b = b1 + b2 + c / 29; c = c % 29;

a = a1 + a2 + b / 17; b = b % 17;

printf("%d.%d.%d\n", a, b, c);

return 0;

}

1027. Colors in Mars (20)-PAT甲级真题

People in Mars represent the colors in their computers in a similar way as the Earth people. That is, a color is represented by a 6-digit number, where the first 2 digits are for Red, the middle 2 digits for Green, and the last 2 digits for Blue. The only difference is that they use radix 13 (0-9 and A-C) instead of 16. Now given a color in three decimal numbers (each between 0 and 168), you are supposed to output their Mars RGB values.

Input

Each input file contains one test case which occupies a line containing the three decimal color values.

Output

For each test case you should output the Mars RGB value in the following format: first output “#, then followed by a 6-digit number where all the English characters must be upper-cased. If a single color is only 1-digit long, you must print a “0” to the left.

Sample Input
15 43 71
Sample Output
#123456

题目大意：给三个十进制的数，把它们转换为十三进制的数输出。要求在前面加上一个”#”号
分析：因为0~168的十进制转换为13进制不会超过两位数，所以这个两位数为(num / 13)(num % 13)构成的数字

#include <cstdio>
using namespace std;
int main() {
    char c[14] = {"0123456789ABC"};
    printf("#");
    for(int i = 0; i < 3; i++) {
        int num;
        scanf("%d", &num);
        printf("%c%c", c[num/13], c[num%13]);
    }
    return 0;
}

#include <cstdio>

using namespace std;

int main() {

char c[14] = {"0123456789ABC"};

printf("#");

for(int i = 0; i < 3; i++) {

int num;

scanf("%d", &num);

printf("%c%c", c[num/13], c[num%13]);

}

return 0;

}

1019. General Palindromic Number (20)-PAT甲级真题

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Although palindromic numbers are most often considered in the decimal system, the concept of palindromicity can be applied to the natural numbers in any numeral system. Consider a number N > 0 in base b >= 2, where it is written in standard notation with k+1 digits ai as the sum of (aibi) for i from 0 to k. Here, as usual, 0 <= ai < b for all i and ak is non-zero. Then N is palindromic if and only if ai = ak-i for all i. Zero is written 0 in any base and is also palindromic by definition.

Given any non-negative decimal integer N and a base b, you are supposed to tell if N is a palindromic number in base b.

Input Specification:

Each input file contains one test case. Each case consists of two non-negative numbers N and b, where 0 <= N <= 109 is the decimal number and 2 <= b <= 109 is the base. The numbers are separated by a space.

Output Specification:

For each test case, first print in one line “Yes” if N is a palindromic number in base b, or “No” if not. Then in the next line, print N as the number in base b in the form “ak ak-1 … a0”. Notice that there must be no extra space at the end of output.

Sample Input 1:
27 2
Sample Output 1:
Yes
1 1 0 1 1
Sample Input 2:
121 5
Sample Output 2:
No
4 4 1

题目大意：给出两个整数a和b，问十进制的a在b进制下是否为回文数。是的话输出Yes，不是输出No。并且输出a在b进制下的表示，以空格隔开

分析：将a转换为b进制形式，保存在int的数组里面，比较数组左右两端是否对称。
注意：如果是0，要输出Yes和0

#include <cstdio>
using namespace std;
int main() {
    int a, b;
    scanf("%d %d", &a, &b);
    int arr[40], index = 0;
    while(a != 0) {
        arr[index++] = a % b;
        a = a / b;
    }
    int flag = 0;
    for(int i = 0; i < index / 2; i++) {
        if(arr[i] != arr[index-i-1]) {
            printf("No\n");
            flag = 1;
            break;
        }
    }
    if(!flag) printf("Yes\n");
    for(int i = index - 1; i >= 0; i--) {
        printf("%d", arr[i]);
        if(i != 0) printf(" ");
    }
    if(index == 0)
        printf("0");
    return 0;
}

#include <cstdio>

using namespace std;

int main() {

int a, b;

scanf("%d %d", &a, &b);

int arr[40], index = 0;

while(a != 0) {

arr[index++] = a % b;

a = a / b;

}

int flag = 0;

for(int i = 0; i < index / 2; i++) {

if(arr[i] != arr[index-i-1]) {

printf("No\n");

flag = 1;

break;

}

if(!flag) printf("Yes\n");

for(int i = index - 1; i >= 0; i--) {

printf("%d", arr[i]);

if(i != 0) printf(" ");

}

if(index == 0)

printf("0");

return 0;

}

1031. Hello World for U (20)-PAT甲级真题

Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, “helloworld” can be printed as:
h d
e l
l r
lowo
That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible — that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 – 2 = N.

Input Specification:
Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

Output Specification:
For each test case, print the input string in the shape of U as specified in the description.

Sample Input:
helloworld!

Sample Output:
h !
e d
l l
lowor

题目大意：用所给字符串按U型输出。n1和n3是左右两条竖线从上到下的字符个数，n2是底部横线从左到右的字符个数。
要求：
1. n1 == n3
2. n2 >= n1
3. n1为在满足上述条件的情况下的最大值

分析：假设n = 字符串长度 + 2，因为2 * n1 + n2 = n，且要保证n2 >= n1, n1尽可能地大，分类讨论：
1. 如果n % 3 == 0，n正好被3整除，直接n1 == n2 == n3;
2. 如果n % 3 == 1，因为n2要比n1大，所以把多出来的那1个给n2
3. 如果n % 3 == 2, 就把多出来的那2个给n2
所以得到公式：n1 = n / 3，n2 = n / 3 + n % 3
把它们存储到二维字符数组中，一开始初始化字符数组为空格，然后按照u型填充进去，最后输出这个数组u～～

#include <iostream>
#include <string.h>
using namespace std;
int main() {
    char c[81], u[30][30];
    memset(u, ' ', sizeof(u));
    scanf("%s", c);
    int n = strlen(c) + 2;
    int n1 = n / 3, n2 = n / 3 + n % 3, index = 0;
    for(int i = 0; i < n1; i++) u[i][0] = c[index++];
    for(int i = 1; i <= n2 - 2; i++) u[n1-1][i] = c[index++];
    for(int i = n1 - 1; i >= 0; i--) u[i][n2-1] = c[index++];
    for(int i = 0; i < n1; i++) {
        for(int j = 0; j < n2; j++) 
            printf("%c", u[i][j]);
        printf("\n");
    }
    return 0;
}

#include <iostream>

#include <string.h>

using namespace std;

int main() {

char c[81], u[30][30];

memset(u, ' ', sizeof(u));

scanf("%s", c);

int n = strlen(c) + 2;

int n1 = n / 3, n2 = n / 3 + n % 3, index = 0;

for(int i = 0; i < n1; i++) u[i][0] = c[index++];

for(int i = 1; i <= n2 - 2; i++) u[n1-1][i] = c[index++];

for(int i = n1 - 1; i >= 0; i--) u[i][n2-1] = c[index++];

for(int i = 0; i < n1; i++) {

for(int j = 0; j < n2; j++)

printf("%c", u[i][j]);

printf("\n");

}

return 0;

}