OJ – 第15页 – liuchuo

1112. Stucked Keyboard (20)-PAT甲级真题

On a broken keyboard, some of the keys are always stucked. So when you type some sentences, the characters corresponding to those keys will appear repeatedly on screen for k times.

Now given a resulting string on screen, you are supposed to list all the possible stucked keys, and the original string.

Notice that there might be some characters that are typed repeatedly. The stucked key will always repeat output for a fixed k times whenever it is pressed. For example, when k=3, from the string “thiiis iiisss a teeeeeest” we know that the keys “i” and “e” might be stucked, but “s” is not even though it appears repeatedly sometimes. The original string could be “this isss a teest”.

Input Specification:

Each input file contains one test case. For each case, the 1st line gives a positive integer k ( 1<k<=100 ) which is the output repeating times of a stucked key. The 2nd line contains the resulting string on screen, which consists of no more than 1000 characters from {a-z}, {0-9} and “_”. It is guaranteed that the string is non-empty.

Output Specification:

For each test case, print in one line the possible stucked keys, in the order of being detected. Make sure that each key is printed once only. Then in the next line print the original string. It is guaranteed that there is at least one stucked key.

Sample Input:
3
caseee1__thiiis_iiisss_a_teeeeeest
Sample Output:
ei
case1__this_isss_a_teest

题目大意：键盘某些键卡住了，按一次重复k次，要求找出可能的键，并且输出正确的字符串顺序。可能的键要求按照被发现的顺序输出。

分析：考察STL的应用～
map<char, bool>存储出现的键是否坏，set<char>存储输出可能坏的键的时候，当前字符是否已经被输出过，输出过的键放在set里面.

寻找坏键：遍历字符串的每个字符的时候，与pre（字符串当前字符s[i]的前一个字符）相比较，如果相等就继续计数cnt++，如果不相等，令cnt = 1表示当前字符出现了一次～如果cnt % k等于0 则令s[i]可能是坏键，置map对应的字符的bool值为true～

输出坏键：由于需要根据坏键发现的顺序输出，所以遍历整个字符串的方式输出～，并且确保不会重复输出～（用set集合确保，输出过了的放在set里面）
输出整个正确的字符串：如果当前s[i]是坏键，在输出一次后，令 i = i + k – 1，再输出，保证坏键出现k次只输出一次～

tips：谢谢在csdn博客的评论里的同学友情提醒，如果出现了先不是坏建后又判断是坏键的情况，这种情况会出现错误，因为前面已经认为它不是坏键了，说明它一定不是坏建，所以要加一个sureNoBroken，把确定不是坏键的键标记出来，在map都设置完成后把确定不是坏键的m标记为false。虽然测试用例里面没有考虑到这种情况，但是如果输入3 aabbaaa，应该输出没有坏键。

#include <iostream>
#include <map>
#include <cstdio>
#include <set>
using namespace std;
bool sureNoBroken[256];
int main() {
    int k, cnt = 1;
    scanf("%d", &k);
    string s;
    cin >> s;
    map<char, bool> m;
    set<char> printed;
    char pre = '#';
    s = s + '#';
    for(int i = 0; i < s.length(); i++) {
        if(s[i] == pre) {
            cnt++;
        } else {
            if(cnt % k != 0) {
                sureNoBroken[pre] = true;
            }
            cnt = 1;
        }
        if(i != s.length() - 1) m[s[i]] = (cnt % k == 0);
        pre = s[i];
    }
    for(int i = 0; i < s.length() - 1; i++) {
        if(sureNoBroken[s[i]] == true)
            m[s[i]] = false;
    }
    for(int i = 0; i < s.length() - 1; i++) {
        if(m[s[i]] && printed.find(s[i]) == printed.end()) {
            printf("%c", s[i]);
            printed.insert(s[i]);
        }
    }
    printf("\n");
    for(int i = 0; i < s.length() - 1; i++) {
        printf("%c", s[i]);
        if(m[s[i]])
            i = i + k - 1;
    }
    return 0;
}

#include <iostream>

#include <map>

#include <cstdio>

#include <set>

using namespace std;

bool sureNoBroken[256];

int main() {

int k, cnt = 1;

scanf("%d", &k);

string s;

cin >> s;

map<char, bool> m;

set<char> printed;

char pre = '#';

s = s + '#';

for(int i = 0; i < s.length(); i++) {

if(s[i] == pre) {

cnt++;

} else {

if(cnt % k != 0) {

sureNoBroken[pre] = true;

}

cnt = 1;

}

if(i != s.length() - 1) m[s[i]] = (cnt % k == 0);

pre = s[i];

}

for(int i = 0; i < s.length() - 1; i++) {

if(sureNoBroken[s[i]] == true)

m[s[i]] = false;

}

for(int i = 0; i < s.length() - 1; i++) {

if(m[s[i]] && printed.find(s[i]) == printed.end()) {

printf("%c", s[i]);

printed.insert(s[i]);

}

printf("\n");

for(int i = 0; i < s.length() - 1; i++) {

printf("%c", s[i]);

if(m[s[i]])

i = i + k - 1;

}

return 0;

}

1105. Spiral Matrix (25)-PAT甲级真题

This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrix is filled in from the first element at the upper-left corner, then move in a clockwise spiral. The matrix has m rows and n columns, where m and n satisfy the following: m*n must be equal to N; m>=n; and m-n is the minimum of all the possible values.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N. Then the next line contains N positive integers to be filled into the spiral matrix. All the numbers are no more than 104. The numbers in a line are separated by spaces.

Output Specification:

For each test case, output the resulting matrix in m lines, each contains n numbers. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.

Sample Input:
12
37 76 20 98 76 42 53 95 60 81 58 93
Sample Output:
98 95 93
42 37 81
53 20 76
58 60 76

题目大意：将给定的N个正整数按非递增的顺序，填入“螺旋矩阵”~所谓“螺旋矩阵”，是指从左上角第1个格子开始，按顺时针螺旋方向填充~要求矩阵的规模为m行n列，满足条件：m*n等于N；m>=n；且m-n取所有可能值中的最小值~

分析：先计算行数m和列数n的值，n从根号N的整数部分开始，往前推一直到1，找到第一个满足N % n== 0的，m的值等于N/n～将N个给定的值输入数组a，并将a数组中的值按非递增排序，接着建立m行n列的数组b，填充时按层数填充，一个包裹矩阵的口字型为一层，计算螺旋矩阵的层数level，如果m的值为偶数，层数为m/2，如果m为奇数，层数为m/2+1，所以level = m / 2 + m % 2；因为是从左上角第1个格子开始，按顺时针螺旋方向填充，所以外层for循环控制层数i从0到level，内层for循环按左上到右上、右上到右下、右下到左下、左下到左上的顺序一层层填充，注意内层for循环中还要控制t <= N – 1，因为如果螺旋矩阵中所有的元素已经都填充完毕，就不能再重复填充～填充完毕后，输出整个矩阵～

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
int cmp(int a, int b) {return a > b;}
int main() {
    int N, m, n, t = 0;
    scanf("%d", &N);
    for (n = sqrt((double)N); n >= 1; n--) {
        if (N % n == 0) {
            m = N / n;
            break;
        }
    }
    vector<int> a(N);
    for (int i = 0; i < N; i++)
        scanf("%d", &a[i]);
    sort(a.begin(), a.end(), cmp);
    vector<vector<int> > b(m, vector<int>(n));
    int level = m / 2 + m % 2;
    for (int i = 0; i < level; i++) {
        for (int j = i; j <= n - 1 - i && t <= N - 1; j++)
                b[i][j] = a[t++];
        for (int j = i + 1; j <= m - 2 - i && t <= N - 1; j++)
                b[j][n - 1 - i] = a[t++];
        for (int j = n - i - 1; j >= i && t <= N - 1; j--)
                b[m - 1 - i][j] = a[t++];
        for (int j = m - 2 - i; j >= i + 1 && t <= N - 1; j--)
                b[j][i] = a[t++];
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0 ; j < n; j++) {
            printf("%d", b[i][j]);
            if (j != n - 1) printf(" ");
        }
        printf("\n");
    }
    return 0;
}

#include <iostream>

#include <algorithm>

#include <cmath>

#include <vector>

using namespace std;

int cmp(int a, int b) {return a > b;}

int main() {

int N, m, n, t = 0;

scanf("%d", &N);

for (n = sqrt((double)N); n >= 1; n--) {

if (N % n == 0) {

m = N / n;

break;

}

vector<int> a(N);

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

scanf("%d", &a[i]);

sort(a.begin(), a.end(), cmp);

vector<vector<int> > b(m, vector<int>(n));

int level = m / 2 + m % 2;

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

for (int j = i; j <= n - 1 - i && t <= N - 1; j++)

b[i][j] = a[t++];

for (int j = i + 1; j <= m - 2 - i && t <= N - 1; j++)

b[j][n - 1 - i] = a[t++];

for (int j = n - i - 1; j >= i && t <= N - 1; j--)

b[m - 1 - i][j] = a[t++];

for (int j = m - 2 - i; j >= i + 1 && t <= N - 1; j--)

b[j][i] = a[t++];

}

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

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

printf("%d", b[i][j]);

if (j != n - 1) printf(" ");

}

printf("\n");

}

return 0;

}

1050. 螺旋矩阵(25)-PAT乙级真题

本题要求将给定的N个正整数按非递增的顺序，填入“螺旋矩阵”。所谓“螺旋矩阵”，是指从左上角第1个格子开始，按顺时针螺旋方向填充。要求矩阵的规模为m行n列，满足条件：m*n等于N；m>=n；且m-n取所有可能值中的最小值。

输入格式：

输入在第1行中给出一个正整数N，第2行给出N个待填充的正整数。所有数字不超过10^4，相邻数字以空格分隔。

输出格式：

输出螺旋矩阵。每行n个数字，共m行。相邻数字以1个空格分隔，行末不得有多余空格。

输入样例：

12
37 76 20 98 76 42 53 95 60 81 58 93

输出样例：

98 95 93
42 37 81
53 20 76
58 60 76

分析：首先计算行数m和列数n的值，n从根号N的整数部分开始，往前推一直到1，找到第一个满足N % n== 0的，m的值等于N/n～将N个给定的值输入数组a，并将a数组中的值按非递增排序，接着建立m行n列的数组b，填充时按层数填充，一个包裹矩阵的口字型为一层，计算螺旋矩阵的层数level，如果m的值为偶数，层数为m/2，如果m为奇数，层数为m/2+1，所以level = m / 2 + m % 2；因为是从左上角第1个格子开始，按顺时针螺旋方向填充，所以外层for循环控制层数i从0到level，内层for循环按左上到右上、右上到右下、右下到左下、左下到左上的顺序一层层填充，注意内层for循环中还要控制t <= N – 1，因为如果螺旋矩阵中所有的元素已经都填充完毕，就不能再重复填充～填充完毕后，输出整个矩阵～

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
int cmp(int a, int b) {return a > b;}
int main() {
    int N, m, n, t = 0;
    scanf("%d", &N);
    for (n = sqrt((double)N); n >= 1; n--) {
        if (N % n == 0) {
            m = N / n;
            break;
        }
    }
    vector<int> a(N);
    for (int i = 0; i < N; i++)
        scanf("%d", &a[i]);
    sort(a.begin(), a.end(), cmp);
    vector<vector<int> > b(m, vector<int>(n));
    int level = m / 2 + m % 2;
    for (int i = 0; i < level; i++) {
        for (int j = i; j <= n - 1 - i && t <= N - 1; j++)
                b[i][j] = a[t++];
        for (int j = i + 1; j <= m - 2 - i && t <= N - 1; j++)
                b[j][n - 1 - i] = a[t++];
        for (int j = n - i - 1; j >= i && t <= N - 1; j--)
                b[m - 1 - i][j] = a[t++];
        for (int j = m - 2 - i; j >= i + 1 && t <= N - 1; j--)
                b[j][i] = a[t++];
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0 ; j < n; j++) {
            printf("%d", b[i][j]);
            if (j != n - 1) printf(" ");
        }
        printf("\n");
    }
    return 0;
}

#include <iostream>

#include <algorithm>

#include <cmath>

#include <vector>

using namespace std;

int cmp(int a, int b) {return a > b;}

int main() {

int N, m, n, t = 0;

scanf("%d", &N);

for (n = sqrt((double)N); n >= 1; n--) {

if (N % n == 0) {

m = N / n;

break;

}

vector<int> a(N);

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

scanf("%d", &a[i]);

sort(a.begin(), a.end(), cmp);

vector<vector<int> > b(m, vector<int>(n));

int level = m / 2 + m % 2;

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

for (int j = i; j <= n - 1 - i && t <= N - 1; j++)

b[i][j] = a[t++];

for (int j = i + 1; j <= m - 2 - i && t <= N - 1; j++)

b[j][n - 1 - i] = a[t++];

for (int j = n - i - 1; j >= i && t <= N - 1; j--)

b[m - 1 - i][j] = a[t++];

for (int j = m - 2 - i; j >= i + 1 && t <= N - 1; j--)

b[j][i] = a[t++];

}

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

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

printf("%d", b[i][j]);

if (j != n - 1) printf(" ");

}

printf("\n");

}

return 0;

}

1062. Talent and Virtue (25)-PAT甲级真题

About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people’s talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a “sage（圣人）”; being less excellent but with one’s virtue outweighs talent can be called a “nobleman（君子）”; being good in neither is a “fool man（愚人）”; yet a fool man is better than a “small man（小人）” who prefers talent than virtue.

Now given the grades of talent and virtue of a group of people, you are supposed to rank them according to Sima Guang’s theory.

Input Specification:

Each input file contains one test case. Each case first gives 3 positive integers in a line: N (<=105), the total number of people to be ranked; L (>=60), the lower bound of the qualified grades — that is, only the ones whose grades of talent and virtue are both not below this line will be ranked; and H (<100), the higher line of qualification — that is, those with both grades not below this line are considered as the “sages”, and will be ranked in non-increasing order according to their total grades. Those with talent grades below H but virtue grades not are cosidered as the “noblemen”, and are also ranked in non-increasing order according to their total grades, but they are listed after the “sages”. Those with both grades below H, but with virtue not lower than talent are considered as the “fool men”. They are ranked in the same way but after the “noblemen”. The rest of people whose grades both pass the L line are ranked after the “fool men”.

Then N lines follow, each gives the information of a person in the format:

ID_Number Virtue_Grade Talent_Grade
where ID_Number is an 8-digit number, and both grades are integers in [0, 100]. All the numbers are separated by a space.
Output Specification:

The first line of output must give M (<=N), the total number of people that are actually ranked. Then M lines follow, each gives the information of a person in the same format as the input, according to the ranking rules. If there is a tie of the total grade, they must be ranked with respect to their virtue grades in non-increasing order. If there is still a tie, then output in increasing order of their ID’s.

Sample Input:
14 60 80
10000001 64 90
10000002 90 60
10000011 85 80
10000003 85 80
10000004 80 85
10000005 82 77
10000006 83 76
10000007 90 78
10000008 75 79
10000009 59 90
10000010 88 45
10000012 80 100
10000013 90 99
10000014 66 60
Sample Output:
12
10000013 90 99
10000012 80 100
10000003 85 80
10000011 85 80
10000004 80 85
10000007 90 78
10000006 83 76
10000005 82 77
10000002 90 60
10000014 66 60
10000008 75 79
10000001 64 90

题目大意：输入第1行给出3个正整数，分别为：N为考生总数；L为录取最低分数线，即德分和才分均不低于L的考生才有资格被考虑录取；H为优先录取线——德分和才分均不低于此线的被定义为“才德全尽”，此类考生按德才总分从高到低排序；才分不到但德分到线的一类考生属于“德胜才”，也按总分排序，但排在第一类考生之后；德才分均低于H，但是德分不低于才分的考生属于“才德兼亡”但尚有“德胜才”者，按总分排序，但排在第二类考生之后；其他达到最低线L的考生也按总分排序，但排在第三类考生之后。输出第1行首先给出达到最低分数线的考生人数M，随后M行，每行按照输入格式输出一位考生的信息，考生按输入中说明的规则从高到低排序。当某类考生中有多人总分相同时，按其德分降序排列；若德分也并列，则按准考证号的升序输出～

分析：用结构体存储～写好cmp函数～结构体数组vector v[4]中v[0]保存第一类考生，v[1]保存第二类考生…以此类推～写好cmp函数很重要，cmp函数中，排序先按照总分排序，然后按照德分排序，最后按照准考证号排序…最后输出符合条件的结果～

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
struct node {
    int num, de, cai;
};
int cmp(struct node a, struct node b) {
    if ((a.de + a.cai) != (b.de + b.cai))
        return (a.de + a.cai) > (b.de + b.cai);
    else if (a.de != b.de)
        return a.de > b.de;
    else
        return a.num < b.num;
}
int main() {
    int n, low, high;
    scanf("%d %d %d", &n, &low, &high);
    vector<node> v[4];
    node temp;
    int total = n;
    for (int i = 0; i < n; i++) {
        scanf("%d %d %d", &temp.num, &temp.de, &temp.cai);
        if (temp.de < low || temp.cai < low)
            total--;
        else if (temp.de >= high && temp.cai >= high)
            v[0].push_back(temp);
        else if (temp.de >= high && temp.cai < high)
            v[1].push_back(temp);
        else if (temp.de < high && temp.cai < high && temp.de >= temp.cai)
            v[2].push_back(temp);
        else
            v[3].push_back(temp);
    }
    printf("%d\n", total);
    for (int i = 0; i < 4; i++) {
        sort(v[i].begin(), v[i].end(), cmp);
        for (int j = 0; j < v[i].size(); j++)
            printf("%d %d %d\n", v[i][j].num, v[i][j].de, v[i][j].cai);
    }
    return 0;
}

#include <iostream>

#include <algorithm>

#include <vector>

using namespace std;

struct node {

int num, de, cai;

};

int cmp(struct node a, struct node b) {

if ((a.de + a.cai) != (b.de + b.cai))

return (a.de + a.cai) > (b.de + b.cai);

else if (a.de != b.de)

return a.de > b.de;

else

return a.num < b.num;

}

int main() {

int n, low, high;

scanf("%d %d %d", &n, &low, &high);

vector<node> v[4];

node temp;

int total = n;

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

scanf("%d %d %d", &temp.num, &temp.de, &temp.cai);

if (temp.de < low || temp.cai < low)

total--;

else if (temp.de >= high && temp.cai >= high)

v[0].push_back(temp);

else if (temp.de >= high && temp.cai < high)

v[1].push_back(temp);

else if (temp.de < high && temp.cai < high && temp.de >= temp.cai)

v[2].push_back(temp);

else

v[3].push_back(temp);

}

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

for (int i = 0; i < 4; i++) {

sort(v[i].begin(), v[i].end(), cmp);

for (int j = 0; j < v[i].size(); j++)

printf("%d %d %d\n", v[i][j].num, v[i][j].de, v[i][j].cai);

}

return 0;

}

1077. Kuchiguse (20)-PAT甲级真题

The Japanese language is notorious for its sentence ending particles. Personal preference of such particles can be considered as a reflection of the speaker’s personality. Such a preference is called “Kuchiguse” and is often exaggerated artistically in Anime and Manga. For example, the artificial sentence ending particle “nyan~” is often used as a stereotype for characters with a cat-like personality:
Itai nyan~ (It hurts, nyan~)
Ninjin wa iyada nyan~ (I hate carrots, nyan~)
Now given a few lines spoken by the same character, can you find her Kuchiguse?

Input Specification:
Each input file contains one test case. For each case, the first line is an integer N (2<=N<=100). Following are N file lines of 0~256 (inclusive) characters in length, each representing a character’s spoken line. The spoken lines are case sensitive.

Output Specification:
For each test case, print in one line the kuchiguse of the character, i.e., the longest common suffix of all N lines. If there is no such suffix, write “nai”.

Sample Input 1:
3
Itai nyan~
Ninjin wa iyadanyan~
uhhh nyan~
Sample Output 1:
nyan~
Sample Input 2:
3
Itai!
Ninjinnwaiyada T_T
T_T
Sample Output 2:
nai

题目大意：给定N给字符串，求他们的公共后缀，如果不存在公共后缀，就输出“nai”
分析：因为是后缀，反过来比较太麻烦，所以每输入一个字符串，就把它逆序过来再比较会比较容易～
首先ans = s；后来每输入的一个字符串，都和ans比较，如果后面不相同的就把它截取掉，最后输出ans即可（要逆序输出～，所以先将ans倒置reverse一下～）

#include <iostream>
#include <algorithm>
using namespace std;
int main() {
    int n;
    scanf("%d\n", &n);
    string ans;
    for(int i = 0; i < n; i++) {
        string s;
        getline(cin, s);
        int lens = s.length();
        reverse(s.begin(), s.end());
        if(i == 0) {
            ans = s;
            continue;
        } else {
            int lenans = ans.length();
            if(lens < lenans) swap(ans, s);
            int minlen = min(lens, lenans);
            for(int j = 0; j < minlen; j++) {
                if(ans[j] != s[j]) {
                    ans = ans.substr(0, j);
                    break;
                }
            }
        }
    }
    reverse(ans.begin(), ans.end());
    if (ans.length() == 0) ans = "nai";
    cout << ans;
    return 0;
}

#include <iostream>

#include <algorithm>

using namespace std;

int main() {

int n;

scanf("%d\n", &n);

string ans;

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

string s;

getline(cin, s);

int lens = s.length();

reverse(s.begin(), s.end());

if(i == 0) {

ans = s;

continue;

} else {

int lenans = ans.length();

if(lens < lenans) swap(ans, s);

int minlen = min(lens, lenans);

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

if(ans[j] != s[j]) {

ans = ans.substr(0, j);

break;

}

reverse(ans.begin(), ans.end());

if (ans.length() == 0) ans = "nai";

cout << ans;

return 0;

}

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;

}