liuchuo – 第11页 – 柳婼の blog

1158 Telefraud Detection – PAT甲级真题

Telefraud（电信诈骗） remains a common and persistent problem in our society. In some cases, unsuspecting victims lose their entire life savings. To stop this crime, you are supposed to write a program to detect those suspects from a huge amount of phone call records.

A person must be detected as a suspect if he/she makes more than K short phone calls to different people everyday, but no more than 20% of these people would call back. And more, if two suspects are calling each other, we say they might belong to the same gang. A makes a short phone call to B means that the total duration of the calls from A to B is no more than 5 minutes.

Input Specification:

Each input file contains one test case. For each case, the first line gives 3 positive integers K (≤500, the threshold（阈值） of the amount of short phone calls), N (≤103, the number of different phone numbers), and M (≤105, the number of phone call records). Then M lines of one day’s records are given, each in the format:

caller receiver duration

where caller and receiver are numbered from 1 to N, and duration is no more than 1440 minutes in a day.

Output Specification:

Print in each line all the detected suspects in a gang, in ascending order of their numbers. The gangs are printed in ascending order of their first members. The numbers in a line must be separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.

If no one is detected, output None instead.

Sample Input 1:

5 15 31
1 4 2
1 5 2
1 5 4
1 7 5
1 8 3
1 9 1
1 6 5
1 15 2
1 15 5
3 2 2
3 5 15
3 13 1
3 12 1
3 14 1
3 10 2
3 11 5
5 2 1
5 3 10
5 1 1
5 7 2
5 6 1
5 13 4
5 15 1
11 10 5
12 14 1
6 1 1
6 9 2
6 10 5
6 11 2
6 12 1
6 13 1

Sample Output 1:

3 5
6

Note: In sample 1, although 1 had 9 records, but there were 7 distinct receivers, among which 5 and 15 both had conversations lasted more than 5 minutes in total. Hence 1 had made 5 short phone calls and didn’t exceed the threshold 5, and therefore is not a suspect.

Sample Input 2:

5 7 8
1 2 1
1 3 1
1 4 1
1 5 1
1 6 1
1 7 1
2 1 1
3 1 1

Sample Output 2:

None

题目大意：编写一个程序，从大量的通话记录中检测出电信诈骗嫌疑人。如果一个人每天给不同的人打超过K个短电话，但这些人中回电的比例不超过20%，那他可能是嫌疑人。而且，如果两个嫌疑人互相打电话，我们认为他们可能属于同一个团伙。A给B打短电话是指A到B的通话总时长不超过5分钟。给三个正整数：短电话数量的阈值K，不同电话号码的数量N，通话记录数量M。接下来M行给出一天的通话记录，格式为呼叫者caller、接收者receiver、时长duration。输出格式：在每一行输出一个团伙中所有检测到的嫌疑人，按数字升序排列。这些帮派按第一个成员的升序排列。一行中的数字必须使用1个空格分隔，并且行首或行尾不得有多余空格。
分析：p为并查集的父集，sc中存储短通话的人数，rec中存储短通话收到回电的人数，mark用来标记是否输出过，record记录两个人的通话总时长，su中保存嫌疑人的编号。我们先记录下来两人之间的通话总时间，以此来统计短通话以及回电的次数，进而判断出某个人是不是嫌疑人。如果没有嫌疑人，则输出None。因为要按照帮派输出，所以我们用并查集把互相通过电话的嫌疑人设置到同一个集合内，然后就可以输出答案了~

#include <iostream>
#include <vector>
using namespace std;
int k, n, m, c, r, d, p[1001], sc[1001], rec[1001], mark[1001], record[1001][1001];
vector<int> su;
int Find(int a) {
    if (p[a] != a) return p[a] = Find(p[a]);
    return a;
}
void add(int a, int b) {
    int f1 = Find(a), f2 = Find(b);
    if (f1 < f2) p[f2] = f1;
    else p[f1] = f2;
}
int main() {
    for (int i = 1; i <= 1000; i++) p[i] = i;
    cin >> k >> n >> m;
    for (int i = 0; i < m; i++) {
        cin >> c >> r >> d;
        record[c][r] += d;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (record[i][j] && record[i][j] <= 5) {
                sc[i]++;
                if (record[j][i]) rec[i]++;
            }
        }
        if (sc[i] > k && rec[i] * 5 <= sc[i]) su.push_back(i);
    }
    if (su.empty()) {
        cout << "None";
        return 0;
    }
    for (int i = 0; i < su.size(); i++) {
        for (int j = i + 1; j < su.size(); j++) {
            if (record[su[i]][su[j]] && record[su[j]][su[i]]) add(su[i], su[j]);
        }
    }
    for (int i = 0; i < su.size(); i++) {
        if (mark[su[i]]) continue;
        cout << su[i];
        for (int j = i + 1; j < su.size(); j++) {
            if (Find(su[i]) == Find(su[j])) {
                cout << ' ' << su[j];
                mark[su[j]] = 1;
            }
        }
        cout << '\n';
    }
    return 0;
}

#include <iostream>

#include <vector>

using namespace std;

int k, n, m, c, r, d, p[1001], sc[1001], rec[1001], mark[1001], record[1001][1001];

vector<int> su;

int Find(int a) {

if (p[a] != a) return p[a] = Find(p[a]);

return a;

}

void add(int a, int b) {

int f1 = Find(a), f2 = Find(b);

if (f1 < f2) p[f2] = f1;

else p[f1] = f2;

}

int main() {

for (int i = 1; i <= 1000; i++) p[i] = i;

cin >> k >> n >> m;

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

cin >> c >> r >> d;

record[c][r] += d;

}

for (int i = 1; i <= n; i++) {

for (int j = 1; j <= n; j++) {

if (record[i][j] && record[i][j] <= 5) {

sc[i]++;

if (record[j][i]) rec[i]++;

}

if (sc[i] > k && rec[i] * 5 <= sc[i]) su.push_back(i);

}

if (su.empty()) {

cout << "None";

return 0;

}

for (int i = 0; i < su.size(); i++) {

for (int j = i + 1; j < su.size(); j++) {

if (record[su[i]][su[j]] && record[su[j]][su[i]]) add(su[i], su[j]);

}

for (int i = 0; i < su.size(); i++) {

if (mark[su[i]]) continue;

cout << su[i];

for (int j = i + 1; j < su.size(); j++) {

if (Find(su[i]) == Find(su[j])) {

cout << ' ' << su[j];

mark[su[j]] = 1;

}

cout << '\n';

}

return 0;

}

1157 Anniversary – PAT甲级真题

Zhejiang University is about to celebrate her 122th anniversary in 2019. To prepare for the celebration, the alumni association （校友会） has gathered the ID’s of all her alumni. Now your job is to write a program to count the number of alumni among all the people who come to the celebration.

Input Specification:

Each input file contains one test case. For each case, the first part is about the information of all the alumni. Given in the first line is a positive integer N (≤105). Then N lines follow, each contains an ID number of an alumnus. An ID number is a string of 18 digits or the letter X. It is guaranteed that all the ID’s are distinct.

The next part gives the information of all the people who come to the celebration. Again given in the first line is a positive integer M (≤105). Then M lines follow, each contains an ID number of a guest. It is guaranteed that all the ID’s are distinct.

Output Specification:

First print in a line the number of alumni among all the people who come to the celebration. Then in the second line, print the ID of the oldest alumnus — notice that the 7th – 14th digits of the ID gives one’s birth date. If no alumnus comes, output the ID of the oldest guest instead. It is guaranteed that such an alumnus or guest is unique.

Sample Input:

5
372928196906118710
610481197806202213
440684198612150417
13072819571002001X
150702193604190912
6
530125197901260019
150702193604190912
220221196701020034
610481197806202213
440684198612150417
370205198709275042

Sample Output:

3
150702193604190912

题目大意：为了准备校庆，校友会收集了所有校友的身份证号。编写程序，根据来参加校庆的所有人士的身份证号，统计来了多少校友。输入在第一行正整数N，随后N行，每行给出一位校友的身份证号（18位由数字和大写字母X组成的字符串）。题目保证身份证号不重复。随后给出前来参加校庆的所有人士的信息：首先是一个不超过正整数M，随后M行，每行给出一位人士的身份证号。题目保证身份证号不重复。首先在第一行输出参加校庆的校友的人数。然后在第二行输出最年长的校友的身份证号——注意身份证第7-14位给出的是yyyymmdd格式的生日。如果没有校友来，则在第二行输出最年长的来宾的身份证号。题目保证这样的校友或来宾必是唯一的。
分析：使用容器set<string> record存储校友的身份证信息，smallx记录校友中最年长的生日，smalla记录所有人中最年长的生日，ansx记录最年长的校友的身份证号，ansa记录所有人中最年长的人的身份证号，cnt记录有多少校友参加。将校友身份证存储在set容器record中，使用record.count来判断参加校庆的人是否是校友，使用substr(6, 8)快速提取身份证号中的生日信息，最后根据cnt的值输出~cnt不等于0，则说明有校友参加，输出最年长的校友的身份证号ansx；如果cnt等于0，说明没有校友参加，输出所有人中最年长的人的身份证号ansa~

#include <iostream>
#include <set>
using namespace std;
int n, m, cnt;
string temp, smallx = "99999999", smalla = "99999999", ansx, ansa;
set<string> record;
int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> temp;
        record.insert(temp);
    }
    cin >> m;
    for (int i = 0; i < m; i++) {
        cin >> temp;
        if (record.count(temp)) {
            cnt++;
            if (smallx > temp.substr(6, 8)) {
                smallx = temp.substr(6, 8);
                ansx = temp;
            }
        }
        if (smalla > temp.substr(6, 8)) {
                smalla = temp.substr(6, 8);
                ansa = temp;
        }
    }
    cout << cnt << endl;
    if (cnt) cout << ansx;
    else cout << ansa;
    return 0;
}

#include <iostream>

#include <set>

using namespace std;

int n, m, cnt;

string temp, smallx = "99999999", smalla = "99999999", ansx, ansa;

set<string> record;

int main() {

cin >> n;

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

cin >> temp;

record.insert(temp);

}

cin >> m;

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

cin >> temp;

if (record.count(temp)) {

cnt++;

if (smallx > temp.substr(6, 8)) {

smallx = temp.substr(6, 8);

ansx = temp;

}

if (smalla > temp.substr(6, 8)) {

smalla = temp.substr(6, 8);

ansa = temp;

}

cout << cnt << endl;

if (cnt) cout << ansx;

else cout << ansa;

return 0;

}

1156 Sexy Primes – PAT甲级真题

Sexy primes are pairs of primes of the form (p, p+6), so-named since “sex” is the Latin word for “six”. (Quoted from http://mathworld.wolfram.com/SexyPrimes.html)

Now given an integer, you are supposed to tell if it is a sexy prime.

Input Specification:

Each input file contains one test case. Each case gives a positive integer N (≤108).

Output Specification:

For each case, print in a line Yes if N is a sexy prime, then print in the next line the other sexy prime paired with N (if the answer is not unique, output the smaller number). Or if N is not a sexy prime, print No instead, then print in the next line the smallest sexy prime which is larger than N.

Sample Input 1:

Sample Output 1:

Yes
41

Sample Input 2:

Sample Output 2:

No
23

题目大意：性感素数是指形如 (p, p+6) 这样的一对素数。给定一个整数，判断其是否为一个性感素数。若N是一个性感素数，则在一行中输出Yes，并在第二行输出与N配对的另一个性感素数（若这样的数不唯一，输出较小的那个）。若N不是性感素数，则在一行中输出No，然后在第二行输出大于N的最小性感素数。
分析：is_prime判断是否为素数。判断p和p-6、p和p+6是否同时为素数，如果是，则为性感素数输出Yes。不是的话就从p+1往后找到第一个满足要求的数ans~

#include <iostream>
#include <cmath>
using namespace std;
int p, ans;
int is_prime(int x) {
    if (x < 2) return 0;
    for (int i = 2; i <= sqrt(x); i++)
        if (x % i == 0) return 0; 
    return 1;
}
int main() {
    cin >> p;
    if (is_prime(p) && is_prime(p - 6)) {
        cout << "Yes\n" << p - 6;
    } else if (is_prime(p) && is_prime(p + 6)) {
        cout << "Yes\n" << p + 6;
    } else {
        for (ans = p + 1; ; ans++) {
            if (is_prime(ans) && is_prime(ans - 6)) break;
            if (is_prime(ans) && is_prime(ans + 6)) break;
        }
        cout << "No\n" << ans;
    }
    return 0;
}

#include <iostream>

#include <cmath>

using namespace std;

int p, ans;

int is_prime(int x) {

if (x < 2) return 0;

for (int i = 2; i <= sqrt(x); i++)

if (x % i == 0) return 0;

return 1;

}

int main() {

cin >> p;

if (is_prime(p) && is_prime(p - 6)) {

cout << "Yes\n" << p - 6;

} else if (is_prime(p) && is_prime(p + 6)) {

cout << "Yes\n" << p + 6;

} else {

for (ans = p + 1; ; ans++) {

if (is_prime(ans) && is_prime(ans - 6)) break;

if (is_prime(ans) && is_prime(ans + 6)) break;

}

cout << "No\n" << ans;

}

return 0;

}

1110 区块反转 – PAT乙级真题

给定一个单链表 L，我们将每 K 个结点看成一个区块（链表最后若不足 K 个结点，也看成一个区块），请编写程序将 L 中所有区块的链接反转。例如：给定 L 为 1→2→3→4→5→6→7→8，K 为 3，则输出应该为 7→8→4→5→6→1→2→3。

输入格式：

每个输入包含 1 个测试用例。每个测试用例第 1 行给出第 1 个结点的地址、结点总个数正整数 N (≤105)、以及正整数 K (≤N)，即区块的大小。结点的地址是 5 位非负整数，NULL 地址用 −1 表示。

接下来有 N 行，每行格式为：

Address Data Next

其中 Address 是结点地址，Data 是该结点保存的整数数据，Next 是下一结点的地址。

输出格式：

对每个测试用例，顺序输出反转后的链表，其上每个结点占一行，格式与输入相同。

输入样例：

00100 8 3
71120 7 88666
00000 4 99999
00100 1 12309
68237 6 71120
33218 3 00000
99999 5 68237
88666 8 -1
12309 2 33218

输出样例：

71120 7 88666
88666 8 00000
00000 4 99999
99999 5 68237
68237 6 00100
00100 1 12309
12309 2 33218
33218 3 -1

分析：L为输入链表，ans为答案链表，二维数组E为区块链表。先将链表数据记录在结构体A中，遍历A将链表正确顺序记录在链表L中，然后需要重新定义链表长度n（因为有输入中有无效的假结点信息）。遍历链表L，将每K个结点所分隔成的区块保存在二维数组E中，再从后往前将区块链表E中的值添加到答案链表ans中，最后根据格式输出ans链表就好啦~

#include <iostream>
#include <vector>
using namespace std;
struct node {
    int data, next;
}A[100001];
vector<int> L, ans, E[100001];
int s, n , a, t, k, mark, cnt, c;
int main() {
    cin >> s >> n >> k;
    for (int i = 0; i < n; i++) {
        cin >> a;
        cin >> A[a].data >> A[a].next;
    }
    t = s;
    while (t != -1) {
        L.push_back(t);
        t = A[t].next;
    }
    n = L.size();
    for (int i = 0; i < n; i++) {
        E[c].push_back(L[i]);
        cnt++;
        if (cnt == k && i != n - 1) {
            cnt = 0;
            c++;
        }
    }
    for (int i = c; i >= 0; i--)
        for (auto it : E[i]) 
            ans.push_back(it);
    for (int i = 1; i < n; i++) 
        printf("%05d %d %05d\n", ans[i - 1], A[ans[i - 1]].data, ans[i]);
    printf("%05d %d -1", ans.back(), A[ans.back()].data);
    return 0;
}

#include <iostream>

#include <vector>

using namespace std;

struct node {

int data, next;

}A[100001];

vector<int> L, ans, E[100001];

int s, n , a, t, k, mark, cnt, c;

int main() {

cin >> s >> n >> k;

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

cin >> a;

cin >> A[a].data >> A[a].next;

}

t = s;

while (t != -1) {

L.push_back(t);

t = A[t].next;

}

n = L.size();

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

E[c].push_back(L[i]);

cnt++;

if (cnt == k && i != n - 1) {

cnt = 0;

c++;

}

for (int i = c; i >= 0; i--)

for (auto it : E[i])

ans.push_back(it);

for (int i = 1; i < n; i++)

printf("%05d %d %05d\n", ans[i - 1], A[ans[i - 1]].data, ans[i]);

printf("%05d %d -1", ans.back(), A[ans.back()].data);

return 0;

}

1109 擅长C – PAT乙级真题

当你被面试官要求用 C 写一个“Hello World”时，有本事像下图显示的那样写一个出来吗？

输入格式：

输入首先给出 26 个英文大写字母 A-Z，每个字母用一个 7×5 的、由 C 和 . 组成的矩阵构成。最后在一行中给出一个句子，以回车结束。句子是由若干个单词（每个包含不超过 10 个连续的大写英文字母）组成的，单词间以任何非大写英文字母分隔。

题目保证至少给出一个单词。

输出格式：

对每个单词，将其每个字母用矩阵形式在一行中输出，字母间有一列空格分隔。单词的首尾不得有多余空格。

相邻的两个单词间必须有一空行分隔。输出的首尾不得有多余空行。

输入样例：

..C..
.C.C.
C...C
CCCCC
C...C
C...C
C...C
CCCC.
C...C
C...C
CCCC.
C...C
C...C
CCCC.
.CCC.
C...C
C....
C....
C....
C...C
.CCC.
CCCC.
C...C
C...C
C...C
C...C
C...C
CCCC.
CCCCC
C....
C....
CCCC.
C....
C....
CCCCC
CCCCC
C....
C....
CCCC.
C....
C....
C....
CCCC.
C...C
C....
C.CCC
C...C
C...C
CCCC.
C...C
C...C
C...C
CCCCC
C...C
C...C
C...C
CCCCC
..C..
..C..
..C..
..C..
..C..
CCCCC
CCCCC
....C
....C
....C
....C
C...C
.CCC.
C...C
C..C.
C.C..
CC...
C.C..
C..C.
C...C
C....
C....
C....
C....
C....
C....
CCCCC
C...C
C...C
CC.CC
C.C.C
C...C
C...C
C...C
C...C
C...C
CC..C
C.C.C
C..CC
C...C
C...C
.CCC.
C...C
C...C
C...C
C...C
C...C
.CCC.
CCCC.
C...C
C...C
CCCC.
C....
C....
C....
.CCC.
C...C
C...C
C...C
C.C.C
C..CC
.CCC.
CCCC.
C...C
CCCC.
CC...
C.C..
C..C.
C...C
.CCC.
C...C
C....
.CCC.
....C
C...C
.CCC.
CCCCC
..C..
..C..
..C..
..C..
..C..
..C..
C...C
C...C
C...C
C...C
C...C
C...C
.CCC.
C...C
C...C
C...C
C...C
C...C
.C.C.
..C..
C...C
C...C
C...C
C.C.C
CC.CC
C...C
C...C
C...C
C...C
.C.C.
..C..
.C.C.
C...C
C...C
C...C
C...C
.C.C.
..C..
..C..
..C..
..C..
CCCCC
....C
...C.
..C..
.C...
C....
CCCCC
HELLO~WORLD!

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

136

137

138

139

140

141

142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157

158

159

160

161

162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178

179

180

181

182

183

184

..C..

.C.C.

C...C

CCCCC

C...C

CCCC.

C...C

CCCC.

C...C

CCCC.

.CCC.

C...C

C....

C...C

.CCC.

CCCC.

C...C

CCCC.

CCCCC

C....

CCCC.

C....

CCCCC

C....

CCCC.

C....

CCCC.

C...C

C....

C.CCC

C...C

CCCC.

C...C

CCCCC

C...C

CCCCC

..C..

CCCCC

....C

C...C

.CCC.

C...C

C..C.

C.C..

CC...

C.C..

C..C.

C...C

C....

CCCCC

C...C

CC.CC

C.C.C

C...C

CC..C

C.C.C

C..CC

C...C

.CCC.

C...C

.CCC.

CCCC.

C...C

CCCC.

C....

.CCC.

C...C

C.C.C

C..CC

.CCC.

CCCC.

C...C

CCCC.

CC...

C.C..

C..C.

C...C

.CCC.

C...C

C....

.CCC.

....C

C...C

.CCC.

CCCCC

..C..

C...C

.CCC.

C...C

.C.C.

..C..

C...C

C.C.C

CC.CC

C...C

.C.C.

..C..

.C.C.

C...C

.C.C.

..C..

CCCCC

....C

...C.

..C..

.C...

C....

CCCCC

HELLO~WORLD!

输出样例：

C...C CCCCC C.... C.... .CCC.
C...C C.... C.... C.... C...C
C...C C.... C.... C.... C...C
CCCCC CCCC. C.... C.... C...C
C...C C.... C.... C.... C...C
C...C C.... C.... C.... C...C
C...C CCCCC CCCCC CCCCC .CCC.

C...C .CCC. CCCC. C.... CCCC.
C...C C...C C...C C.... C...C
C...C C...C CCCC. C.... C...C
C.C.C C...C CC... C.... C...C
CC.CC C...C C.C.. C.... C...C
C...C C...C C..C. C.... C...C
C...C .CCC. C...C CCCCC CCCC.

C...C CCCCC C.... C.... .CCC.

C...C C.... C.... C.... C...C

CCCCC CCCC. C.... C.... C...C

C...C C.... C.... C.... C...C

C...C CCCCC CCCCC CCCCC .CCC.

C...C .CCC. CCCC. C.... CCCC.

C...C C...C C...C C.... C...C

C...C C...C CCCC. C.... C...C

C.C.C C...C CC... C.... C...C

CC.CC C...C C.C.. C.... C...C

C...C C...C C..C. C.... C...C

C...C .CCC. C...C CCCCC CCCC.

分析：三维数组a中储存26个字母对应的图形矩阵，out中储存每行要输出的图形矩阵，同时初始化out所有元素为空格。字符串中可能存在乱七八糟的字符，所以用getline做句子的输入。句子的单词间以任何非大写字母分隔，用while循环遍历找到下标j为非大写字母所在下标，i为当前单词首下标，然后根据单词中每一个字母，将输出图形记录到out中，最后输出out数组~flag用来标记之前是否输出过一个单词，如果输出过，flag=1，则当再次需要输出单词之前需要先输出一个’\n’~

#include <iostream>
using namespace std;
char a[26][7][5], out[7][100];
string s;
int main() {
    for (int i = 0; i < 7; i++)
        for (int j = 0; j < 100; j++)
            out[i][j] = ' ';
    for (int i = 0; i < 26; i++)
        for (int j = 0; j < 7; j++)
            for (int k = 0; k < 5; k++)
                cin >> a[i][j][k];
    getchar();
    getline(cin, s);
    for (int i = 0, j, flag = 0; i < s.size(); i++) {
        j = i;
        while (j < s.size() && s[j] >= 'A' && s[j] <= 'Z') j++;
        if (i == j) continue;
        for (int k = i; k < j; k++)
            for (int l = 0; l < 7; l++)
                for (int m = 0; m < 5; m++)
                    out[l][m + (k - i) * 6] = a[s[k] - 'A'][l][m];
        if (flag) cout << '\n';
        for (int k = 0; k < 7; k++) {
            flag = 1;
            for (int l = 0; l < 6 * (j - i) - 1; l++) cout << out[k][l];
            cout << '\n';
        }
        i = j;
    }
    return 0;
}

#include <iostream>

using namespace std;

char a[26][7][5], out[7][100];

string s;

int main() {

for (int i = 0; i < 7; i++)

for (int j = 0; j < 100; j++)

out[i][j] = ' ';

for (int i = 0; i < 26; i++)

for (int j = 0; j < 7; j++)

for (int k = 0; k < 5; k++)

cin >> a[i][j][k];

getchar();

getline(cin, s);

for (int i = 0, j, flag = 0; i < s.size(); i++) {

j = i;

while (j < s.size() && s[j] >= 'A' && s[j] <= 'Z') j++;

if (i == j) continue;

for (int k = i; k < j; k++)

for (int l = 0; l < 7; l++)

for (int m = 0; m < 5; m++)

out[l][m + (k - i) * 6] = a[s[k] - 'A'][l][m];

if (flag) cout << '\n';

for (int k = 0; k < 7; k++) {

flag = 1;

for (int l = 0; l < 6 * (j - i) - 1; l++) cout << out[k][l];

cout << '\n';

}

i = j;

}

return 0;

}

1108 String复读机 – PAT乙级真题

给定一个长度不超过 10⁴ 的、仅由英文字母构成的字符串。请将字符重新调整顺序，按 StringString.... （注意区分大小写）这样的顺序输出，并忽略其它字符。当然，六种字符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 String 的顺序打印，直到所有字符都被输出。例如 gnirtSSs 要调整成 StringS 输出，其中 s 是多余字符被忽略。

输入格式：

输入在一行中给出一个长度不超过 10⁴ 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：

sTRidlinSayBingStrropriiSHSiRiagIgtSSr

输出样例：

StringStringSrigSriSiSii

分析：字母的ASCII不会超过128，将输入的字符串in中所有字母出现的次数存储在数组a[128]中，然后重复遍历”String”字母，如果a数组中还有剩余某个对应字符就输出，并将该字符的剩余次数-1，flag = 1表示这轮循环还有输出，直到不再有输出（flag == 0）为止~

#include <iostream>
using namespace std;
string s = "String", in;
int a[200], flag = 1;
int main() {
    cin >> in;
    for (auto it : in) a[it]++;
    while (flag) {
        flag = 0;
        for (int i = 0; i < 6; i++) {
            if (a[s[i]]) {
                cout << s[i];
                a[s[i]]--;
                flag = 1;
            }
        }
    }
    return 0;
}

#include <iostream>

using namespace std;

string s = "String", in;

int a[200], flag = 1;

int main() {

cin >> in;

for (auto it : in) a[it]++;

while (flag) {

flag = 0;

for (int i = 0; i < 6; i++) {

if (a[s[i]]) {

cout << s[i];

a[s[i]]--;

flag = 1;

}

return 0;

}