【PAT甲级】字符串处理及进制转换专题
2024-05-30 12:53:05
目录
- 字符串处理
- PAT甲级 1001 A+B Format (20 分)
- PAT甲级1005 Spell It Right (20 分)
- PAT甲级1035 Password (20 分)
- PAT甲级 1061 Dating (20 分) / PAT 乙级 1014 福尔摩斯的约会 (20分)
- PAT甲级 1038 Recover the Smallest Number (30 分)
- PAT甲级 1060 Are They Equal (25 分)
- PAT甲级 1073 Scientific Notation (20 分) / PAT 乙级 1024 科学计数法
- PAT甲级 1077 Kuchiguse (20 分)
- PAT甲级 1108 Finding Average (20 分) / PAT乙级 1054 求平均值 (20分)
- PAT甲级 1140 Look-and-say Sequence (20 分) / PAT乙级 1084 外观数列
- PAT甲级 1152 Google Recruitment (20 分) / PAT 乙级 1094 谷歌的招聘 (20分)
- PAT甲级 1082
- PAT乙级 1052 卖个萌
- 进制转换
- PAT甲级 1015 Reversible Primes (20 分)
- PAT甲级 1019 General Palindromic Number (20分)
- PAT甲级 1023 Have Fun with Numbers (20 分)
- PAT甲级 1024 Palindromic Number (25 分)
- PAT甲级 1136 A Delayed Palindrome (20分) / PAT 乙级 1079 延迟的回文数 (20 分)
- PAT甲级 1069 The Black Hole of Numbers (20 分) / PAT乙级 1019 数字黑洞
- PAT乙级 1074 宇宙无敌加法器
- PAT甲级 1027 Colors in Mars (20 分)
- PAT甲级 1100 Mars Numbers (20 分) / PAT乙级 1044 火星数字 (20分)
- PAT甲级 1058 A+B in Hogwarts (20 分)
字符串处理
插入：insert(pos, s) insert(it, s.begin(), s.end())
删除：erase(it) erase(s.begin(), s.end()) = clear()
截取：substr(pos, len)
寻找子串：find(s)
替换子串：replace(pos, len, s) replace(s.begin(), s.end(), s1)
// 消前导0
while(s.length() != 0 && s[0] == '0') s.erase(s.begin());
if(s.length() == 0) cout << 0;
cout << s;
PAT甲级 1001 A+B Format (20 分)
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main() {int a, b;cin >> a >> b;string s = to_string(a + b);string r = s; if (s[0] == '-') { // 处理负号 cout << "-"; r = s.substr(1);} string result; // 设置新字符串，以免添加逗号字符串变长 reverse(r.begin(), r.end()); // 倒序添加逗号 for (int i = 0; i < r.size(); i++) {result.push_back(r[i]); if ((i+1) % 3 == 0 && (i+1) != r.size()) result.push_back(',');}reverse(result.begin(), result.end());cout << t << endl;return 0;
}
PAT甲级1005 Spell It Right (20 分)
#include <iostream>
#include <string>
#include <vector>
#include <cstring>
#include <cstdio>
using namespace std;// 字符串数组
string num[]={"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}; int main() {string s;cin >> s;int sum = 0;for (int i = 0; s[i] != '\0'; i++) sum += s[i] - '0';s = to_string(sum);if (s[0] == '0') cout << num[0]; // 结果为0的情况else {for (int i = 0; i < s.size(); i++) {cout << num[s[i]-'0'];if (i != s.size()-1) cout << " ";} }return 0;
}
PAT甲级1035 Password (20 分)
#include <iostream>
#include <string>
#include <vector>
#include <map>
using namespace std;map<char, char> change = {{'1', '@'}, {'0', '%'}, {'l', 'L'}, {'O', 'o'}
};vector<string> v;int main() {int n = 0;cin >> n;for (int i = 0; i < n; i++) {string name, s;cin >> name >> s;bool flag = false;for (int j = 0; j < s.size(); j++) {if (change[s[j]]) { // 字符替换映射flag = true; s[j] = change[s[j]];}}if (flag) v.push_back(name + ' ' + s); // 整行保存存字符串}if (v.size() > 0) {cout << v.size() << endl;for (int i = 0; i < v.size(); i++)cout << v[i] << endl;}else if (n == 1) cout << "There is 1 account and no account is modified" << endl;else cout << "There are " << n << " accounts and no account is modified" << endl;return 0;
}
PAT甲级 1061 Dating (20 分) / PAT 乙级 1014 福尔摩斯的约会 (20分)
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;string week[] = { "MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN" };int main() {string a, b, c, d;cin >> a >> b >> c >> d;int i;for (i = 0; i < a.size() && i < b.size(); i++) {if (a[i] == b[i] && (a[i] >= 'A' && a[i] <= 'G')) {cout << week[a[i] - 'A'] << " ";break;}}for (i++; i < a.size() && i < b.size(); i++) {if (a[i] == b[i] && (a[i] >= '0' && a[i] <= '9')) {printf("%02d:", a[i] - '0');break;} else if (a[i] == b[i] && (a[i] >= 'A' && a[i] <= 'N')) {printf("%02d:", a[i] - 'A' + 10);break;}}for (i = 0; i < c.size() && i < d.size(); i++) {if (c[i] == d[i] && isalpha(c[i])) {printf("%02d", i); break;}}return 0;
}
PAT甲级 1038 Recover the Smallest Number (30 分)
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;bool cmp(string a, string b) {return a+b < b+a;
}string str[10010];
string s;int main() {int n = 0;cin >> n;for (int i = 0; i < n; i++)cin >> str[i];sort(str, str+n,cmp);for (int i = 0; i < n; i++)s += str[i];while(s.length() != 0 && s[0] == '0') s.erase(s.begin());if(s.length() == 0) cout << 0;cout << s;return 0;
}
PAT甲级 1060 Are They Equal (25 分)
自然数 →\to→ 科学计数法：搜索尾数和指数
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;string format(string s, int n) {while (s.length() > 0 && s[0] == '0') s.erase(s.begin());int e = 0;if (s[0] == '.') {s.erase(s.begin());while (s.length() > 0 && s[0] == '0') {s.erase(s.begin()); e--;}}else {int k = 0;while (k < s.length() && s[k] != '.') {k++; e++;} if (k < s.length()) s.erase(s.begin() + k);}if (s.length() == 0) e = 0;int len = 0;string res = "0.";for (int i = 0; i < n; i++) {if (len < s.length()) res += s[len++];else res += '0';}res = res + "*10^" + to_string(e);return res;
}int main() {int n = 0;string a, b, s1, s2;cin >> n >> a >> b;s1 = format(a, n);s2 = format(b, n);if (s1 == s2) {cout << "YES " << s1 << endl; }else {cout << "NO " << s1 << " " << s2 << endl; }return 0;
}
PAT甲级 1073 Scientific Notation (20 分) / PAT 乙级 1024 科学计数法
科学计数法 →\to→ 自然数： s.substr() 截取指数和尾数
#include <iostream>
#include <string>
using namespace std;int main() {string s;cin >> s;if (s[0] == '-') cout << s[0];int pos = s.find('E');string integer = s.substr(1,1);string point = s.substr(3,pos-3);int expon = stoi(s.substr(pos+1));if (expon == 0) {cout << integer << "." << point << endl;}else if (expon < 0) {cout << "0.";for (int i = 0; i < abs(expon)-1; i++)cout << "0";cout << integer << point << endl;}else {cout << integer;int length = point.size();if (length < expon) { // 尾数位数 < 指数cout << point;for (int i = 0; i < expon - length; i++)cout << "0";}else {for (int i = 0; i < length; i++) { // 尾数位数 > 指数cout << point[i];if (length-expon > 0 && i == expon-1)cout << ".";}}}return 0;
}
PAT甲级 1077 Kuchiguse (20 分)
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;int main() { // 最长公共后缀int n;cin >> n;getchar();vector<string> v(n);int minn = 256; for (int i = 0; i < n; i++) { getline(cin, v[i]); // 按行存储字符串 if (v[i].size() < minn) minn = v[i].size(); // 应该比较的最短长度 reverse(v[i].begin(), v[i].end()); // 反转以便比较后缀 } int cnt = 0;for (int j = 0; j < minn; j++) {bool flag = true;for (int i = 1; i < n; i++) { if (v[0][j] != v[i][j]) { // 逐个比较 flag = false; break;}}if (!flag) break;cnt++;}if (cnt == 0) cout << "nai" << endl;else {string s = v[0].substr(0, cnt);reverse(s.begin(), s.end());cout << s << endl;}return 0;
}
PAT甲级 1108 Finding Average (20 分) / PAT乙级 1054 求平均值 (20分)
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;int main() {int n = 0, cnt = 0;scanf("%d", &n);char a[50], b[50];double temp = 0.0, sum = 0.0;for (int i = 0; i < n; i++) {scanf("%s", a);sscanf(a, "%lf", &temp); // 字符数组->浮点数sprintf(b, "%.2lf", temp); // 浮点数->字符数组bool flag = false;for (int j = 0; j < strlen(a); j++) if (a[j] != b[j]) flag = true; if (flag || temp < -1000 || temp > 1000) { // 非法输入printf("ERROR: %s is not a legal number\n", a);continue;} else {sum += temp; cnt++; // 总和及个数}}if(cnt == 1)printf("The average of 1 number is %.2f", sum);else if(cnt > 1)printf("The average of %d numbers is %.2f", cnt, sum / cnt);elseprintf("The average of 0 numbers is Undefined");return 0;
}
PAT甲级 1140 Look-and-say Sequence (20 分) / PAT乙级 1084 外观数列
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;int main() {int d = 0, n = 0;cin >> d >> n;string s = to_string(d); // 首字符for (int i = 0; i < n-1; i++) { // 当 n>=2 处理字符串string t;int same = 1, len = s.size();for (int j = 0; j < len; j++) {if (j+1 == len) { // 末尾字符单独处理 t += s[j];t += to_string(same);} else {if (s[j] == s[j+1]) same++; // 段内连续相等字符数 else {t += s[j]; t += to_string(name); same = 1; // 还原下一段相等计数器}}}s = t; // 上一轮生成字符串作为下一轮检查的字符串 }cout << s << endl;return 0;
}
PAT甲级 1152 Google Recruitment (20 分) / PAT 乙级 1094 谷歌的招聘 (20分)
#include <iostream>
#include <string>
#include <cmath>
using namespace std;bool IsPrime(int prime){if(prime < 2) return false;int limit = int(sqrt(prime*1.0));for(int i=2; i <= limit; i++){if(prime % i == 0) return false;}return true;
}int main() {int l = 0, k = 0;string n;cin >> l >> k >> n;int prime = 0;string s;for (int i = 0; i <= l-k; i++) {s = n.substr(i,k);prime = stoi(s);if (IsPrime(prime)) {cout << s << endl;return 0;}}cout << "404" << endl;return 0;
}
PAT甲级 1082
PAT乙级 1052 卖个萌
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;int main() {string s[3];for (int i = 0; i < 3; i++) // 原字符串行getline(cin, s[i]); vector<string> v[3];for (int i = 0; i < 3; i++) {int start = 0, stop = 0;for (int j = 0; j < s[i].size(); j++) {if (s[i][j] == '[') {start = j+1; stop = 0;}else if (s[i][j] == ']') {v[i].push_back(s[i].substr(start, stop)); // [表情]向量}else {stop++;}}}int n = 0;cin >> n;for (int i = 0; i < n; i++) {int arr[5] = {0};bool flag = false;for (int i = 0; i < 5; i++) {cin >> arr[i]; if (arr[i] < 1) flag = true; // 序号过小}int len1 = v[0].size(), len2 = v[1].size(), len3 = v[2].size();if (arr[0] > len1 || arr[1] > len2 || arr[2] > len3 || arr[3] > len2 || arr[4] > len1) // 序号过大flag = true;if (flag) cout << "Are you kidding me? @\\/@" << endl;else {cout << v[0][arr[0]-1] << "(" << v[1][arr[1]-1] << v[2][arr[2]-1] << v[1][arr[3]-1] << ")" << v[0][arr[4]-1] << endl;}}return 0;
}
进制转换
PAT甲级 1015 Reversible Primes (20 分)
#include <iostream>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;bool IsPrime(int n) {if (n < 2) return false;int limit = int(sqrt(n*1.0));for (int i = 2; i <= limit; i++)if (n % i == 0) return false;return true;
}int reverse(int n, int base) {vector<int> v;while (n != 0) { // 除基取余获取反转数字v.push_back(n % base);n /= base;}int result = 0, temp = 1; // 反转数字的值for (int i = v.size()-1; i >= 0; i--) {result += v[i] * temp;temp *= base; }return result;
}int main() {int n = 0, base = 0;while (cin >> n && n >= 0) {cin >> base;int r = reverse(n, base);if (IsPrime(n) && IsPrime(r)) cout << "Yes" << endl;else cout << "No" << endl; }return 0;
}
PAT甲级 1019 General Palindromic Number (20分)
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;vector<int> palin(int n, int radix) { // 进制转换 vector<int> v; // 十进制以上需要保存多位，所以不能使用 string 存储 while (n != 0) {v.push_back( n % radix );n /= radix; }return v;
}int main() {int n = 0, radix = 0;cin >> n >> radix;vector<int> v = palin(n, radix);if (v.size() == 0) {cout << "Yes" << endl;cout << 0 << endl;return 0;}vector<int> vec = v;reverse(vec.begin(), vec.end());if (vec == v) cout << "Yes" << endl;else cout << "No" << endl;for (int i = 0; i < vec.size(); i++) {cout << vec[i];if (i != vec.size()-1) cout << " ";}return 0;
}
PAT甲级 1023 Have Fun with Numbers (20 分)
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;string compute(string s, string r) {string k;int temp = 0, carry = 0;for (int i = 0; i < s.size(); i++) { // 模拟竖式加法temp = (s[i]-'0') + (r[i]-'0')+ carry; // 相加k += to_string(temp % 10); // 低位余数carry = temp / 10; // 高位进位}if (carry != 0) k += to_string(carry); // 最高位进位reverse(k.begin(),k.end());return k;
}int main() {string s, r;cin >> s;reverse(s.begin(), s.end());r = compute(s, s);string a = s, b = r;sort(a.begin(), a.end());sort(b.begin(), b.end());if (a == b) cout << "Yes" <<endl;else cout << "No" << endl;cout << r << endl;return 0;
}
PAT甲级 1024 Palindromic Number (25 分)
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;string compute(string s, string r) {string k;int temp = 0, carry = 0;for (int i = 0; i < s.size(); i++) { // 模拟竖式加法temp = (s[i]-'0') + (r[i]-'0')+ carry; // 相加k += to_string(temp % 10); // 低位余数carry = temp / 10; // 高位进位}if (carry != 0) k += to_string(carry); // 最高位进位reverse(k.begin(),k.end());return k;
}int main() {int k = 0;string s;cin >> s >> k;string r = s;reverse(r.begin(), r.end());if (s == r) {cout << r << endl << 0 << endl;return 0;}int cnt = 0;while (cnt < k) {s= compute(s, r);r = s;reverse(s.begin(), s.end());cnt++;if (s == r) break;}cout << r << endl << cnt << endl;return 0;
}
PAT甲级 1136 A Delayed Palindrome (20分) / PAT 乙级 1079 延迟的回文数 (20 分)
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;string compute(string s, string r) {string k;int temp = 0, carry = 0;for (int i = 0; i < s.size(); i++) { // 模拟竖式加法temp = (s[i]-'0') + (r[i]-'0')+ carry; // 相加k += to_string(temp % 10); // 低位余数carry = temp / 10; // 高位进位}if (carry != 0) k += to_string(carry); // 最高位进位reverse(k.begin(),k.end());return k;
}int main() {string s;cin >> s;string r = s;reverse(r.begin(), r.end());if (s == r) {cout << s << " is a palindromic number." << endl;return 0;} int cnt = 0;while (cnt++ < 10) {string k = compute(s, r); // string k = to_string( stoll(s) + stoll(r) ); 最后一个样例超时cout << s << " + " << r << " = " << k << endl;s = k; r = s;reverse(r.begin(), r.end());if (s == r) {cout << s << " is a palindromic number." << endl;break;}}if (cnt == 11) cout << "Not found in 10 iterations." << endl;return 0;
}
PAT甲级 1069 The Black Hole of Numbers (20 分) / PAT乙级 1019 数字黑洞
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;bool cmp(char a, char b) {return a > b;
}int main() {string s;cin >> s;s.insert(0, 4 - s.size(), '0');do {string a = s, b = s;sort(a.begin(), a.end(), cmp);sort(b.begin(), b.end());int result = stoi(a) - stoi(b);s = to_string(result);s.insert(0, 4 - s.size(), '0'); // 添加前缀0cout << a << " - " << b << " = " << s << endl; } while (s != "0000" && s != "6174");return 0;
}
PAT乙级 1074 宇宙无敌加法器
PAT甲级 1060 模拟竖式加法 + PAT甲级1136 清除前导零 + PAT甲级 1069 添加前导零
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;vector<int> v;string compute(string s, string r) {string k;int i, temp = 0, carry = 0;for (i = 0; i < s.size(); i++) { // 竖式加法temp = (s[i]-'0') + (r[i]-'0')+ carry; k += to_string(temp % v[i]);carry = temp / v[i]; }if (carry != 0) k += to_string(carry);reverse(k.begin(),k.end());return k;
}int main() {string s, a, b;cin >> s >> a >> b;while (a.length() != 0 && a[0] == '0') a.erase(a.begin()); // 清除无用前缀零while (b.length() != 0 && b[0] == '0') b.erase(b.begin());if (a.length() == 0 && b.length() == 0) { // 计算结果为零cout << 0 << endl; return 0;}if (a.length() > b.length()) b.insert(0, a.length()-b.length(), '0'); // 添加前缀零对齐if (b.length() > a.length()) a.insert(0, b.length()-a.length(), '0');reverse(s.begin(), s.end());reverse(a.begin(), a.end());reverse(b.begin(), b.end());for (int i = 0; i < s.size(); i++) { if (s[i] == '0') v.push_back(10); // 0表示十进制else v.push_back(s[i] - '0');}cout << compute(a,b) << endl;return 0;
}
PAT甲级 1027 Colors in Mars (20 分)
#include <iostream>
using namespace std;int main() {int a[3];for (int i = 0; i < 3; i++)scanf("%d", &a[i]);printf("#");for (int i = 0; i < 3; i++) printf("%X%X", a[i]/13, a[i]%13); // 十六进制输出printf("\n");return 0;
}
PAT甲级 1100 Mars Numbers (20 分) / PAT乙级 1044 火星数字 (20分)
#include<iostream>
#include<string>
using namespace std;string units[13] = {"tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"
};
string tens[13] = {" ", "tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"
};void etom(int e) {if (e / 13) cout << tens[e / 13];if ((e / 13) && (e % 13)) cout << " ";if (e % 13 || e == 0) cout << units[e % 13];
}void mtoe(string s) {string s1 = s.substr(0,3), s2;if (s.size() > 4) s2 = s.substr(4,3);int num1 = 0, num2 = 0;for (int i = 1; i <= 12; i++) {if (s1 == tens[i]) num1 = i;if (s1 == units[i] || s2 == units[i]) num2 = i;}cout << num1*13 + num2;
}int main() {int n;cin >> n;getchar();string s;for (int i = 0; i < n; i++) {getline(cin, s);isdigit(s[0]) ? etom(stoi(s)) : mtoe(s);cout << endl;}return 0;
}
PAT甲级 1058 A+B in Hogwarts (20 分)
#include <iostream>
using namespace std;const int Galleon = 29*17;
const int Sickle = 29;
int main() {long a1, b1, c1; // 用int可能会溢出long a2, b2, c2;scanf("%ld.%ld.%ld %ld.%ld.%ld", &a1, &b1, &c1, &a2, &b2, &c2);long long c = (a1+a2)*Galleon + (b1+b2)*Sickle + c1 + c2;printf("%ld.%ld.%ld", c/Galleon, c%Galleon/Sickle, c%Sickle);return 0;
}
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