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RSS订阅原创 PAT (Basic Level) Practice （中文）1058 选择题
PAT (Basic Level) Practice （中文） 1058
2022-06-02 23:38:58
125
原创 PAT (Basic Level) Practice （中文）1103 缘分数
#include<iostream> #include<cmath> #include<stack> using namespace std; // 记录答案 struct solution { int a, b; solution(int a1, int b1) :a(a1), b(b1) {}; friend ostream& operator<<(ostream& out, const solution&...
2022-05-26 20:59:12
134
原创 PAT (Basic Level) Practice （中文） 1064 朋友数
map<long, int> company; // 记录种类 map自动实现排序 void deal(const string& num) { long sum = 0; for (int i = 0; i < num.size(); i++)sum += num[i] - '0'; if (company.find(sum) != company.end()) company[sum]++; else company.insert({ sum,1 }); } i.
2022-05-24 17:03:30
145
原创 ＜ POJ 1002 : 方便记忆的电话号码＞
#include<iostream> #include<string> #include<map> #include<ctype.h> using namespace std; int main() { map<char, char>phonenumber{ {'a','2'},{'b','2'},{'c','2'} ,{'d','3'},{'e','3'},{'f','3'},{'g','4'},{'h','4'}, {'i','4'},.
2022-05-17 14:22:44
267
原创 ＜POJ 1005 I Think I Need a Houseboat ＞
#include<iostream> #include<cmath> using namespace std; int main() { const double pai = 3.1415926; int n; cin >> n; for (int Z = 1; Z <= n; Z++) { long squre = 50; int N = 1; double R = sqrt(2 * squre / pai); double X, Y; c.
2022-05-17 13:25:39
96
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