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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
122
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
130
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
142
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
260
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
93
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