课评: 转出学生归档（王子墨/张立坤/黄俊博）+ 周四班级总结 + 新生章梓皓档案 + C++示例代码

C++代码/GESP考级/一级/202609新龟兔赛跑.cpp

@@ -0,0 +1,26 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+using namespace std;
+
+int n;
+int arr[100100];
+
+int main(){
+	double wut, tuv, tut; // 乌龟到终点的时间  兔子的速度  兔子休息的时间 
+	cin >> wut >> tuv >> tut;
+	
+	// 先求出兔子到终点的时间
+	double t = 1000.0 / tuv + tut;
+	
+	if(t == wut){
+		cout << "tie";
+	} else if(t < wut){
+		cout << "rabbit";
+	} else {
+		cout << "turtle";
+	}
+	cout << endl;
+	cout << fixed << setprecision(2) << t;
+	return 0;
+}
+

C++代码/GESP考级/一级/202609棋盘上的奖赏.cpp

@@ -0,0 +1,19 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+using namespace std;
+
+int n;
+int arr[100100];
+
+int main(){
+	cin >> n;
+	long long cnt = 1;
+	
+	for(int i = 1; i <= n; i++){
+		cnt = cnt * 2;
+	}
+	
+	cout << cnt - 1;
+	return 0;
+}
+

C++代码/GESP考级/三级/202609公共二进制位.cpp

@@ -0,0 +1,53 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+using namespace std;
+
+int n, num;
+int s[100]; 
+
+string str = "01";
+
+// 10进制转换成 2进制
+string f(int n){
+	if(n == 0) return "0";
+	string re = "";
+	while(n > 0){
+		int g = n % 2;
+		
+		re = str[g] + re;
+		
+		n /= 2;
+	} 
+	return re;
+} 
+
+
+int main(){
+	cin >> n;
+	int maxx = 0; 
+	for(int i = 1; i <= n; i++){
+		cin >> num;
+		string curr = f(num);
+		int len = curr.size();
+		int idx = 0;
+		for(int j = len - 1; j >= 0; j--){
+			char c = curr[j];
+			if(c == '1'){
+				s[idx]++;
+			}
+			idx++;
+		}
+		maxx = max(maxx, idx); // 统计最长字符串
+//		cout << "当前这个数二进制是" << curr << "    " << idx << endl; 
+	}
+	
+	int cnt = 0;
+	for(int i = 0; i < maxx; i++){
+		if(s[i] == n) {
+			cnt++;
+		}
+	}
+	cout << cnt;
+	return 0;
+}
+

C++代码/GESP考级/三级/202609分割字符串.cpp

@@ -0,0 +1,46 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+using namespace std;
+
+string s;
+
+int main(){
+	getline(cin, s);
+	
+	int len = s.size();
+	string curr = ""; // 当前这个字符串
+	int k = 1; // 需要多少个空格 
+	int idx = 0; // 当前的下标
+	
+	while(idx < len){
+		char c = s[idx];
+		if(c == ' '){
+			// 当遇到空格
+			string z = s.substr(idx, k);
+			bool ok = true;
+			for(char c1 : z){
+				if(c1 != ' ') {
+					ok = false;
+					break;
+				} 
+			} 
+			if(ok){
+				cout << curr << endl;
+				idx += k;
+				k++; // 下一次的空格要加一 
+				curr = "";
+				continue; 
+			}
+		}
+		
+		curr += c;
+		if(idx == len - 1){
+			cout << curr << endl;
+		}
+		idx++;
+	} 
+	
+	
+	return 0;
+}
+

C++代码/GESP考级/三级/202609分割字符串解法2.cpp

@@ -0,0 +1,36 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+using namespace std;
+
+string s;
+
+int main() {
+	getline(cin, s);
+
+	int len = s.size();
+	string curr = ""; // 要输出的这个字符串
+	int kg = 1; // 中断的空格数量
+	int cnt = 0; // 当前空格的数量
+
+	for(char c : s) {
+		curr += c;
+		if(c == ' ') {
+			cnt++;
+		} else {
+			cnt = 0;
+		}
+
+		if(cnt == kg) {
+			cout << curr << endl;
+			cnt = 0;
+			kg++;
+			curr = "";
+		}
+
+	}
+	if(curr.size() >= 1) {
+		cout << curr << endl;
+	}
+	return 0;
+}
+

C++代码/GESP考级/三级/202609分割字符串解法3.cpp

@@ -0,0 +1,30 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+using namespace std;
+
+string s;
+
+
+int main(){
+	getline(cin, s);
+	
+	string kg = "";
+	int l = 0, r = 0;
+	
+	while(true){
+		kg = kg + ' ';
+		if(s.find(kg, l) == string::npos){
+			// 如果没有找到
+			r = s.size() - 1;
+			cout << s.substr(l, r - l) << endl;
+			break; 
+		} else {
+			// 找到了
+			r = s.find(kg, l);
+			cout << s.substr(l, r - l) << endl;
+			l = r + kg.size();
+		}
+	}
+	return 0;
+}
+

C++代码/GESP考级/二级/202609字符变换.cpp

@@ -0,0 +1,23 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+using namespace std;
+
+int a, b, c;
+
+int main(){
+	cin >> a >> b >> c;
+	int cnt = 0;
+	for(int i = 1; i <= 8; i++){
+		for(int j = 1 ;j <= 16; j++){
+			cnt++;
+			if(cnt == a || cnt == b || cnt == c){
+				cout << (char)cnt;
+			}else {
+				cout << '*';
+			}
+		}
+		cout << endl;
+	}
+	return 0;
+}
+

C++代码/GESP考级/二级/202609直角三角形.cpp

@@ -0,0 +1,32 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+using namespace std;
+
+double a, b, t;
+int cnt = 3;
+
+int main(){
+	cin >> a >> b >> t;
+	
+	double curr = 0;
+	
+	if(a > t){
+		cout << 1;
+		return 0;
+	} else if(b > t){
+		cout << 2;
+		return 0; 
+	}
+	
+	curr = sqrt(a * a + b * b); // µÚÈýÏî 
+	while(curr <= t){
+		cnt++;
+		a = b;
+		b = curr;
+		curr = sqrt(a * a + b * b);
+	}
+	
+	cout << cnt;
+	return 0;
+}
+