🎨 style: improve format of the code

AdvancedLevel_C++/1070. Mooncake (25).cpp

@@ -12,12 +12,9 @@ int main() {
     int n, need;
     cin >> n >> need;
     vector<mooncake> a(n);
-    for (int i = 0; i < n; i++)
-        cin >> a[i].mount;
-    for (int i = 0; i < n; i++)
-        cin >> a[i].price;
-    for (int i = 0; i < n; i++)
-        a[i].unit = a[i].price / a[i].mount;
+    for (int i = 0; i < n; i++) scanf("%f", &a[i].mount);
+    for (int i = 0; i < n; i++) scanf("%f", &a[i].price);
+    for (int i = 0; i < n; i++) a[i].unit = a[i].price / a[i].mount;
     sort(a.begin(), a.end(), cmp);
     float result = 0.0;
     for (int i = 0; i < n; i++) {

AdvancedLevel_C++/1110. Complete Binary Tree (25).cpp

@@ -1,56 +1,42 @@
 #include <iostream>
-#include <queue>
-#include <vector>
-#include <string>
 using namespace std;
-struct TREE {
-    int left, right;
-};
+struct node{
+    int l, r;
+}a[100];
+int maxn = -1, ans;
+void dfs(int root, int index) {
+    if(index > maxn) {
+        maxn = index;
+        ans = root;
+    }
+    if(a[root].l != -1) dfs(a[root].l, index * 2);
+    if(a[root].r != -1) dfs(a[root].r, index * 2 + 1);
+}
 int main() {
-    int n, root = 0;
-    scanf("%d", &n);
-    vector<TREE> tree(n);
-    vector<int> book(n);
-    for(int i = 0; i < n; i++) {
+    int n, root = 0, have[100] = {0};
+    cin >> n;
+    for (int i = 0; i < n; i++) {
         string l, r;
         cin >> l >> r;
-        if(l == "-") {
-            tree[i].left = -1;
+        if (l == "-") {
+            a[i].l = -1;
         } else {
-            tree[i].left = stoi(l);
-            book[tree[i].left] = 1;
+            a[i].l = stoi(l);
+            have[stoi(l)] = 1;
         }
-        if(r == "-"){
-            tree[i].right = -1;
+        if (r == "-") {
+            a[i].r = -1;
         } else {
-            tree[i].right = stoi(r);
-            book[tree[i].right] = 1;
-        }   
-    }
-    for(int i = 0; i < n; i++) {
-        if(book[i] == 0) {
-            root = i;
-            break;
-        }
-    }
-    queue<int> q;
-    q.push(root);
-    int cnt = 0, lastnode = 0;
-    while(!q.empty()) {
-        int node = q.front();
-        q.pop();
-        if(node != -1) {
-            lastnode = node;
-            cnt++;
-        }else {
-            if(cnt != n)
-                printf("NO %d", root);
-            else
-                printf("YES %d", lastnode);
-            return 0;
+            a[i].r = stoi(r);
+            have[stoi(r)] = 1;
         }
-        q.push(tree[node].left);
-        q.push(tree[node].right);
     }
+    while (have[root] != 0) root++;
+    dfs(root, 1);
+    if (maxn == n)
+        cout << "YES " << ans;
+    else
+        cout << "NO " << root;
     return 0;
-}
+}
+

BasicLevel_C++/1027. 打印沙漏(20).cpp

@@ -1,31 +1,25 @@
 #include <iostream>
 using namespace std;
 int main() {
-    int N;
+    int N, row = 0;
     char c;
     cin >> N >> c;
-    int row = 0;
     for (int i = 0; i < N; i++) {
         if ((2 * i * (i + 2) + 1) > N) {
             row = i - 1;
             break;
         }
     }
     for (int i = row; i >= 1; i--) {
-        for (int k = row - i; k >= 1; k--)
-            cout << " ";
-        for (int j = i * 2 + 1; j >= 1; j--)
-            cout << c;
+        for (int k = row - i; k >= 1; k--) cout << " ";
+        for (int j = i * 2 + 1; j >= 1; j--) cout << c;
         cout << endl;
     }
-    for (int i = 0; i < row; i++)
-        cout << " ";
+    for (int i = 0; i < row; i++) cout << " ";
     cout << c << endl;
     for (int i = 1; i <= row; i++) {
-        for (int k = row - i; k >= 1; k--)
-            cout << " ";
-        for (int j = i * 2 + 1; j >= 1; j--)
-            cout << c;
+        for (int k = row - i; k >= 1; k--) cout << " ";
+        for (int j = i * 2 + 1; j >= 1; j--) cout << c;
         cout << endl;
     }
     cout << (N - (2 * row * (row + 2) + 1));

BasicLevel_C++/1031. 查验身份证(15).cpp

@@ -3,26 +3,21 @@
 using namespace std;
 bool func(string a);
 int main() {
-    int n;
-    int count = 0;
+    int n, count = 0;
     cin >> n;
     for (int i = 0; i < n; i++) {
         string a;
         cin >> a;
         if (func(a)) count++;
     }
-    if (count == 0) 
-        cout << "All passed";
+    if (count == 0) cout << "All passed";
     return 0;
 }
-
 bool func(string s) {
-    int sum = 0;
-    int a[18];
+    int sum = 0, a[18];
     for (int i = 0; i < 17; i++)
         a[i] = s[i] - '0';
-    if (s[17] == 'X')
-        a[17] = 10;
+    if (s[17] == 'X') a[17] = 10;
     else
         a[17] = s[17] - '0';
     int b[17] = {7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2};

CCCC-GPLT/L1-002. 打印沙漏 .cpp

@@ -1,44 +1,27 @@
-include <iostream>
+#include <iostream>
 using namespace std;
 int main() {
-    int N;
-    cin >> N;
+    int N, row = 0;
     char c;
-    cin >> c;
-    //计算行数row，只计算一半的行数，不包括最中间的那个一个符号
-    int row = 0;
-    for (int i = 1; i < N; i++) {
+    cin >> N >> c;
+    for (int i = 0; i < N; i++) {
         if ((2 * i * (i + 2) + 1) > N) {
             row = i - 1;
             break;
         }
     }
-    // 打印上半部分
     for (int i = row; i >= 1; i--) {
-        for (int k = row - i; k >= 1; k--) {
-            cout << " ";
-        }
-        for (int j = i * 2 + 1; j >= 1; j--) {
-            cout << c;
-        }
+        for (int k = row - i; k >= 1; k--) cout << " ";
+        for (int j = i * 2 + 1; j >= 1; j--) cout << c;
         cout << endl;
     }
-    // 打印中间的那个符号
-    for (int i = 0; i < row; i++) {
-        cout << " ";
-    }
+    for (int i = 0; i < row; i++) cout << " ";
     cout << c << endl;
-    // 打印下半部分
     for (int i = 1; i <= row; i++) {
-        for (int k = row - i; k >= 1; k--) {
-            cout << " ";
-        }
-        for (int j = i * 2 + 1; j >= 1; j--) {
-            cout << c;
-        }
+        for (int k = row - i; k >= 1; k--) cout << " ";
+        for (int j = i * 2 + 1; j >= 1; j--) cout << c;
         cout << endl;
     }
-    // 输出还剩下几个符号没有用
-        cout << (N - (2 * row * (row + 2) + 1));
+    cout << (N - (2 * row * (row + 2) + 1));
     return 0;
 }

CCCC-GPLT/L1-003. 个位数统计 .cpp

@@ -6,13 +6,9 @@ int main() {
     cin >> s;
     int len = s.length();
     int a[10] = {0};
-    for (int i = 0; i < len; i++) {
-        a[s[i] - '0']++;
-    }
+    for (int i = 0; i < len; i++) a[s[i] - '0']++;
     for (int i = 0; i < 10; i++) {
-        if (a[i] != 0) {
-            cout << i << ":" << a[i] << endl;
-        }
+        if (a[i] != 0) cout << i << ":" << a[i] << endl;
     }
     return 0;
 }

CCCC-GPLT/L1-005. 考试座位号 .cpp

@@ -1,29 +1,23 @@
 #include <iostream>
+#include <vector>
 using namespace std;
 struct stu {
     string s;
-    int one;
-    int two;
+    int one, two;
 };
-
 int main() {
-    int n;
+    int n, m, temp;
     cin >> n;
-    stu *a = new stu [n];
+    vector<stu> a(n);
     for (int i = 0; i < n; i++) {
-        cin >> a[i].s;
-        cin >> a[i].one;
-        cin >> a[i].two;
+        cin >> a[i].s >> a[i].one >> a[i].two;
     }
-    int m;
     cin >> m;
-    int temp;
     for (int i = 0; i < m; i++) {
         cin >> temp;
         for (int j = 0; j < n; j++) {
             if (a[j].one == temp) {
-                cout << a[j].s << " ";
-                cout << a[j].two << endl;
+                cout << a[j].s << " " << a[j].two << endl;
                 break;
             }
         }

CCCC-GPLT/L1-006. 连续因子.cpp

@@ -1,33 +1,3 @@
-#include <iostream>
-#include <cstdio>
-#include <cmath>
-using namespace std;
-int main() {
-    int n;
-    scanf("%d", &n);
-    int max = sqrt(n);
-    for(int len = 12; len >= 1; len--) {
-        for(int start = 2; start <= max; start++) {
-            long long int ans = 1;
-            for(int i = start; i - start <= len - 1; i++) {
-                ans *= i;
-            }
-            if(n % ans == 0) {
-                printf("%d\n%d", len, start);
-                for(int i = start + 1; i - start <= len - 1; i++) {
-                    printf("*%d", i);
-                }
-                return 0;
-            }
-        }
-    }
-    printf("1\n%d", n);
-    return 0;
-}
-
-[Update v2.0] 由github用户littlesevenmo提供的更高效的解法
-不用算连续因子最多不会超过12个。也不需要三重循环，两重循环即可，直接去计算当前部分乘积能不能整除N，代码如下：
-
 #include <iostream>
 #include <cmath>
 using namespace std;

CCCC-GPLT/L1-008. 求整数段和.cpp

@@ -4,8 +4,7 @@ using namespace std;
 int main() {
     int a, b;
     cin >> a >> b;
-    int temp = a;
-    int sum = 0;
+    int temp = a, sum = 0;
     while(temp <= b) {
         for(int j = 0; j < 5 && temp <= b; j++) {
             printf("%5d", temp);

CCCC-GPLT/L1-010. 比较大小.cpp

@@ -3,15 +3,9 @@ using namespace std;
 int main() {
     int a, b, c;
     cin >> a >> b >> c;
-    if(b > c) {
-        swap(b, c);
-    }
-    if(a > b) {
-        swap(a, b);
-    }
-    if(b > c) {
-        swap(b, c);
-    }
+    if(b > c) swap(b, c);
+    if(a > b) swap(a, b);
+    if(b > c) swap(b, c);
     cout << a << "->" << b << "->" << c;
     return 0;
 }

CCCC-GPLT/L1-011. A-B.cpp

@@ -5,13 +5,9 @@ int main() {
     string s, a;
     getline(cin, s);
     getline(cin, a);
-    for(int i = 0; i < a.length(); i++) {
-        book[a[i]] = 1;
-    }
+    for(int i = 0; i < a.length(); i++) book[a[i]] = 1;
     for(int i = 0; i < s.length(); i++) {
-        if(book[s[i]] == 1) {
-            continue;
-        }
+        if(book[s[i]] == 1) continue;
         cout << s[i];
     }
     return 0;

CCCC-GPLT/L1-013. 计算阶乘和.cpp

@@ -1,14 +1,11 @@
 #include <iostream>
 using namespace std;
 int main() {
-	int n;
+	int n, ans = 0;
 	cin >> n;
-	int ans = 0;
 	for(int i = 1; i <= n; i++) {
 		int temp = 1;
-		for(int j = 1; j <= i; j++) {
-			temp *= j;
-		}
+		for(int j = 1; j <= i; j++) temp *= j;
 		ans += temp;
 	}
 	cout << ans;