Refactor C++ solution in 1.cpp by improving code readability, adding a function to calculate the number of digits, and enhancing logic for counting valid pairs A and B based on their digit lengths.

Author: tkzzzzzz6

niuke/daily_problem/25_11-21小苯的计算式/1.cpp

@@ -1,46 +1,82 @@
-#include<bits/stdc++.h>
+#include <bits/stdc++.h>
 #define il inline
 using namespace std;
 
 #define pb push_back
-#define fastio \
-  ios::sync_with_stdio(false); \
-  cin.tie(0);
+#define fastio                   \
+    ios::sync_with_stdio(false); \
+    cin.tie(0);
 
 typedef long long ll;
 typedef unsigned long long ull;
 
-const ll N = 5e5+5, mod = 1e9+7, inf = 2e18;
+const ll N = 5e5 + 5, mod = 1e9 + 7, inf = 2e18;
 const double eps = 1e-9;
 const double PI = 3.1415926;
 
-il void solve(){
-    int n,c;
+// 获取数字的位数
+int getNumlen(int num)
+{
+    if (num == 0)
+    {
+        return 1;
+    }
+    int len = 0;
+    while (num > 0)
+    {
+        ++len;
+        num /= 10;
+    }
+    return len;
+}
+
+il void solve()
+{
+    int n, c;
     cin >> n >> c;
-    n = n - 3;
+
+    // 计算 C 的位数，得到 A+B 的位数和
+    int len_c = getNumlen(c);
+    n = n - len_c - 2; // 减去 C 的位数和两个符号('+' 和 '=')
+
+    // 边界判断：A 和 B 至少各占 1 位
+    if (n < 2)
+    {
+        cout << 0 << endl;
+        return;
+    }
+
     int cnt = 0;
-    for (int i = 0; i < n;++i)
+    // 枚举 A 从 0 到 C，计算对应的 B，判断位数是否匹配
+    for (int a = 0; a <= c; ++a)
     {
-        for (int j = 0; j < n-i;++j)
+        int b = c - a;
+        if (b < 0)
+            continue;
+
+        int len_a = getNumlen(a);
+        int len_b = getNumlen(b);
+
+        // 检查 A 和 B 的位数之和是否等于目标长度
+        if (len_a + len_b == n)
         {
-            if(i + j == c)
-            {
-                ++cnt;
-            }
+            ++cnt;
         }
     }
+
+    cout << cnt << endl;
 }
 
 int main()
 {
     fastio
-    
-    int t = 1;
-    cin >> t;
-    while(t--)
+
+        int t = 1;
+    // cin >> t;
+    while (t--)
     {
         solve();
     }
-    
+
     return 0;
 }