codechef div4

Author: coderhare

Diff for: codechef/June Long One 2022 Division 4 (Rated).md

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+> 这个比赛有72个小时，我刚上去看的时候发现还剩下7个小时，就做了一下
+> 感觉后面还有3道全是数学题，有点做不下去，希望有空能补上
+
+### A
+```c++
+void solve() {
+    int n;
+    cin >> n;
+    while(n--){
+        int a, b;
+        cin >> a >> b;
+        cout << max(0, a - b) << endl;
+    }
+}
+
+```
+
+### B
+```c++
+void solve() {
+    int n;
+    cin >> n;
+    while(n--){
+        int a, b;
+        cin >> a >> b;
+        cout << ((a - 1)/6 + 1) * 1LL * b << endl;
+    }
+}
+```
+
+### C
++1, +2, +1, +2...这样每2次为一回合会漏掉一个，规律来看就是以2位首项的等差数列
+```c++
+void solve() {
+    int n;
+    cin >> n;
+    while(n--){
+        int a, b;
+        cin >> a >> b;
+        //1 3 4 6 7 9 10 12
+        //2 5 8 11
+        int rest = b - a;
+        if(rest < 0 || (rest % 3 == 2)) cout << "NO" << endl;
+        else cout << "YES" << endl;
+    }
+}
+```
+
+### D
+模式匹配，主要是看某个串是否完全匹配模式串，如果不能则答案+1
+```c++
+void solve() {
+    int n;
+    cin >> n;
+    string a, b;
+    cin >> a >> b;
+    vector<vector<int>> pos(256);
+    for (int i = 0; i < b.size(); i++) pos[b[i] - 'a'].push_back(i);
+    int ans = 0;
+    for (int i = 0; i < 26; i++) {
+        bool ok = 0;
+        for (int j: pos[i]) {
+            if (a[j] != b[j]) {
+                ok = 1;
+                break;
+            }
+        }
+        ans += ok;
+    }
+    cout << ans << endl;
+}
+
+```
+
+### E
+构造题，事实上更好的构造方法可以将[1,..., N]两段分开，然后交错拼接
+下面的做法主要是观察到最明显的性质：只有后面为`[1,N]`满足N-1的diff，
+```c++
+void solve() {
+    bitset<100010> st;
+    int n;
+    cin >> n;
+    vector <int> ans(n + 1);
+    ans[n] = n, ans[n - 1] = 1;
+    st[n] = st[1] = 1;
+    for (int i = n - 2; i > 0; i--){
+        int last = ans[i + 1], t;
+        if(last + i > 1 && last + i < n && !st[last + i]) st[last + i] = 1, t = last + i;
+        else if(last - i > 1 && last - i < n && !st[last - i]) st[last - i] = 1, t = last - i;
+        ans[i] = t;
+    }
+    for (int i = 1; i <= n; i++) {
+        cout << ans[i] << " \n"[i == n];
+    }
+}
+```
+### F
+一点点数学，先找到它们的差值，两者的gcd是有办法达到这个差值的，让次小值达到这个值。然后思考，此时这两个数实际上是倍数关系，能有方案
+使得他们减去某个值，也达到某个该最大公约数的约数的情况。
+用O(sqrt(n))的筛法即可得到约数分布情况
+```c++
+void solve() {
+    int a, b;
+    cin >> a >> b;
+    int d = abs(a - b);
+    int ans = 0;
+    for (int i = 1; i <= sqrt(d); i++){
+        if(d % i == 0) {
+            ans++;
+            if(d != i * i) ans++;
+        }
+    }
+    cout << ans << endl;
+}
+```