update Starter72B (Rated for div2).md

Author: coderhare

Diff for: codechef/Starter72B (Rated for div2).md

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+# Starter72B (Rated for div2)
+
+这场打得很窘迫。
+
+## A
+
+1可以作为除数，接着我们考虑它的两个不同因数是否存在即可
+
+```cpp
+void solve() {
+    int n;
+    cin >> n;
+    for (int i = 2; i <= sqrt(n); i++) {
+        if (n % i==0 && n/i != i){
+            cout << i << " " << n/i << " " << 1 << "\n";
+            return;
+        }
+    }
+    cout << -1 << "\n";
+}
+```
+
+
+
+## B
+
+公式：∑(Ai * K^{i-1}) = S
+
+没有想到证明，只是一开始想用背包做，但是`1≤S≤1e18`实在太大了，于是转而思考别的做法，最后尝试了贪心的做法，
+
+先将S减至为≤0，接着从大到小依次减去相应的数字，看看是否能使其为0.
+
+```cpp
+void solve() {
+    ll n, k, s;
+    cin >> n >> k >> s;
+    int ed = 0;
+    ll base = 1;
+    vector <int> ans(n);
+    for (int i = 0; i < n; i++) {
+        s -= base;
+        ed = i;
+        if (s <= 0) break;
+        base *= k;
+    }
+    s = -s;
+    fill(begin(ans), begin(ans)+ed+1,1);
+    for (int i = ed; i >= 0; i--) {
+        if (s >= base) {
+            if (s >= 2 * base) ans[i] = -1, s -= 2 * base;
+            else ans[i] = 0, s -= base;
+        }
+        base /= k;
+    }
+    if (s != 0) {
+        cout << -2 << "\n";
+        return;
+    }
+    for (int x: ans) cout << x << " ";
+    cout << "\n";
+}
+```
+
+
+
+## C
+
+推公式，可以发现，题目要求的条件可以转化为`2*s[r] - r - 1 = 2 * s[l] - l`，其中`s[i]`为前缀`[1..i]`的1的数目。
+
+于是可以做一次DP，最后根据DP值回溯即可
+
+```cpp
+void solve() {
+    int n;
+    cin >> n;
+    vector <int> a(n);
+    for (int i = 0; i < n; i++) {
+        cin >> a[i];
+    }
+    vector <int> s(n+10);
+    for (int i = 1; i <= n; i++) {
+        s[i] = s[i-1] + a[i-1];
+        //cout << s[i] << " " << s[i-1] << " " << a[i] << "\n";
+    }
+    unordered_map <int, int> m;
+    int ans = 0;
+    vector <int> f(n+10);
+    int mn = 2000;
+    for (int i = 1; i <= n; i++) {
+        int now = 2 * s[i] - i-1;
+        if (a[i-1]==0) continue;
+        int q = m[now];
+        f[i] = q+1;
+        m[now+1] = f[i];
+        ans = max(ans, q+1);
+    }
+    int idx = max_element(begin(f), end(f)) - begin(f);
+    vector <int> res;
+    res.push_back(idx+1);
+    for (int i = idx; i >= 0; i--) {
+        if (ans && ans == f[i]) {
+            res.push_back(i);
+            ans--;
+        }
+    }
+    reverse(all(res));
+    cout << res.size() << "\n";
+    for (int x: res) cout << x << " ";
+    cout << "\n";
+}
+```
+