《如 何 速 通 一 套 题》4.0

A sprial

找规律。

直接做。

#include <bits/stdc++.h>
#define int long long
using namespace std;
 
int t, n;
 
int sqrtll(int n) {
  int l = 1, r = 1000000, ans = 0;
  for(; l <= r; ) {
    int mid = (l + r) >> 1;
    if(mid * mid >= n) {
      ans = mid, r = mid - 1;
    }else {
      l = mid + 1;
    }
  }
  return ans;
}
 
signed main() {
  freopen("spiral.in", "r", stdin);
  freopen("spiral.out", "w", stdout);
  for(cin >> t; t--; ) {
    cin >> n;
    int i = sqrtll(n);
    int tmp = n - (i - 1) * (i - 1);
    if((i & 1)) {
      if(tmp <= i) {
        cout << i << ' ' << tmp << '\n';
      }else {
        cout << 2 * i - tmp << ' ' << i << '\n';
      }
    }else {
      if(tmp <= i) {
        cout << tmp << ' ' << i << '\n';
      }else {
        cout << i << ' ' << 2 * i - tmp << '\n';
      }
    }
  }
  return 0;
}

B write

C curve

$$\because \frac{x}{3}\le \lfloor \frac{x}{2}\rfloor \le \frac{x}{2}$$

$$\therefore \frac{x}{3^{r-l+1}}+\sum _{i=l}^r\frac{a_i}{3^{r-i+1}}\le \lfloor \frac{x}{2^{r-l+1}}+\sum _{i=l}^r\frac{a_i}{2^{r-i+1}}\rfloor \le \frac{x}{2^{r-l+1}}+\sum _{i=l}^r\frac{a_i}{2^{r-i+1}}$$

$$\because a_i,x\le {10}^9$$

$$\therefore \text{当 }r-i+1\ge 30\ge {\log }_2{10}^9,\frac{x}{3^{r-i+1}}\to 0,\frac{x}{2^{r-i+1}}\to 0,$$

$$\therefore \lfloor \frac{x}{2^{r-i+1}}\rfloor \to 0$$

直接暴力。

#include <bits/stdc++.h>
using namespace std;
 
int n, arr[200020], q, x, l, r;
 
int main() {
  freopen("curve.in", "r", stdin);
  freopen("curve.out", "w", stdout);
  cin >> n;
  for(int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for(cin >> q; q--; ) {
    cin >> x >> l >> r;
    l = max(l, r - 50);
    for(int i = l; i <= r; i++) {
      x = (x + arr[i]) / 2;
    }
    cout << x << '\n';
  }
  return 0;
}

D coin

乱搞

我们充分发挥人类智慧，根据数学直觉，不能凑出的数必然在 $2.69\times {10}^7$ 以上。

直接对于 $2.69\times {10}^7$ 下面的数暴力 dp，上面的都凑的出来。

/**
 * @brief "According" to my direst sence,
 * @brief the maximum number which "will output -1" is not greater than 2.69*(10^7).
*/
#include <bits/stdc++.h>
#define int long long
using namespace std;
 
const int kCstf = 2.69e7;
 
int n, l, arr[22], dp[kCstf + 20];
 
signed main() {
  freopen("coin.in", "r", stdin);
  freopen("coin.out", "w", stdout);
  cin >> n >> l;
  for(int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  if(l <= kCstf) {
    dp[0] = 1;
    for(int i = 1; i <= n; i++) {
      for(int j = arr[i]; j <= l; j++) {
        dp[j] |= dp[j - arr[i]];
      }
    }
    int ans = 0;
    for(int i = 1; i <= l; i++) {
      ans += dp[i];
    }
    cout << ans << '\n';
  }else {
    dp[0] = 1;
    for(int i = 1; i <= n; i++) {
      for(int j = arr[i]; j <= kCstf; j++) {
        dp[j] |= dp[j - arr[i]];
      }
    }
    int ans = 0;
    for(int i = 1; i <= kCstf; i++) {
      ans += dp[i];
    }
    cout << ans + (l - kCstf) << '\n';
  }
  return 0;
}

正解

同余最短路，模数是最大值。