Core Logic and Mathematical Principles
Counting Inversion Pairs with Merge Sort
An inversion pair is defined as a pair $(i, j)$ satisfying $i < j$ and $A[i] > A[j]$. Using the divide-and-conquer approach, the sequence $A[l..r]$ is divided into $A[l..mid]$ and $A[mid+1..r]$. During the merge step with two pointers, if the element pointed to by the left sequence pointer $p_1$ is greater than the element pointed to by the right sequence pointer $p_2$, since the left sequence is already sorted, all elements in $A[p_1..mid]$ are greater than $A[p_2]$. At this point, the contribution to the count of inversion pairs across the two parts is:
$$ \Delta = mid - p_1 + 1 $$
The overall time complexity is $O(N \log N)$, and the space complexity is $O(N)$.
Fast Exponentiation Using Bit Manipulation
To compute $a^b \pmod m$, if the linear recursive complexity is $O(b)$, it will definitely lead to TLE when $b \le 10^{18}$. By using binary decomposition, $b$ can be expressed as:
$$ b = \sum_{i=0}^{k} c_i \cdot 2^i \quad (c_i \in {0, 1}) $$
Thus:
$$ a^b = \prod_{i=0}^{k} a^{c_i \cdot 2^i} $$
Using the doubling method, we update $a \leftarrow a^2 \pmod m$ in each iteration. If the current least significant bit of $b$ is $1$ ($b \& 1 = 1$), we multiply the current value into the answer. The time complexity is $O(\log b)$, and the space complexity is $O(1)$.
Algorithm Derivation and State Design
Derivation of Inversion Pairs with Merge Sort
The divide-and-conquer process satisfies the recurrence relation:
$$ T(N) = 2T\left(\frac{N}{2}\right) + O(N) $$
Solving this gives $T(N) = O(N \log N)$. During the merge, we set two pointers $i = l, j = mid + 1$.
- If $A[i] \le A[j]$: $A[i]$ is correctly positioned, and no inversion pair is created, so $i \leftarrow i + 1$.
- If $A[i] > A[j]$: $A[j]$ is correctly positioned, implying that all elements from $A[i]$ to $mid$ form inversion pairs with $A[j]$. We update the counter as $ans \leftarrow ans + (mid - i + 1)$ and increment $j \leftarrow j + 1$.
Derivation of Fast Exponentiation Using Bit Manipulation
Let $ans = 1 \pmod m$. The loop continues while $b > 0$:
- Check the least significant bit:
if (b & 1)$\Rightarrow$ $ans = ans \cdot a \pmod m$. - Square the base: $a = a \cdot a \pmod m$.
- Right shift the exponent:
b >>= 1.
C++ Standard Code
#include <iostream>
#include <vector>
using namespace std;
// Counting inversion pairs with merge sort, using global variables or passing by reference to avoid copying overhead
long long inverse_pairs_count = 0;
void merge_sort(vector<int>& a, vector<int>& temp, int l, int r) {
if (l >= r) return;
int mid = l + ((r - l) >> 1); // Prevent overflow, use bit manipulation for speed
merge_sort(a, temp, l, mid);
merge_sort(a, temp, mid + 1, r);
int i = l, j = mid + 1, k = l;
while (i <= mid && j <= r) {
if (a[i] <= a[j]) {
temp[k++] = a[i++];
} else {
temp[k++] = a[j++];
// Critical pitfall: must convert to long long to prevent overflow during summation; accumulate remaining left interval count at once
inverse_pairs_count += (mid - i + 1);
}
}
while (i <= mid) temp[k++] = a[i++];
while (j <= r) temp[k++] = a[j++];
for (i = l; i <= r; ++i) {
a[i] = temp[i];
}
}
// Fast exponentiation using bit manipulation
long long quick_pow(long long a, long long b, long long m) {
long long ans = 1 % m; // Critical pitfall: if m=1, the answer must be 0, 1%1=0
a %= m; // Prevent initial a from exceeding m to avoid overflow in subsequent multiplications
while (b > 0) {
if (b & 1) {
ans = (__int128)ans * a % m; // Use __int128 to prevent overflow when multiplying two long long values
}
a = (__int128)a * a % m;
b >>= 1;
}
return ans;
}
int main() {
// Optimize standard input/output stream, disconnect from stdin/stdout synchronization to improve IO efficiency
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
if (!(cin >> n)) return 0;
vector<int> a(n);
vector<int> temp(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
merge_sort(a, temp, 0, n - 1);
cout << inverse_pairs_count << "\n";
return 0;
}NOIP Practical Pitfalls Guide
- Inversion pair counter overflow with
intIn the extreme case of a sequence of length $N$, the number of inversion pairs can be as high as $\frac{N(N-1)}{2}$. For the NOIP data range $N = 5 \times 10^5$, the maximum number of inversion pairs is about $1.25 \times 10^{11}$, far exceeding theintlimit of $2 \times 10^9$. It is essential to uselong longto store the counter, or it will overflow and become negative. - Fast exponentiation multiplication overflow with
long longand not taking modulus of $1$ When the modulus $m \approx 10^{18}$, multiplying twolong longvalues can reach $10^{36}$, causing overflow in a 64-bit register. In a Linux 64-bit environment, it is necessary to use__int128for forced type conversion before taking modulus. Additionally, when $b=0$ and $m=1$, directly returning $1$ will lead to incorrect judgment; the initial answer should be written as1 % m.
Classic NOIP/Luogu Problems
Luogu P1908 Inversion Pairs
- Problem Description: Given a sequence of length $N$, find the total number of inversion pairs. $N \le 5 \times 10^5$, $A[i] \le 10^9$.
- Core Idea and Thought Process: This is a pure inversion pair template problem. Due to the large data range, a brute-force enumeration of $O(N^2)$ is not feasible. The core idea is to utilize the divide-and-conquer nature of merge sort to calculate inversion pairs across two halves in $O(N \log N)$ time complexity.
Luogu P1226 [Template] Fast Exponentiation
- Problem Description: Given three integers $a, b, p$, find the value of $a^b \pmod p$. $a, b, p \le 2 \times 10^9$.
- Core Idea and Thought Process: This problem examines modular arithmetic under exponential growth. The key idea is to decompose the large exponent $b$ into binary representation, transforming the exponentiation into $O(\log b)$ multiplications. Each time, we check the current binary bit with $b \ & \ 1$ to dynamically maintain the squared base values.