题解思路

这是一道典型的递推动态规划结合组合数学的问题，核心在于通过卡特兰数建模有效的栈操作序列，并利用组合数学计算特定元素贡献的方案数。

核心算法分析

问题本质理解： 题目要求计算所有有效操作序列的收益总和。每个有效序列都是一个合法的入栈出栈序列，这正对应于卡特兰数的经典应用场景。

关键数学建模：

卡特兰数应用： 长度为2n的合法括号序列（入栈出栈序列）的数量为第n个卡特兰数： Cn=1n+1(2nn)C_n = \frac{1}{n+1}\binom{2n}{n}

组合计数核心： 对于元素a[i]在栈大小为k时被弹出的贡献，需要计算满足以下条件的序列数量：

1. a[i]入栈前有(i-1)次入栈和(i-k)次出栈
2. a[i]入栈到出栈之间的子序列是合法的
3. a[i]出栈后剩余操作构成合法序列

状态分解策略：

前缀有效序列数：

// p次入栈，q次出栈的有效前缀序列数（栈高度>=0）
private static long validPrefixCount(int pushCount, int popCount) {
    if (pushCount < popCount) return 0;
    return binomialCoeff[pushCount + popCount][pushCount] - 
           binomialCoeff[pushCount + popCount][pushCount + 1];
}

后缀完整序列数：

// P个新元素，栈中初始s个元素的完整有效序列数
private static long validSuffixCount(int newElements, int initialStack) {
    if (newElements < 0) return 0;
    int total = 2 * newElements + initialStack;
    return binomialCoeff[total][newElements] - 
           binomialCoeff[total][newElements + initialStack + 1];
}

中间子序列处理： 对于a[i]入栈到出栈之间处理t个后续元素的情况，使用卡特兰数： ways(t)=Ct=1t+1(2tt)\text{ways}(t) = C_t = \frac{1}{t+1}\binom{2t}{t}

详细算法设计

预计算组合数和卡特兰数：

// 预计算二项式系数
long[][] binomialCoeff = new long[2*n+1][2*n+1];
for (int i = 0; i <= 2*n; i++) {
    binomialCoeff[i][0] = 1;
    for (int j = 1; j <= i; j++) {
        binomialCoeff[i][j] = binomialCoeff[i-1][j-1] + binomialCoeff[i-1][j];
    }
}

// 预计算卡特兰数
long[] catalan = new long[n+1];
catalan[0] = 1;
for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
        catalan[i+1] += catalan[j] * catalan[i-j];
    }
}

贡献计算核心逻辑：

long totalContribution = 0;

for (int elementIndex = 1; elementIndex <= n; elementIndex++) {
    for (int stackSize = 1; stackSize <= elementIndex; stackSize++) {
        
        // 计算中间子序列的总方案数
        long intermediateSum = 0;
        for (int processedCount = 0; processedCount <= n - elementIndex; processedCount++) {
            long remainingCount = n - elementIndex - processedCount;
            intermediateSum += catalan[processedCount] * 
                             validSuffixCount(remainingCount, stackSize - 1);
        }
        
        // 总方案数 = 前缀方案数 × 中间方案数
        long totalWays = validPrefixCount(elementIndex - 1, elementIndex - stackSize) * 
                        intermediateSum;
        
        // 累加贡献
        totalContribution += array_A[elementIndex-1] * array_B[stackSize-1] * totalWays;
    }
}

复杂度分析：

• 预计算：O(n²)
• 主循环：O(n³)
• 总体复杂度：O(n³)
• 空间复杂度：O(n²)

Java代码实现

import java.io.*;
import java.util.*;

public class Main {
    
    private static long[][] binomialCoefficients;
    private static long[] catalanNumbers;
    
    private static void precomputeCombinatorics(int maxN) {
        int size = 2 * maxN + 1;
        binomialCoefficients = new long[size][size];
        
        // 计算二项式系数 C(n,k)
        for (int i = 0; i < size; i++) {
            binomialCoefficients[i][0] = 1;
            for (int j = 1; j <= i; j++) {
                binomialCoefficients[i][j] = binomialCoefficients[i-1][j-1] + 
                                           binomialCoefficients[i-1][j];
            }
        }
        
        // 计算卡特兰数
        catalanNumbers = new long[maxN + 1];
        catalanNumbers[0] = 1;
        
        for (int i = 0; i < maxN; i++) {
            catalanNumbers[i + 1] = 0;
            for (int j = 0; j <= i; j++) {
                catalanNumbers[i + 1] += catalanNumbers[j] * catalanNumbers[i - j];
            }
        }
    }
    
    private static long getBinomialCoeff(int n, int k) {
        if (k < 0 || k > n || n < 0) return 0;
        return binomialCoefficients[n][k];
    }
    
    // 计算有效前缀序列的数量（栈操作序列的前缀，保证栈高度非负）
    private static long computeValidPrefixSequences(int pushOps, int popOps) {
        if (pushOps < popOps) return 0;
        return getBinomialCoeff(pushOps + popOps, pushOps) - 
               getBinomialCoeff(pushOps + popOps, pushOps + 1);
    }
    
    // 计算完整有效序列的数量（给定初始栈状态和新元素数量）
    private static long computeValidCompleteSequences(int newElementCount, int initialStackSize) {
        if (newElementCount < 0) return 0;
        int totalOperations = 2 * newElementCount + initialStackSize;
        return getBinomialCoeff(totalOperations, newElementCount) - 
               getBinomialCoeff(totalOperations, newElementCount + initialStackSize + 1);
    }
    
    private static long calculateTotalBenefit(int n, int[] arrayA, int[] arrayB) {
        precomputeCombinatorics(n);
        
        long totalBenefit = 0;
        
        // 遍历每个元素和每个可能的栈大小
        for (int elementPos = 1; elementPos <= n; elementPos++) {
            for (int stackDepth = 1; stackDepth <= elementPos; stackDepth++) {
                
                // 计算元素a[elementPos]在栈深度为stackDepth时被弹出的总方案数
                long totalSchemesSum = 0;
                
                // 枚举在该元素入栈到出栈之间处理的后续元素数量
                for (int intermediateProcessed = 0; intermediateProcessed <= n - elementPos; intermediateProcessed++) {
                    int remainingToProcess = n - elementPos - intermediateProcessed;
                    
                    // 中间处理intermediateProcessed个元素的方案数（卡特兰数）
                    long intermediateWays = catalanNumbers[intermediateProcessed];
                    
                    // 后续处理remainingToProcess个元素的完整方案数
                    long suffixWays = computeValidCompleteSequences(remainingToProcess, stackDepth - 1);
                    
                    totalSchemesSum += intermediateWays * suffixWays;
                }
                
                // 前缀部分的有效方案数
                long prefixWays = computeValidPrefixSequences(elementPos - 1, elementPos - stackDepth);
                
                // 总的有效操作方案数
                long totalValidSchemes = prefixWays * totalSchemesSum;
                
                // 累加该元素的贡献
                totalBenefit += (long)arrayA[elementPos - 1] * arrayB[stackDepth - 1] * totalValidSchemes;
            }
        }
        
        return totalBenefit;
    }
    
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        
        int n = Integer.parseInt(reader.readLine());
        
        String[] aInput = reader.readLine().split(" ");
        int[] arrayA = new int[n];
        for (int i = 0; i < n; i++) {
            arrayA[i] = Integer.parseInt(aInput[i]);
        }
        
        String[] bInput = reader.readLine().split(" ");
        int[] arrayB = new int[n];
        for (int i = 0; i < n; i++) {
            arrayB[i] = Integer.parseInt(bInput[i]);
        }
        
        long result = calculateTotalBenefit(n, arrayA, arrayB);
        System.out.println(result);
        
        reader.close();
    }
}

C++代码实现

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class StackBenefitCalculator {
private:
    vector<vector<long long>> combinationTable;
    vector<long long> catalanSequence;
    int maxSize;
    
    void initializeCombinatorics(int n) {
        maxSize = 2 * n + 1;
        combinationTable.assign(maxSize, vector<long long>(maxSize, 0));
        
        // 构建二项式系数表
        for (int i = 0; i < maxSize; i++) {
            combinationTable[i][0] = 1;
            for (int j = 1; j <= i; j++) {
                combinationTable[i][j] = combinationTable[i-1][j-1] + combinationTable[i-1][j];
            }
        }
        
        // 构建卡特兰数序列
        catalanSequence.assign(n + 1, 0);
        catalanSequence[0] = 1;
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= i; j++) {
                catalanSequence[i + 1] += catalanSequence[j] * catalanSequence[i - j];
            }
        }
    }
    
    long long getCombination(int n, int k) {
        if (k < 0 || k > n || n < 0 || n >= maxSize) return 0;
        return combinationTable[n][k];
    }
    
    // 计算部分栈操作序列的有效数量
    long long countValidPartialSequences(int pushCount, int popCount) {
        if (pushCount < popCount) return 0;
        return getCombination(pushCount + popCount, pushCount) - 
               getCombination(pushCount + popCount, pushCount + 1);
    }
    
    // 计算完整栈操作序列的有效数量
    long long countValidFullSequences(int newElements, int initialDepth) {
        if (newElements < 0) return 0;
        int totalOps = 2 * newElements + initialDepth;
        return getCombination(totalOps, newElements) - 
               getCombination(totalOps, newElements + initialDepth + 1);
    }
    
public:
    long long solve(int n, const vector<int>& valuesA, const vector<int>& valuesB) {
        initializeCombinatorics(n);
        
        long long totalRevenue = 0;
        
        // 枚举每个元素位置和对应的栈深度
        for (int pos = 1; pos <= n; pos++) {
            for (int depth = 1; depth <= pos; depth++) {
                
                // 计算该元素在指定深度被弹出时的所有方案
                long long aggregatedWays = 0;
                
                // 枚举中间处理的元素数量
                for (int midProcessed = 0; midProcessed <= n - pos; midProcessed++) {
                    int laterProcessed = n - pos - midProcessed;
                    
                    // 中间处理方案数（卡特兰数）
                    long long midWays = catalanSequence[midProcessed];
                    
                    // 后续处理方案数
                    long long laterWays = countValidFullSequences(laterProcessed, depth - 1);
                    
                    aggregatedWays += midWays * laterWays;
                }
                
                // 前序处理方案数
                long long prefixWays = countValidPartialSequences(pos - 1, pos - depth);
                
                // 该配置下的总有效方案数
                long long configurationWays = prefixWays * aggregatedWays;
                
                // 累加收益贡献
                totalRevenue += static_cast<long long>(valuesA[pos - 1]) * 
                               valuesB[depth - 1] * configurationWays;
            }
        }
        
        return totalRevenue;
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;
    
    vector<int> arrayA(n), arrayB(n);
    
    for (int i = 0; i < n; i++) {
        cin >> arrayA[i];
    }
    
    for (int i = 0; i < n; i++) {
        cin >> arrayB[i];
    }
    
    StackBenefitCalculator calculator;
    long long result = calculator.solve(n, arrayA, arrayB);
    
    cout << result << endl;
    
    return 0;
}

时间复杂度：O(n³)，主要由三重循环决定 空间复杂度：O(n²)，用于存储组合数表