题解思路

这是一道典型的数位动态规划问题，核心在于处理超大整数的约束计数，通过tight状态精确控制搜索边界。

核心算法分析

数位DP状态设计： 我们需要维护两个关键状态：

• tight状态：当前构造的数字前缀是否与n的前缀完全相等
• 约束满足：每一位是否满足给定的约束条件

状态转移方程： 设dp[i][tight]表示处理到第i位（从高位开始），tight状态下的合法数字个数。

dp[i+1][new_tight] += dp[i][tight]

其中new_tight的计算规则：

• 如果当前位选择的数字小于n的对应位，则new_tight = false
• 如果当前位选择的数字等于n的对应位，则new_tight保持原值
• 如果当前位选择的数字大于n的对应位，则这种选择非法

约束处理策略：

预处理约束：

int[] constraints = new int[bitLength];
Arrays.fill(constraints, -1); // -1表示无约束

for (int i = 0; i < m; i++) {
    int position = constraints_input[i][0];
    int value = constraints_input[i][1];
    
    if (position >= bitLength) {
        // 约束位置超出n的位数，只有约束为0才可能满足
        if (value == 1) return 0;
        continue;
    }
    
    int highBitIndex = bitLength - 1 - position;
    if (constraints[highBitIndex] != -1 && constraints[highBitIndex] != value) {
        return 0; // 约束冲突
    }
    constraints[highBitIndex] = value;
}

状态转移核心逻辑：

for (int bit = 0; bit < bitLength; bit++) {
    int nBit = nBinary.charAt(bit) - '0';
    Set<Integer> allowedDigits = new HashSet<>();
    
    if (constraints[bit] == -1) {
        allowedDigits.add(0);
        allowedDigits.add(1);
    } else {
        allowedDigits.add(constraints[bit]);
    }
    
    long newNotTight = 0, newTight = 0;
    
    // 从不紧约束状态转移
    if (notTightCount > 0) {
        newNotTight = (newNotTight + (long)notTightCount * allowedDigits.size()) % MOD;
    }
    
    // 从紧约束状态转移
    if (tightCount > 0) {
        for (int digit : allowedDigits) {
            if (digit < nBit) {
                newNotTight = (newNotTight + tightCount) % MOD;
            } else if (digit == nBit) {
                newTight = (newTight + tightCount) % MOD;
            }
        }
    }
    
    notTightCount = (int)newNotTight;
    tightCount = (int)newTight;
}

算法正确性保证：

1. 完备性：所有可能的数字构造路径都被考虑
2. 互斥性：tight和not-tight状态互不重叠
3. 约束满足：每一位的选择都严格遵守约束条件

详细算法设计

预处理阶段：

• 将输入的二进制字符串转换为位约束数组
• 检查约束一致性，及早发现冲突情况
• 处理超出位数范围的约束

动态规划阶段：

• 从最高位开始逐位构造
• 维护tight/not-tight两种状态的计数
• 根据约束条件筛选合法的数字选择

边界处理：

• 正确处理n的最高位
• 处理约束位置超出n位数的特殊情况
• 模运算防止整数溢出

Java代码实现

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

public class Main {
    private static final int MODULUS = 998244353;
    
    private static long solveBinaryConstraints(String nBinary, List<int[]> restrictionList) {
        int binaryLength = nBinary.length();
        int[] positionConstraints = new int[binaryLength];
        Arrays.fill(positionConstraints, -1); // -1 表示该位无约束
        
        // 预处理所有位置约束
        for (int[] restriction : restrictionList) {
            int lowBitPosition = restriction[0];
            int requiredValue = restriction[1];
            
            // 检查约束位置是否超出n的位数
            if (lowBitPosition >= binaryLength) {
                if (requiredValue == 1) {
                    return 0; // 要求高位为1，但n的该位为0，无解
                }
                continue; // 要求高位为0，自动满足
            }
            
            // 转换为从高位开始的索引
            int highBitIndex = binaryLength - 1 - lowBitPosition;
            
            // 检查约束一致性
            if (positionConstraints[highBitIndex] != -1 && 
                positionConstraints[highBitIndex] != requiredValue) {
                return 0; // 约束冲突
            }
            
            positionConstraints[highBitIndex] = requiredValue;
        }
        
        // 数位DP状态：[当前位][是否紧贴上界]
        long countNotTight = 0; // 前缀已经小于n的对应前缀
        long countTight = 1;    // 前缀等于n的对应前缀
        
        // 从最高位开始逐位处理
        for (int bitPosition = 0; bitPosition < binaryLength; bitPosition++) {
            int nCurrentBit = nBinary.charAt(bitPosition) - '0';
            
            // 确定当前位的合法选择
            List<Integer> validChoices = new ArrayList<>();
            if (positionConstraints[bitPosition] == -1) {
                validChoices.add(0);
                validChoices.add(1);
            } else {
                validChoices.add(positionConstraints[bitPosition]);
            }
            
            long nextCountNotTight = 0;
            long nextCountTight = 0;
            
            // 从"已经小于"状态转移
            if (countNotTight > 0) {
                nextCountNotTight = (nextCountNotTight + 
                                   countNotTight * validChoices.size()) % MODULUS;
            }
            
            // 从"紧贴上界"状态转移
            if (countTight > 0) {
                for (int choice : validChoices) {
                    if (choice < nCurrentBit) {
                        nextCountNotTight = (nextCountNotTight + countTight) % MODULUS;
                    } else if (choice == nCurrentBit) {
                        nextCountTight = (nextCountTight + countTight) % MODULUS;
                    }
                    // choice > nCurrentBit 的情况直接跳过（非法）
                }
            }
            
            countNotTight = nextCountNotTight;
            countTight = nextCountTight;
        }
        
        return (countNotTight + countTight) % MODULUS;
    }
    
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        
        String nBinaryStr = reader.readLine().trim();
        int constraintCount = Integer.parseInt(reader.readLine());
        
        List<int[]> constraintList = new ArrayList<>();
        
        for (int i = 0; i < constraintCount; i++) {
            String[] parts = reader.readLine().split(" ");
            int position = Integer.parseInt(parts[0]);
            int value = Integer.parseInt(parts[1]);
            constraintList.add(new int[]{position, value});
        }
        
        long result = solveBinaryConstraints(nBinaryStr, constraintList);
        System.out.println(result);
        
        reader.close();
    }
}

C++代码实现

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

const int MOD = 998244353;

class BinaryCountingSolver {
private:
    static long long computeValidNumbers(const string& upperBound, 
                                       const vector<pair<int, int>>& bitConstraints) {
        int totalBits = upperBound.length();
        vector<int> bitRequirements(totalBits, -1);
        
        // 解析并验证所有位约束
        for (const auto& constraint : bitConstraints) {
            int lowOrderPosition = constraint.first;
            int mandatoryValue = constraint.second;
            
            // 处理超出上界位数的约束
            if (lowOrderPosition >= totalBits) {
                if (mandatoryValue == 1) {
                    return 0; // 不可能满足的约束
                }
                continue; // 自动满足的约束
            }
            
            // 转换为高位优先的位置索引
            int highOrderPosition = totalBits - 1 - lowOrderPosition;
            
            // 约束一致性检查
            if (bitRequirements[highOrderPosition] != -1 && 
                bitRequirements[highOrderPosition] != mandatoryValue) {
                return 0; // 约束矛盾
            }
            
            bitRequirements[highOrderPosition] = mandatoryValue;
        }
        
        // 数位DP：维护两种状态的计数
        long long looseBoundCount = 0;  // 前缀严格小于上界
        long long tightBoundCount = 1;  // 前缀等于上界前缀
        
        // 逐位构造合法数字
        for (int currentBit = 0; currentBit < totalBits; currentBit++) {
            int upperBoundBit = upperBound[currentBit] - '0';
            
            // 收集当前位的所有合法选择
            vector<int> feasibleDigits;
            if (bitRequirements[currentBit] == -1) {
                feasibleDigits.push_back(0);
                feasibleDigits.push_back(1);
            } else {
                feasibleDigits.push_back(bitRequirements[currentBit]);
            }
            
            long long newLooseCount = 0;
            long long newTightCount = 0;
            
            // 状态转移：从loose状态
            if (looseBoundCount > 0) {
                newLooseCount = (newLooseCount + 
                               looseBoundCount * (long long)feasibleDigits.size()) % MOD;
            }
            
            // 状态转移：从tight状态
            if (tightBoundCount > 0) {
                for (int digit : feasibleDigits) {
                    if (digit < upperBoundBit) {
                        newLooseCount = (newLooseCount + tightBoundCount) % MOD;
                    } else if (digit == upperBoundBit) {
                        newTightCount = (newTightCount + tightBoundCount) % MOD;
                    }
                    // digit > upperBoundBit 情况被自动排除
                }
            }
            
            looseBoundCount = newLooseCount;
            tightBoundCount = newTightCount;
        }
        
        return (looseBoundCount + tightBoundCount) % MOD;
    }
    
public:
    static long long solve(const string& nBinary, 
                          const vector<pair<int, int>>& constraints) {
        return computeValidNumbers(nBinary, constraints);
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    string nBinaryRepresentation;
    cin >> nBinaryRepresentation;
    
    int numberOfConstraints;
    cin >> numberOfConstraints;
    
    vector<pair<int, int>> allConstraints;
    
    for (int i = 0; i < numberOfConstraints; i++) {
        int bitPosition, requiredBit;
        cin >> bitPosition >> requiredBit;
        allConstraints.emplace_back(bitPosition, requiredBit);
    }
    
    long long finalAnswer = BinaryCountingSolver::solve(nBinaryRepresentation, allConstraints);
    cout << finalAnswer << endl;
    
    return 0;
}

时间复杂度：O(L + m)，其中L为二进制位数，m为约束数量 空间复杂度：O(L)，用于存储位约束信息