leetcode 2266. 统计打字方案数

2266. 统计打字方案数

题目挺简单的，就是溢出、取余特别令人抓狂

class Solution {
public:
    const int MOD = 1'000'000'007;
    int count(const int &choices,const int &num){
        if(num <= 2)  return num;
        if(num == 3)  return 4;
        vector<long> dp(num+1);//dp[i]表示 i 个字符的组合个数
        dp[0] = 1;dp[1] = 1;dp[2] = 2;dp[3] = 4;
        for(int i = 4;i <= num;++i){
            dp[i] += (dp[i-1] + dp[i-2] + dp[i-3]);
            if(choices == 4)  dp[i] += dp[i-4];
            if(dp[i] >= MOD)  dp[i] %= MOD;
        }
        return dp[num];
    }
    int countTexts(string pressedKeys) {
        int n = pressedKeys.size();
        long res = 1;
        for(int i = 0,j = 0;i < n && j < n;){
            while(j < n && pressedKeys[i] == pressedKeys[j])  ++j;
            if(pressedKeys[i] == '7' || pressedKeys[i] == '9'){
                res = res * count(4,j-i) % MOD;//如果写成res *= count(4,j-i)就会溢出
            }
            else  res = res * count(3,j-i) % MOD;
            i = j;
        }
        if(res >= MOD)  return res & MOD;
        return res;
    }
};