表达式求值 [思维题, 矩阵快速幂]

$表达式求值$

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$\color{red}{正解部分}$

按顺序扫表达式, 记录四个状态,

$S: 上一个加号前的总和$
$M: 上一个加号到最后一个乘号为止的乘积$
$T: 上一个加号之后的运算结果$
$e: 1$

当前位置的字符为 $c_i$,

$c_i =\ '+':\ \ \ \ \ (S, M, T, e) \rightarrow (S+T, e, 0, e)$
$c_i =\ '\times':\ \ \ \ \ (S,M,T,e) \rightarrow (S, T, 0, e)$
$c_i =\ 数字k: (S, M, T, e) \rightarrow (S, M, 10T+kM, e)$

矩阵快速幂 加速转移, 最后 $S+T$ 即为答案 .

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$\color{red}{实现部分}$

#include<bits/stdc++.h>
#define reg register

const int mod = 998244353;

struct Matrix{
        int C[5][5];
        Matrix(){ memset(C, 0, sizeof C); }
} add, mul, num, Ans;

Matrix modify(const Matrix &a, const Matrix &b){
        Matrix res;
        for(reg int i = 1; i <= 4; i ++)
                for(reg int j = 1; j <= 4; j ++)
                        for(reg int k = 1; k <= 4; k ++)
                                res.C[i][j] = (1ll*res.C[i][j] + (1ll*a.C[i][k]*b.C[k][j]%mod)) % mod;
        return res;
}

Matrix Ksm(Matrix a, int b){
        Matrix res;
        for(reg int i = 1; i <= 4; i ++) res.C[i][i] = 1;
        while(b){
                if(b & 1) res = modify(res, a);
                a = modify(a, a); b >>= 1;
        }
        return res;
}

int N;
int r;

char s[20];

void Work(){
        scanf("%d%s", &r, s+1);
        int len = strlen(s+1);
        Matrix Mmp;
        for(reg int i = 1; i <= 4; i ++) Mmp.C[i][i] = 1;
        for(reg int i = 1; i <= len; i ++){
                if(s[i] == '+') Mmp = modify(Mmp, add);
                else if(s[i] == '*') Mmp = modify(Mmp, mul);
                else num.C[2][3] = s[i]-'0', Mmp = modify(Mmp, num);
        }
        Mmp = Ksm(Mmp, r); Ans = modify(Ans, Mmp);
}

int main(){
        add.C[1][1] = add.C[3][1] = add.C[4][2] = add.C[4][4] = 1;
        mul.C[1][1] = mul.C[3][2] = mul.C[4][4] = 1;
        num.C[1][1] = num.C[2][2] = num.C[4][4] = 1, num.C[3][3] = 10;
        Ans.C[1][2] = Ans.C[1][4] = 1; //
        scanf("%d", &N);
        for(reg int i = 1; i <= N; i ++) Work();
        Ans = modify(Ans, add);
        printf("%d\n", Ans.C[1][1]);
        return 0;
}