知識の証明 [ACM-ICPC JAG 模擬国内予選 2018年 C]

https://onlinejudge.u-aizu.ac.jp/challenges/sources/JAG/Prelim/2883

前提知識

• 構文解析

解法

https://onlinejudge.u-aizu.ac.jp/solutions/problem/2883/review/2994318/hamayanhamayan/C++14

構文解析を頑張る。
実装例として下に載せておく。

string S, P;
//---------------------------------------------------------------------------------------------------
int i, a, b, c, d;
int f() {
    if (S[i] == '[') {
        i++;
        int l, r, res;
        if (S[i] == '+') {
            i++;
            l = f();
            r = f();
            res = l | r;
        }
        else if (S[i] == '*') {
            i++;
            l = f();
            r = f();
            res = l & r;
        }
        else {
            i++;
            l = f();
            r = f();
            res = l ^ r;
        }
        i++;
        return res;
    }

    int res;
    if (S[i] == 'a') res = a;
    else if (S[i] == 'b') res = b;
    else if (S[i] == 'c') res = c;
    else res = d;

    i++;
    return res;
}
//---------------------------------------------------------------------------------------------------
void _main() {
    while (cin >> S) {
        if (S == ".") return;
        cin >> P;

        i = 0;
        a = P[0] - '0';
        b = P[1] - '0';
        c = P[2] - '0';
        d = P[3] - '0';
        int ha = f();
        int cnt = 0;
        rep(j, 0, 10000) {
            i = 0;
            a = j / 1000;
            b = (j / 100) % 10;
            c = (j / 10) % 10;
            d = j % 10;

            if (ha == f()) cnt++;
        }

        printf("%d %d\n", ha, cnt);
    }
}