ACM-ICPC 2018 沈阳赛区网络预赛 I. Lattice's basics in digital electronics 阅读题加模拟题

#include<bitset>
#include <cstdio>
#include <cmath>
#include <complex>
#include <algorithm>
#include <iostream>
#include<string.h>
#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mmm(a,b) memset(a,b,sizeof(a))
typedef long long ll;
using namespace std;
const int maxn = 8e5;

int tot = 0;
const int MAXN = 10000009;
bitset<maxn>bi;
bitset<4>buff[200000+5];
char s[200000 + 5];
char tree[1024 * 4];
int n;
string ss;
string cd;
int main()
{
    
    int t; cin >> t; while (t--) {
        ss.clear();
        cd.clear();
        mmm(tree, 0);
        int n,m;
        cin >> m >> n;
        rep(i, 1, n) {
            int x;
            char op[11];
            cin >> x >> op;
            int len = strlen(op);
            int now = 1;
            rep(j, 0, len - 1) {
                if (op[j] == ‘1‘) now = now * 2 + 1;
                else now = now * 2;
            }
            tree[now] = (char)x;
        }
        scanf("%s", s);
        int len = strlen(s);
        int tot = 0;
        rep(i, 0, len - 1) {
            if (isdigit(s[i]))buff[tot++] = s[i] - 0;
            else if (s[i] >= ‘a‘&&s[i] <= ‘z‘)buff[tot++] = s[i] - ‘a‘ + 10;
            else buff[tot++] = s[i] - ‘A‘ + 10;
        }
        //rep(i, 0, tot - 1)cout << buff[i] ;cout << endl;
        
        rep(i, 0, tot - 1)
            ss += buff[i].to_string();
        len = ss.length();
        int cnt = 0;
        rep(i, 0, len-1) {
            if (i % 9 == 8) {
                if (cnt % 2 == 0 && ss[i] == ‘1‘|| cnt % 2 == 1 && ss[i] == ‘0‘) {
                    string temp = ss.substr(i - 8, 8);
                    cd += temp;
                }
            }
            if (i % 9 == 0)cnt = 0;
            if (ss[i] == ‘1‘)cnt++;
        }
        len = cd.length();
        rep(i, 0, len - 1) {
            int now = 1;
            while (now == 1 || !tree[now]) {
                if (cd[i] == ‘1‘) now = now * 2 + 1;
                else now = now * 2;
                i++;
            }
            printf("%c", tree[now]); m--;
            if (m == 0)break;
            i--;
        }
        cout << endl;
    }
    cin >> t;
    return 0;
}
/*
2
15 9
32 0100
33 11
100 1011
101 0110
104 1010
108 00
111 100
114 0111
119 0101
A6Fd021171c562Fde1
8 3
49 0001
50 01001
51 011
14DB24722698
*/