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

题意：https://nanti.jisuanke.com/t/31450

题解：题目很长的模拟，有点uva的感觉

分成四步

part1 16进制转为二进制string 用bitset的to_string()

part2 parity check 校对，将处理结果pushback到另一个string

part3 建字典树，用形如线段树的数组存

part4 遍历字典树

1A 233

#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 = ;
const int MAXN = ;
bitset<maxn>bi;
bitset<>buff[+];
char s[ + ];
char tree[ * ];
int n;
string ss;
string cd;
int main()
{

    int t; cin >> t; while (t--) {
        ss.clear();
        cd.clear();
        mmm(tree, );
        int n,m;
        cin >> m >> n;
        rep(i, , n) {
            int x;
            char op[];
            cin >> x >> op;
            int len = strlen(op);
            int now = ;
            rep(j, , len - ) {
                if (op[j] == '') now = now *  + ;
                else now = now * ;
            }
            tree[now] = (char)x;
        }
        scanf("%s", s);
        int len = strlen(s);
        int tot = ;
        rep(i, , len - ) {
            if (isdigit(s[i]))buff[tot++] = s[i] - ;
            else if (s[i] >= 'a'&&s[i] <= 'z')buff[tot++] = s[i] - 'a' + ;
            else buff[tot++] = s[i] - 'A' + ;
        }
        //rep(i, 0, tot - 1)cout << buff[i] ;cout << endl;

        rep(i, , tot - )
            ss += buff[i].to_string();
        len = ss.length();
        int cnt = ;
        rep(i, , len-) {
            if (i %  == ) {
                if (cnt %  ==  && ss[i] == ''|| cnt %  ==  && ss[i] == '') {
                    string temp = ss.substr(i - , );
                    cd += temp;
                }
            }
            if (i %  == )cnt = ;
            if (ss[i] == '')cnt++;
        }
        len = cd.length();
        rep(i, , len - ) {
            int now = ;
            while (now ==  || !tree[now]) {
                if (cd[i] == '') now = now *  + ;
                else now = now * ;
                i++;
            }
            printf("%c", tree[now]); m--;
            if (m == )break;
            i--;
        }
        cout << endl;
    }
    cin >> t;
    return ;
}
/*
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
*/