原题链接:https://nanti.jisuanke.com/t/31450

附上队友代码:(感谢队友带飞)

#include <bits/stdc++.h>
using namespace std; #define ll long long
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define DOR(i,a,b) for(int i=(a);i>=(b);--i)
const int maxN=2e5+5,inf=0x3f3f3f3f;
int N, M, K, T;
int g[maxN];
unordered_map<string,int> mp;
char S1[maxN * 4], S2[maxN * 4], S3[maxN * 4]; struct ND{
ND *lch, *rch;
char x;
}; char cg[20][5] = {"0000","0001","0010", "0011", "0100",
"0101", "0110", "0111", "1000", "1001", "1010",
"1011", "1100", "1101", "1110", "1111"}; bool judge(char *s, int st) {
int i = 0, one = 0;
while (i < 9 && s[st + i] != 0) {
if (s[st + i] == '1')
++one;
++i;
}
if (i < 9) return 0;
if (one & 1) return 1;
else return 0;
} int main () {
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
#endif
scanf("%d", &T);
while (T--) {
S1[0] = S2[0] = S3[0] = 0;
mp.clear();
scanf("%d%d", &M, &N);
int a;
char s[15];
FOR(i, 1, N) {
scanf("%d %s", &a, s);
mp[s] = a;
}
scanf("%s", S1);
for (int i = 0; S1[i] != 0; ++i) {
int cnt;
if (isdigit(S1[i])) cnt = S1[i] - '0';
else {
S1[i] = toupper(S1[i]);
cnt = S1[i] - 'A' + 10;
}
FOR(j, 0, 3) {
S2[i * 4 + j] = cg[cnt][j];
}
}
// puts(S1);
// puts(S2);
int L = strlen(S2), cnt = 0;
for (int i = 0; i < L; i += 9) {
bool ok = judge(S2, i);
if (ok) {
for (int j = i; j < i + 8; ++j)
S3[cnt++] = S2[j];
}
}
S3[cnt] = 0;
// cout << "cnt" << cnt << endl; /*
for (int i = 0; i < cnt; ++i) {
if (i && i % 4 == 0) printf(" ");
printf("%c",S3[i]);
}
*/
//puts(S3);
string str = S3;
int L3 = str.length();
int idx = 0;
int all = 0;
while (idx < L3) {
for (int len = 1; len <= 10; ++len) {
if (L3 - idx + 1 < len) {
idx = L3;
break;
}
string subs = str.substr(idx, len);
if (mp.count(subs)) {
printf("%c", mp[subs]);
++all; if (all == M) {
idx = L3;
break;
} idx += subs.length();
break;
}
}
}
puts(""); }
return 0;
}

ACM-ICPC 2018 沈阳赛区网络预赛 I 题 Lattice's basics in digital electronics的更多相关文章

  1. 【ACM-ICPC 2018 沈阳赛区网络预赛 I】Lattice's basics in digital electronics

    [链接] 我是链接,点我呀:) [题意] [题解] 每个单词的前缀都不同. 不能更明示了... 裸的字典树. 模拟一下.输出一下就ojbk了. [代码] #include <bits/stdc+ ...

  2. Fantastic Graph 2018 沈阳赛区网络预赛 F题

    题意: 二分图 有k条边,我们去选择其中的几条 每选中一条那么此条边的u 和 v的度数就+1,最后使得所有点的度数都在[l, r]这个区间内 , 这就相当于 边流入1,流出1,最后使流量平衡 解析: ...

  3. ACM-ICPC 2018 沈阳赛区网络预赛 K题

    题目链接: https://nanti.jisuanke.com/t/31452 AC代码(看到不好推的定理就先打表!!!!): #include<bits/stdc++.h> using ...

  4. Made In Heaven 2018 沈阳赛区网络预赛 D题

    求第k短路 模板题 套模板即可 #include <iostream> #include <cstring> #include <cstdio> #include ...

  5. ACM-ICPC 2018 沈阳赛区网络预赛 K Supreme Number(规律)

    https://nanti.jisuanke.com/t/31452 题意 给出一个n (2 ≤ N ≤ 10100 ),找到最接近且小于n的一个数,这个数需要满足每位上的数字构成的集合的每个非空子集 ...

  6. ACM-ICPC 2018 沈阳赛区网络预赛-K:Supreme Number

    Supreme Number A prime number (or a prime) is a natural number greater than 11 that cannot be formed ...

  7. ACM-ICPC 2018 沈阳赛区网络预赛-D:Made In Heaven(K短路+A*模板)

    Made In Heaven One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with her. ...

  8. 图上两点之间的第k最短路径的长度 ACM-ICPC 2018 沈阳赛区网络预赛 D. Made In Heaven

    131072K   One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with her. Howe ...

  9. ACM-ICPC 2018 沈阳赛区网络预赛 J树分块

    J. Ka Chang Given a rooted tree ( the root is node 11 ) of NN nodes. Initially, each node has zero p ...

随机推荐

  1. 【xlwings】 wps 和 office 的excel creat_sheet区别

    最近在学习 xlwings,参考学习的网址:https://www.jianshu.com/p/b534e0d465f7 写得很棒,学到了很多. 在新建sheet表单, 发现一个问题. import ...

  2. pidof---找寻PID

    pidof---找寻pid 1.根据守护进程找到pid [root@localhost ~]# pidof sshd 2542 1622 [root@localhost ~]# ps -ef | gr ...

  3. element-ui table float类型数据排序失败

    背景:对于16.88这样的数据,点击表头排序无效,仍然是乱序 解决办法:自定义排序方法,:sortable="true" :sort-mothod="xxxx" ...

  4. QT 安卓动态获取权限

    一:在AndroidManifest.xml文件中赋予相关权限 二: package ckdz.Appproject; import android.Manifest; import android. ...

  5. O026、Nova组件详解

    参考https://www.cnblogs.com/CloudMan6/p/5436855.html   本节开始,我们将详细讲解 Nova 的各个子服务.   前面架构概览一节知道 Nova 有若干 ...

  6. Django-DRF-视图的演变(二)

    Django-DRF-视图的演变   版本一(基于类视图APIView类) views.py: APIView是继承的Django View视图的. 1 from .serializers impor ...

  7. Nginx作为静态资源web服务之文件读取

    Nginx作为静态资源web服务之文件读取 文件读取会使用到以下几个配置 1. sendfile 使用nginx作为静态资源服务时,通过配置sendfile可以有效提高文件读取效率,设置为on表示启动 ...

  8. Samba passwd smbpasswd and tdbsam

    ome commands to convert samba backend password-databases. If you use "passdb backend = smbpassw ...

  9. Migrating authentication of Samba from smbpasswd to tdb

    Was addicted various After you upgrade the OS of old Samba server. Put it also was using a set of 2. ...

  10. c++ 一些注意事项

    1.long int的字节信息:int在32位系统下是4字节,long在32位也是4字节,在64位Int不变,但是long变成8字节,所以我们的编译器不同可能会导致我们处理int,long不同 2.注 ...