ACM-ICPC 2018 沈阳赛区网络预赛 I 题 Lattice's basics in digital electronics
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;
}
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