B. Minimum Ternary String (這個B有點狠)
You are given a ternary string (it is a string which consists only of characters ‘0‘, ‘1‘ and ‘2‘).
You can swap any two adjacent (consecutive) characters ‘0‘ and ‘1‘ (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters ‘1‘ and ‘2‘ (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" →→ "100210";
- "010210" →→ "001210";
- "010210" →→ "010120";
- "010210" →→ "010201".
Note than you cannot swap "02" →→ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String aa is lexicographically less than string bb (if strings aa and bb have the same length) if there exists some position ii (1≤i≤|a|1≤i≤|a|, where|s||s| is the length of the string ss) such that for every j<ij<i holds aj=bjaj=bj, and ai<biai<bi.
InputThe first line of the input contains the string
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
Examples input Copy100210output Copy
001120input Copy
11222121output Copy
11112222input Copy
20output Copy
20
這場CF abcd題 我就覺得B題最難
這個題目卡了我好久好久 ,全是寫BUG
代碼有點毒瘤
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e5 + 10; 4 const int INF = 0x3fffffff; 5 typedef long long LL; 6 string s; 7 int sum0[maxn], sum1[maxn], sum2[maxn], vis[maxn]; 8 vector<int>a; 9 int main() { 10 cin >> s; 11 if (s[0] == ‘0‘) sum0[1] = 1; 12 if (s[0] == ‘1‘) sum1[1] = 1; 13 if (s[0] == ‘2‘) sum2[1] = 1; 14 for (int i = 1 ; i < s.size() ; i++) { 15 if (s[i] == ‘0‘) sum0[i + 1] = sum0[i] + 1, sum1[i + 1] = sum1[i], sum2[i + 1] = sum2[i]; 16 if (s[i] == ‘1‘) sum1[i + 1] = sum1[i] + 1, sum0[i + 1] = sum0[i], sum2[i + 1] = sum2[i]; 17 if (s[i] == ‘2‘) sum2[i + 1] = sum2[i] + 1, sum0[i + 1] = sum0[i], sum1[i + 1] = sum1[i]; 18 } 19 int cnt = 0; 20 int len = s.size(); 21 for (int i = 0 ; i < len ; i++) { 22 if (s[i] == ‘2‘) { 23 for (int j = 0 ; j < sum0[i]; j++) a.push_back(0); 24 for (int j = 0 ; j < sum1[len] ; j++) a.push_back(1); 25 for (int j = i ; j < s.size(); j++) { 26 if (s[j] == ‘1‘) continue; 27 if (s[j] == ‘2‘) a.push_back(2); 28 if (s[j] == ‘0‘)a.push_back(0); 29 } 30 cnt = 1; 31 break; 32 } 33 } 34 if (cnt == 0) { 35 for (int i = 0 ; i < sum0[len] ; i++) printf("0"); 36 for (int i = 0 ; i < sum1[len] ; i++) printf("1"); 37 } else { 38 for (int i = 0 ; i < a.size() ; i++) printf("%d", a[i]); 39 } 40 return 0; 41 }
B. Minimum Ternary String (這個B有點狠)