Codeforces Round #478 (Div. 2) ABC
A. Aramic script
In Aramic language words can only represent objects.
Words in Aramic have special properties:
- A word is a root if it does not contain the same letter more than once.
- A root and all its permutations represent the same object.
- The root xx of a word yy is the word that contains all letters that appear in yy in a way that each letter appears once. For example, the rootof "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab".
- Any word in Aramic represents the same object as its root.
You have an ancient script in Aramic. What is the number of different objects mentioned in the script?
The first line contains one integer nn (1≤n≤1031≤n≤103) — the number of words in the script.
The second line contains nn words s1,s2,…,sns1,s2,…,sn — the script itself. The length of each string does not exceed 103103.
It is guaranteed that all characters of the strings are small latin letters.
Output one integer — the number of different objects mentioned in the given ancient Aramic script.
5
a aa aaa ab abb
2
3
amer arem mrea
1
In the first test, there are two objects mentioned. The roots that represent them are "a","ab".
In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".
查找有多少个root,英语渣渣的我看了好久才看到,ab,ba 是一个单词,它的原根是ab,求原根的话是将一个单词重复的字符去掉,比如aaabbb的原根是ab,因为只出现a字符和b字符。懂意思就好做了。
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const int N = 1010; 5 set<int> st; 6 string s; 7 int n; 8 int main() { 9 cin >> n; 10 int ans = 0; 11 for(int i = 0; i < n; i ++) { 12 cin >> s; 13 int cnt = 0; 14 for(int j = 0; j < s.length(); j ++) { 15 cnt |= (1<<(s[j]-'a')); 16 } 17 st.insert(cnt); 18 } 19 cout << st.size() << endl; 20 return 0; 21 }
B. Mancala
Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes.
Initially, each hole has aiai stones. When a player makes a move, he chooses a hole which contains a positive number of stones. He takes all the stones inside it and then redistributes these stones one by one in the next holes in a counter-clockwise direction.
Note that the counter-clockwise order means if the player takes the stones from hole ii, he will put one stone in the (i+1)(i+1)-th hole, then in the (i+2)(i+2)-th, etc. If he puts a stone in the 1414-th hole, the next one will be put in the first hole.
After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli.
Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move.
The only line contains 14 integers a1,a2,…,a14a1,a2,…,a14 (0≤ai≤1090≤ai≤109) — the number of stones in each hole.
It is guaranteed that for any ii (1≤i≤141≤i≤14) aiai is either zero or odd, and there is at least one stone in the board.
Output one integer, the maximum possible score after one move.
0 1 1 0 0 0 0 0 0 7 0 0 0 0
4
5 1 1 1 1 0 0 0 0 0 0 0 0 0
8
In the first test case the board after the move from the hole with 77 stones will look like 1 2 2 0 0 0 0 0 0 0 1 1 1 1. Then the player collects the even numbers and ends up with a score equal to 44.
模拟,14个洞,从第i个洞取出全部石头,然后在第i+1上放一个,第i+2上放一个,循环直到全部放完,然后把14个洞里偶数的数相加,求最大值。
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const int N = 110; 5 ll a[15], b[15]; 6 void init() { 7 for(int i = 0; i < 14; i ++) b[i] = a[i]; 8 } 9 int main() { 10 for(int i = 0; i < 14; i ++) { 11 cin >> a[i]; 12 } 13 ll MAX = 0; 14 for(int i = 0; i < 14; i ++) { 15 init(); 16 ll x = b[i]; 17 b[i] = 0; 18 for(int j = 0; j < 14; j ++) { 19 b[j] += x/14; 20 } 21 x = x%14; 22 for(int j = 0; j < x; j ++) { 23 b[(i+j+1)%14] ++; 24 } 25 ll cnt = 0; 26 for(int j = 0; j < 14; j ++) { 27 if(b[j]%2==0) cnt += b[j]; 28 } 29 MAX = max(MAX, cnt); 30 } 31 // for(int i = 0; i < 14; i ++) printf("%lld ",b[i]);printf("\n"); 32 cout << MAX << endl; 33 return 0; 34 }
C. Valhalla Siege
Ivar the Boneless is a great leader. He is trying to capture Kattegat from Lagertha. The war has begun and wave after wave Ivar's warriors are falling in battle.
Ivar has nn warriors, he places them on a straight line in front of the main gate, in a way that the ii-th warrior stands right after (i−1)(i−1)-th warrior. The first warrior leads the attack.
Each attacker can take up to aiai arrows before he falls to the ground, where aiai is the ii-th warrior's strength.
Lagertha orders her warriors to shoot kiki arrows during the ii-th minute, the arrows one by one hit the first still standing warrior. After all Ivar's warriors fall and all the currently flying arrows fly by, Thor smashes his hammer and all Ivar's warriors get their previous strengths back and stand up to fight again. In other words, if all warriors die in minute tt, they will all be standing to fight at the end of minute tt.
The battle will last for qq minutes, after each minute you should tell Ivar what is the number of his standing warriors.
The first line contains two integers nn and qq (1≤n,q≤2000001≤n,q≤200000) — the number of warriors and the number of minutes in the battle.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) that represent the warriors' strengths.
The third line contains qq integers k1,k2,…,kqk1,k2,…,kq (1≤ki≤10141≤ki≤1014), the ii-th of them represents Lagertha's order at the ii-th minute: kiki arrows will attack the warriors.
Output qq lines, the ii-th of them is the number of standing warriors after the ii-th minute.
5 5
1 2 1 2 1
3 10 1 1 1
3
5
4
4
3
4 4
1 2 3 4
9 1 10 6
1
4
4
1
In the first example:
- after the 1-st minute, the 1-st and 2-nd warriors die.
- after the 2-nd minute all warriors die (and all arrows left over are wasted), then they will be revived thus answer is 5 — all warriors are alive.
- after the 3-rd minute, the 1-st warrior dies.
- after the 4-th minute, the 2-nd warrior takes a hit and his strength decreases by 1.
- after the 5-th minute, the 2-nd warrior dies.
二分查找。
有n个战士,每个战士都有个值ai,表示需要射ai只箭才死,从第一个战士开始射,全部死后就会全部复活。
每射k只箭,求还活着的战士。
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const int N = 2e5+10; 5 ll a[N], n, q, sum[N]; 6 int main() { 7 cin >> n >> q >> a[0]; 8 for(int i = 1; i < n; i ++) { 9 cin >> a[i]; 10 a[i] += a[i-1]; 11 } 12 ll id = 0, ans = 0; 13 while(q--) { 14 ll k; 15 cin >> k; 16 ans += k; 17 id = upper_bound(a,a+n,ans)-a; 18 if(id == n) id = 0, ans = 0; 19 cout << n - id << endl; 20 } 21 return 0; 22 }