Codeforces Round #590 (Div. 3)

A. Equalize Prices Again

题目链接https://codeforces.com/contest/1234/problem/A

题意:给你 n 个数 , 你需要改变这些数使得这 n 个数的值相等 , 并且要求改变后所有数的和需大于等于原来的所有数字的和 , 然后输出满足题意且改变后最小的数值

分析:

签到题。记原来 n 个数的和为 sum , 先取这些数的平均值 ave , 然后每次判断 ave * n >= sum 是否成立成立则直接输出 , 不成立将 ave ++

Codeforces Round #590 (Div. 3)
 1 #include<bits/stdc++.h>
 2 #define ios std::ios::sync_with_stdio(false) , std::cin.tie(0) , std::cout.tie(0)
 3 #define sd(n) scanf("%d",&n)
 4 #define sdd(n,m) scanf("%d%d",&n,&m)
 5 #define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
 6 #define pd(n) printf("%d\n", (n))
 7 #define pdd(n,m) printf("%d %d\n", n, m)
 8 #define pld(n) printf("%lld\n", n)
 9 #define pldd(n,m) printf("%lld %lld\n", n, m)
10 #define sld(n) scanf("%lld",&n)
11 #define sldd(n,m) scanf("%lld%lld",&n,&m)
12 #define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
13 #define sf(n) scanf("%lf",&n)
14 #define sff(n,m) scanf("%lf%lf",&n,&m)
15 #define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
16 #define rep(i,a,n) for (int i=a;i<=n;i++)
17 #define per(i,n,a) for (int i=n;i>=a;i--)
18 #define mm(a,n) memset(a, n, sizeof(a))
19 #define pb push_back
20 #define all(x) (x).begin(),(x).end()
21 #define fi first
22 #define se second
23 #define ll long long
24 #define numm ch - 48
25 #define INF 0x3f3f3f3f
26 #define pi 3.14159265358979323
27 #define debug(x) cout << #x << ": " << x << endl
28 #define debug2(x, y)          cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;
29 #define debug3(x, y, z)       cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;
30 #define debug4(a, b, c, d)    cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
31 using namespace std;
32 template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);
33 for(res=numm;isdigit(ch=getchar());res=(res<<1)+(res<<3)+numm);flag&&(res=-res);}
34 template<typename T>void Out(T x){if(x<0)putchar('-'),x=-x;if(x>9)Out(x/10);putchar(x%10+'0');}
35 ll pow_mod(ll x, ll n , ll mod){ll res=1;while(n){if(n&1)res=res*x%mod;x=x*x%mod;n>>=1;}return res;}
36 const int N = 1e3+10;
37  
38 int main()
39 {
40     int q , n;
41     sd(q);
42     while(q --)
43     {
44         sd(n);
45         int t = 0,ans = 0;
46         rep(i , 1, n)
47         {
48             int x;
49             sd(x);
50             t += x;
51         }
52         ans = t / n;
53         while(ans * n < t)
54         {
55             ans  = ans + 1;
56         }
57     pd(ans);
58     }
59     return 0;
60 }
View Code

 

 

 

B1. Social Network (easy version)

题目链接:https://codeforces.com/contest/1234/problem/B1

题意:给你个长度为 n 的数组和一个队列 , 队列最多可以同时存在 k 个数。遍历这个数组 , 如果当前数组对应的数在队列中则不做改动 , 如果不在则将它插入队首 , 并且将队尾弹出。遍历完后按照队列顺序输出

分析:

签到题。 直接用 set 判断当前数是否已经在队列中 、 deque 进行头插尾删和储存

Codeforces Round #590 (Div. 3)
 1 #include<bits/stdc++.h>
 2 #define ll long long 
 3 #define ios std::ios::sync_with_stdio(false) , std::cin.tie(0) , std::cout.tie(0)
 4 using namespace std;
 5 const int N = 2e5+10;
 6 int n , k;
 7 ll a[N];
 8 set<ll>qq;
 9 deque<ll>txc;
10 deque<ll>::iterator it;
11 int main()
12 {
13     ios;
14     cin >> n >> k;
15     cin >> a[1];
16     qq.insert(a[1]);
17     txc.push_front(a[1]);
18     for(int i = 2 ; i <= n ; i++)
19     {
20         cin >> a[i];
21         if(qq.count(a[i])) continue;
22         if(qq.count(a[i]) == 0 && qq.size() < k)
23         {
24             qq.insert(a[i]);
25             txc.push_front(a[i]);
26         }
27         else if(qq.count(a[i]) == 0 && qq.size() == k)
28         {
29             it = txc.end(); it--;
30             qq.erase(*it);
31             txc.pop_back();
32             txc.push_front(a[i]);
33             qq.insert(a[i]);
34         }
35     }
36     cout << txc.size() << endl;
37     for(it = txc.begin() ; it != txc.end(); it ++)
38     {
39         cout << *it << " ";
40     }
41     return 0;
42 }
43  
View Code

 

 

 

B2. Social Network (hard version)

题目链接:https://codeforces.com/contest/1234/problem/B2

改变了 n 和 k 的范围 , 因为 n 和 k 最大也还是只有 2e5 + 10 , 所以对于set + deque的解法是不会造成影响的同B1一样的代码

 

 

 

C. Pipes

题目链接:https://codeforces.com/contest/1234/problem/C

题意:有六种管子 , 其中1、2可以互相转换 , 3、4、5、6可以互相转换  , 然后给你两行管道 , 每行有 n 列问水能不能从左上角(第1行第1列)流到右下角(第2行第n列)

分析:

因为管子之间可以互相转换 , 所以我们可以先将1、2类型的管子记为1 , 3、4、5、6类型的管子记为2 , 然后仔细思考我们会发现水的流向肯定是固定的:假设水是从第k列第j行流向第k + 1列, 那么第k+1列肯定是用第j行的管子接水 , 如果第k+1列第j行的管子类型为1 , 那么水将直接流向第k+2列 ; 如果第k+1列第j行的管子类型为2 , 那么水只能传给 j 的上一行或者下一行 , 然后再传向第k+2列 , 因为每行每列的管子类型已经确定了, 所以水的流向也就是固定的了。而水能从当前列流向下一列的条件只有几个 :

①接收水的管子类型为 1 ,所在行为 j 直接流向下一列的第 j 行

②接收水的管子类型为 2 , 则如果 j ^ 1 行管子的类型也为 2  , 则流向下一列的第 j ^ 1行

剩下情况水皆不能流通(水不能倒流) , 所以直接dfs跑一遍就可以了

Codeforces Round #590 (Div. 3)
  1 #include<bits/stdc++.h>
  2 #define ios std::ios::sync_with_stdio(false) , std::cin.tie(0) , std::cout.tie(0)
  3 #define sd(n) scanf("%d",&n)
  4 #define sdd(n,m) scanf("%d%d",&n,&m)
  5 #define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
  6 #define pd(n) printf("%d\n", (n))
  7 #define pdd(n,m) printf("%d %d\n", n, m)
  8 #define pld(n) printf("%lld\n", n)
  9 #define pldd(n,m) printf("%lld %lld\n", n, m)
 10 #define sld(n) scanf("%lld",&n)
 11 #define sldd(n,m) scanf("%lld%lld",&n,&m)
 12 #define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
 13 #define sf(n) scanf("%lf",&n)
 14 #define sff(n,m) scanf("%lf%lf",&n,&m)
 15 #define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
 16 #define rep(i,a,n) for (int i=a;i<=n;i++)
 17 #define per(i,n,a) for (int i=n;i>=a;i--)
 18 #define mm(a,n) memset(a, n, sizeof(a))
 19 #define pb push_back
 20 #define all(x) (x).begin(),(x).end()
 21 #define fi first
 22 #define se second
 23 #define ll long long
 24 #define numm ch - 48
 25 #define INF 0x3f3f3f3f
 26 #define pi 3.14159265358979323
 27 #define debug(x) cout << #x << ": " << x << endl
 28 #define debug2(x, y)          cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;
 29 #define debug3(x, y, z)       cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;
 30 #define debug4(a, b, c, d)    cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
 31 using namespace std;
 32 template<typename T>void read(T &res)
 33 {
 34     bool flag=false;
 35     char ch;
 36     while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);
 37     for(res=numm; isdigit(ch=getchar()); res=(res<<1)+(res<<3)+numm);
 38     flag&&(res=-res);
 39 }
 40 template<typename T>void Out(T x)
 41 {
 42     if(x<0)putchar('-'),x=-x;
 43     if(x>9)Out(x/10);
 44     putchar(x%10+'0');
 45 }
 46 ll pow_mod(ll x, ll n , ll mod)
 47 {
 48     ll res=1;
 49     while(n)
 50     {
 51         if(n&1)res=res*x%mod;
 52         x=x*x%mod;
 53         n>>=1;
 54     }
 55     return res;
 56 }
 57 const int N = 1e3+10;
 58 int n;
 59 string s1 , s2 ,s[2];
 60 bool dfs(int i, int j)
 61 {
 62     if(i == n && j == 1)
 63         return 1;
 64     if(i == n) return 0;
 65     if(s[0][i] == '1' && s[1][i] == '2')
 66     {
 67         if(j == 0)  return dfs(i + 1 , 0);
 68         else return 0;
 69     }
 70     if(s[0][i] == '2' && s[1][i] == '1')
 71     {
 72         if(j == 1) return dfs(i + 1 , 1);
 73         else return 0;
 74     }
 75     if(s[0][i] == '1' && s[1][i] == '1')
 76     {
 77         return dfs(i + 1 , j);
 78     }
 79     if(s[0][i] == '2' && s[1][i] == '2')
 80     {
 81         return dfs(i + 1 , j ^ 1);
 82     }
 83 }
 84 int main()
 85 {
 86     ios;
 87     int t;
 88     cin >> t;
 89     while(t--)
 90     {
 91  
 92         cin >> n;
 93         cin >> s1 >> s2;
 94         s[0] = s1 , s[1] = s2;
 95         rep(i ,0 ,n - 1)
 96         {
 97             if(s[0][i] == '2' || s[0][i] == '1') s[0][i] = '1';
 98             else s[0][i] = '2';
 99             if(s[1][i] == '1' || s[1][i] == '2') s[1][i] = '1';
100             else s[1][i] = '2';
101         }
102         if(dfs(0 , 0)) cout << "YES" << endl;
103         else cout << "NO" << endl;
104     }
105     return 0;
106 }
107  
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D. Distinct Characters Queries

题目链接:https://codeforces.com/contest/1234/problem/D

题意:给你一个字符串 , 有q个操作:
①、 将 pos 位置的字符改为 c

②、查询 L~ R 区间不同字符的个数

分析:

挺水的一题。因为全是小写字符 , 所以我们可以对每个单独字符开个线段树, 那么一共就开了26个线段树 ,然后预处理:将母串中第 i 个位置的字符对应的线段树的第 i 个区间的值 + 1。那么当操作为 ① 的时候我们只要将母串pos位置的字符对应线段树的pos区间值 -1 , 然后c字符对应线段树的pos区间 +1 ;  当操作为 ②的时候我们只要判断每个字符是否有出现在L ~ R区间 , 即遍历 26 颗线段树 L ~ R 的区间和是否为 0 .若不为 0 , ans++ , 遍历完后输出ans即可

Codeforces Round #590 (Div. 3)
  1 #include<bits/stdc++.h>
  2 #define ios std::ios::sync_with_stdio(false) , std::cin.tie(0) , std::cout.tie(0)
  3 #define sd(n) scanf("%d",&n)
  4 #define sdd(n,m) scanf("%d%d",&n,&m)
  5 #define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
  6 #define pd(n) printf("%d\n", (n))
  7 #define pdd(n,m) printf("%d %d\n", n, m)
  8 #define pld(n) printf("%lld\n", n)
  9 #define pldd(n,m) printf("%lld %lld\n", n, m)
 10 #define sld(n) scanf("%lld",&n)
 11 #define sldd(n,m) scanf("%lld%lld",&n,&m)
 12 #define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
 13 #define sf(n) scanf("%lf",&n)
 14 #define sff(n,m) scanf("%lf%lf",&n,&m)
 15 #define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
 16 #define rep(i,a,n) for (int i=a;i<=n;i++)
 17 #define per(i,n,a) for (int i=n;i>=a;i--)
 18 #define mm(a,n) memset(a, n, sizeof(a))
 19 #define pb push_back
 20 #define all(x) (x).begin(),(x).end()
 21 #define fi first
 22 #define se second
 23 #define ll long long
 24 #define numm ch - 48
 25 #define INF 0x3f3f3f3f
 26 #define sz(x) ((int)x.size())
 27 #define pi 3.14159265358979323
 28 #define debug(x) cout << #x << ": " << x << endl
 29 #define debug2(x, y)          cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;
 30 #define debug3(x, y, z)       cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;
 31 #define debug4(a, b, c, d)    cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
 32 using namespace std;
 33 template<typename T>void read(T &res)
 34 {
 35     bool flag=false;
 36     char ch;
 37     while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);
 38     for(res=numm; isdigit(ch=getchar()); res=(res<<1)+(res<<3)+numm);
 39     flag&&(res=-res);
 40 }
 41 template<typename T>void Out(T x)
 42 {
 43     if(x<0)putchar('-'),x=-x;
 44     if(x>9)Out(x/10);
 45     putchar(x%10+'0');
 46 }
 47 ll pow_mod(ll x, ll n , ll mod)
 48 {
 49     ll res=1;
 50     while(n)
 51     {
 52         if(n&1)res=res*x%mod;
 53         x=x*x%mod;
 54         n>>=1;
 55     }
 56     return res;
 57 }
 58 #define lson l,mid,rt << 1
 59 #define rson mid + 1,r,rt << 1 | 1
 60 #define ll long long
 61 using namespace std;
 62  
 63 const int N = 4e5 + 5000;
 64 const ll mod = 1e9 + 7;
 65  
 66 ll n,k;
 67 ll a[N];
 68 int t[N][26];
 69  
 70 void update(int l,int r,int rt,int id,int k,int val)
 71 {
 72     if(k < l || k > r) return ;
 73     if(l == r)
 74     {
 75         t[rt][id] += val;
 76         return ;
 77     }
 78     int mid = l + r >> 1;
 79     if(k <= mid) update(lson,id,k,val);
 80     else update(rson,id,k,val);
 81     t[rt][id] = t[rt << 1][id] + t[rt << 1 | 1][id];
 82  
 83 }
 84 int qu(int l,int r,int rt,int ql,int qr,int id)
 85 {
 86     if(r < ql || l > qr) return 0;
 87     int res = 0;
 88     if(ql <= l && qr >= r) return t[rt][id];
 89     int mid = l + r >> 1;
 90     if(ql <= mid)  res += qu(lson,ql,qr,id);
 91     if(qr > mid)  res += qu(rson,ql,qr,id);
 92     return res;
 93 }
 94 int main()
 95 {
 96     string str = " ";
 97     string strr;
 98     cin >> strr;
 99     str += strr;
100     n = sz(strr);
101     rep(i ,1 ,n)
102     {
103         update(1,n,1,str[i] - 'a',i,1);
104     }
105     int q,op,l,r;
106     sd(q);
107     while(q--)
108     {
109         sd(op);
110         if(op == 1)
111         {
112             int pos;
113             char c;
114             sd(pos);
115             cin  >> c;
116             update(1,n,1,str[pos] - 'a',pos ,-1);
117             str[pos] = c;
118             update(1,n,1,c - 'a',pos,1);
119         }
120         else
121         {
122             sdd(l , r);
123             ll ans = 0;
124             rep(i,0,25)
125             {
126                 if(0 < (qu(1,n,1,l,r,i))) ans++;
127             }
128             pd(ans);
129         }
130     }
131     return 0;
132 }
View Code

 

赛后听说有人是用26个set做的 ,因为set占用的空间比较小,所以我也用set写了一遍

思路:每个字符对应set存每个字符在母串中的位置 , 查询的时候只要对 L 进行二分查找判断找到的位置是否 <= R

因为可能是找不到的 , 所以我们可以用两种方法避免:

①、 再加个条件——判断lower_bound(L) 是否 >= L

②、 向每个set插入一个大于母串长度的数 

两种方法我用注释区分

Codeforces Round #590 (Div. 3)
 1 #include<bits/stdc++.h>
 2 #define ios std::ios::sync_with_stdio(false) , std::cin.tie(0) , std::cout.tie(0)
 3 #define sd(n) scanf("%d",&n)
 4 #define sdd(n,m) scanf("%d%d",&n,&m)
 5 #define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
 6 #define pd(n) printf("%d\n", (n))
 7 #define pdd(n,m) printf("%d %d\n", n, m)
 8 #define pld(n) printf("%lld\n", n)
 9 #define pldd(n,m) printf("%lld %lld\n", n, m)
10 #define sld(n) scanf("%lld",&n)
11 #define sldd(n,m) scanf("%lld%lld",&n,&m)
12 #define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
13 #define sf(n) scanf("%lf",&n)
14 #define sff(n,m) scanf("%lf%lf",&n,&m)
15 #define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
16 #define rep(i,a,n) for (int i=a;i<=n;i++)
17 #define per(i,n,a) for (int i=n;i>=a;i--)
18 #define mm(a,n) memset(a, n, sizeof(a))
19 #define pb push_back
20 #define all(x) (x).begin(),(x).end()
21 #define fi first
22 #define se second
23 #define ll long long
24 #define numm ch - 48
25 #define INF 0x3f3f3f3f
26 #define pi 3.14159265358979323
27 #define debug(x) cout << #x << ": " << x << endl
28 #define debug2(x, y)          cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<< endl;
29 #define debug3(x, y, z)       cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;
30 #define debug4(a, b, c, d)    cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
31 using namespace std;
32 template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);
33 for(res=numm;isdigit(ch=getchar());res=(res<<1)+(res<<3)+numm);flag&&(res=-res);}
34 template<typename T>void Out(T x){if(x<0)putchar('-'),x=-x;if(x>9)Out(x/10);putchar(x%10+'0');}
35 ll pow_mod(ll x, ll n , ll mod){ll res=1;while(n){if(n&1)res=res*x%mod;x=x*x%mod;n>>=1;}return res;}
36 const int N = 1e3+10;
37 string str = " ";
38 set<int>haha[27];
39 int main()
40 {
41     ios;
42     string a;
43     cin >> a;
44     str += a;
45     rep(i , 1 , 26)
46     haha[i].insert(999999999);
47     rep(i , 1 , str.size() - 1)
48     {
49         haha[str[i] - 'a' + 1].insert(i);
50     }
51     int q;
52     cin >> q;
53     while(q--)
54     {
55         int x;
56         cin >> x;
57         if(x == 1)
58         {
59             int pos ;char ch;
60             cin >> pos >> ch;
61             haha[ str[pos] - 'a' + 1 ].erase( pos );
62             haha[ch - 'a' + 1].insert(pos);
63             str[pos] = ch;
64         }
65         else if(x == 2)
66         {
67             int l , r , ans = 0;;
68             cin >> l >> r;
69             rep(i ,1 ,26)
70             {
71                 if(*haha[i].lower_bound(l) <= r/* && *haha[i].lower_bound(l) >= l*/)
72                 {
73                     ans ++;
74                 }
75             }
76             cout << ans << endl;
77         }
78     }
79     return 0;
80 }
81  
View Code

 

 

 

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