Codeforces Round #450 (Div. 2)

Posted 2021-02-13 fighting-sh

Codeforces Round #450 (Div. 2)

http://codeforces.com/contest/900

A

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define eb emplace_back
 8 #define maxn 100005
 9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef pair<int,int> pii;
14 typedef pair<long long,int>pli;
15 typedef pair<int,char> pic;
16 typedef pair<pair<int,string>,pii> ppp;
17 typedef unsigned long long ull;
18 const long long MOD=1e9+7;
19 /*#ifndef ONLINE_JUDGE
20         freopen("1.txt","r",stdin);
21 #endif */
22 
23 int a[2];
24 int main(){
25     #ifndef ONLINE_JUDGE
26      //   freopen("1.txt","r",stdin);
27     #endif
28     std::ios::sync_with_stdio(false);   
29     int n;
30     cin>>n;
31     int u,v;
32     for(int i=0;i<n;i++){
33         cin>>u>>v;
34         if(u>0) a[0]++;
35         else a[1]++;
36     }
37     if(a[0]<=1||a[1]<=1) cout<<"YES"<<endl;
38     else cout<<"NO"<<endl;
39 }

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B

模拟除法

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define eb emplace_back
 8 #define maxn 100005
 9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef pair<int,int> pii;
14 typedef pair<long long,int>pli;
15 typedef pair<int,char> pic;
16 typedef pair<pair<int,string>,pii> ppp;
17 typedef unsigned long long ull;
18 const long long MOD=1e9+7;
19 /*#ifndef ONLINE_JUDGE
20         freopen("1.txt","r",stdin);
21 #endif */
22 
23 int a[2];
24 ll fac[15];
25 int main(){
26     #ifndef ONLINE_JUDGE
27      //   freopen("1.txt","r",stdin);
28     #endif
29     int a,b,c;
30     cin>>a>>b>>c;
31     int co=1;
32     while(1){
33         a*=10;
34         int k=a/b;
35         a%=b;
36         if(k==c){
37             cout<<co<<endl;
38             return 0;
39         }
40         co++;
41         if(co>=100000) break;
42     }
43     cout<<-1<<endl;
44 
45 }

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C

找规律

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define eb emplace_back
 8 #define maxn 100005
 9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef pair<int,int> pii;
14 typedef pair<long long,int>pli;
15 typedef pair<int,char> pic;
16 typedef pair<pair<int,string>,pii> ppp;
17 typedef unsigned long long ull;
18 const long long MOD=1e9+7;
19 /*#ifndef ONLINE_JUDGE
20         freopen("1.txt","r",stdin);
21 #endif */
22 
23 int n;
24 int a[maxn],cnt[maxn];
25 
26 int main(){
27     #ifndef ONLINE_JUDGE
28      //   freopen("1.txt","r",stdin);
29     #endif
30     cin>>n;
31     int max1=0,max2=0;
32     for(int i=1;i<=n;i++) cin>>a[i];
33     for(int i=1;i<=n;i++){
34         if(a[i]>max1){
35             max2=max1;
36             max1=a[i];
37             cnt[a[i]]--;
38         }
39         else if(a[i]>max2){
40             cnt[max1]++;
41             max2=a[i];
42         }
43     }
44     int max3=-maxn,ans;
45     for(int i=1;i<=n;i++){
46         if(cnt[i]>max3){
47             max3=cnt[i];
48             ans=i;
49         }
50     }
51     cout<<ans<<endl;
52 
53 }

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D

记忆化搜索+容斥

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define eb emplace_back
 8 #define maxn 100005
 9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef pair<int,int> pii;
14 typedef pair<long long,int>pli;
15 typedef pair<int,char> pic;
16 typedef pair<pair<int,string>,pii> ppp;
17 typedef unsigned long long ull;
18 const long long MOD=1e9+7;
19 /*#ifndef ONLINE_JUDGE
20         freopen("1.txt","r",stdin);
21 #endif */
22 
23 int x,y;
24 map<int,int>mp;
25 
26 ll pow_mul(ll a,ll b){
27     ll ans=1;
28     while(b){
29         if(b&1) ans=(ans*a)%MOD;
30         b>>=1;
31         a=(a*a)%MOD;
32     }
33     return ans;
34 }
35 
36 ll dp(int x){
37     if(x==1) return 1;
38     if(mp.count(x)) return mp[x];
39     mp[x]=pow_mul(2,x-1);
40     for(int i=2;i*i<=x;i++){
41         if(x%i==0){
42             mp[x]=(mp[x]-dp(x/i)+MOD)%MOD;
43             if(i!=x/i){
44                 mp[x]=(mp[x]-dp(i)+MOD)%MOD;
45             }
46         }
47     }
48     return mp[x]=(mp[x]-dp(1)+MOD)%MOD;
49 }
50 
51 int main(){
52     #ifndef ONLINE_JUDGE
53      //   freopen("1.txt","r",stdin);
54     #endif
55     cin>>x>>y;
56     if(y%x){
57         cout<<0<<endl;
58     }
59     else{
60         cout<<dp(y/x)<<endl;
61     }
62 
63 }

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E

DP

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define eb emplace_back
 8 #define maxn 100005
 9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef pair<int,int> pii;
14 typedef pair<long long,int>pli;
15 typedef pair<int,char> pic;
16 typedef pair<pair<int,string>,pii> ppp;
17 typedef unsigned long long ull;
18 const long long MOD=1e9+7;
19 /*#ifndef ONLINE_JUDGE
20         freopen("1.txt","r",stdin);
21 #endif */
22 pii d[maxn];
23 int f[3][maxn], n, m, b[maxn];
24 char a[maxn];
25 
26 int main(){
27     #ifndef ONLINE_JUDGE
28      //   freopen("1.txt","r",stdin);
29     #endif
30      cin >> n;
31     for (int i = 1; i <= n; i++)
32         cin >> a[i];
33     cin >> m;
34     for (int i = 1; i <= n; i++)
35     {
36         if (a[i] == ‘?‘)
37             b[i] = b[i-1] + 1;
38         else b[i] = b[i-1];
39         if (a[i] != ‘a‘)
40             f[1][i] = f[0][i-1]+1;
41         if (a[i] != ‘b‘)
42             f[0][i] = f[1][i-1]+1;
43         if (f[!(m&1)][i] >= m)
44             d[i] = {d[i-m].first + 1, -(b[i]-b[i-m]-d[i-m].second)};
45         d[i] = max(d[i-1], d[i]);
46     }
47     cout << - d[n].second;
48 
49 }

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