Educational Codeforces Round 107 (Rated for Div. 2)

Educational Codeforces Round 107 (Rated for Div. 2)

思路:数1和3的个数
#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=250+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void init(){

}
void solve(){
    int n;cin>>n;
    int ans=0;
    for(int i=0,x;i<n;++i){
        cin>>x;
        if(x==1||x==3)ans++;
    }cout<<ans<<'\n';
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    cin>>t;
    init();
    while(t--){
        solve();
    }
    return 0;
}
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B - GCD Length

思路1:令gcd(x,y)=pow(10,c-1),x和y分别一直乘3和7,直到达到a和b位数

思路2:x=pow(10,a-1)+pow(10,c-1),y=pow(10,b-1)

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=250+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void init(){

}
void solve(){
    int a,b,c;cin>>a>>b>>c;
    int x,y,z;
    z=pow(10,c-1);
    x=y=z;
    while(x<pow(10,a-1))x*=3;
    while(y<pow(10,b-1))y*=7;
    cout<<x<<' '<<y<<'\n';
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    cin>>t;
    init();
    while(t--){
        solve();
    }
    return 0;
}
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C - Yet Another Card Deck

思路:由于数的范围很小,维护每个数的最小位置即可,更新时枚举每个数,若位置在其前面则后移

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=250+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void init(){

}
void solve(){
    int n,q;cin>>n>>q;
    vector<int>c(55,n+5);
    for(int i=1,x;i<=n;++i){
        cin>>x;
        c[x]=min(c[x],i);
    }
    for(int i=0,x;i<q;++i){
        cin>>x;
        cout<<c[x]<<' ';
        for(int j=1;j<=50;++j){
            if(c[j]<c[x])c[j]++;
        }
        c[x]=1;
    }
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    init();
    while(t--){
        solve();
    }
    return 0;
}
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D - Min Cost String

思路:可以看出规律 a ab ac ad... b bc bd... 

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=250+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void init(){

}
void solve(){
    int n,k;cin>>n>>k;
    vector<int>f(30,0);
    string s;

    for(int i=0;i<k;++i){
        s.push_back('a'+i);
        for(int j=i+1;j<k;++j){
            s.push_back('a'+i);
            s.push_back('a'+j);
        }
    }
    string ans;
    while(ans.size()<n){
        ans+=s;
    }
    for(int i=0;i<n;++i)cout<<ans[i];
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    init();
    while(t--){
        solve();
    }
    return 0;
}
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E - Colorings and Dominoes

思路:将求每个方案的数量和的问题,转化为求每个多骨牌的贡献;由于染蓝色和染红色的多骨牌不同,可以分别求两种牌的贡献;

对于一行\一列中连续的k个'o'的多骨牌摆放方案数其实一样,那么预处理出f[i],表示连续i个'o'的所有摆放方案能放的多骨牌数;

对于每个连续的'o'串,贡献为f[i]*2all-i

f[i]的状态转移:

当第i个为蓝色,没有贡献,f[i]+=f[i-1]

当第i个为红色:第i-1个为蓝色时,第i个和第i-1个无法放多骨牌,f[i]+=f[i-2]

        第i-1个为红色时,给前i-2个都新增一个方案,f[i]+=f[i-2];第i个和第i-1个放一个多骨牌,前i-2个有2i-2个方案数,f[i]+=2i-2

即f[i]=f[i-1]+2f[i-2]+2i-2

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=3e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void init(){

}
void solve(){
    int n,m;cin>>n>>m;
    vector<vector<char>>g(n+1,vector<char>(m+1));
    int all=0;
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j)cin>>g[i][j],all+=(g[i][j]=='o');
    int ans=0;
    vector<int>fact(N),f(N);
    fact[0]=1;
    for(int i=1;i<=all;++i)fact[i]=fact[i-1]*2%mod;
    f[0]=f[1]=0;
    for(int i=2;i<=max(n,m);++i)f[i]=((f[i-1]+2*f[i-2])%mod+fact[i-2])%mod;
    for(int i=1;i<=n;++i)
        for(int j=1,k;j<=m;++j){
            k=j;
            while(j<=m&&g[i][j]=='o')j++;
            ans=(ans+f[j-k]*fact[all-j+k]%mod)%mod;
        }
    for(int j=1;j<=m;++j)
        for(int i=1;i<=n;++i){
            int k=i;
            while(i<=n&&g[i][j]=='o')i++;
            ans=(ans+f[i-k]*fact[all-i+k]%mod)%mod;
        }
    cout<<ans;
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    init();
    while(t--){
        solve();
    }
    return 0;
}
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posted @ 2023-08-16 16:11  bible_w  阅读(3)  评论(0编辑  收藏  举报