Educational Codeforces Round 107 (Rated for Div. 2)
A - Review Site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; }View Code
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; }View Code
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; }View Code
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; }View Code
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; }View Code
标签:Educational,Rated,const,int,double,typedef,long,Codeforces,define From: https://www.cnblogs.com/bible-/p/17635392.html