Codeforces Beta Round #16 (Div. 2 Only)
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Codeforces Beta Round #16 (Div. 2 Only)
http://codeforces.com/contest/16
A
水题
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 maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9 freopen("1.txt","r",stdin); 10 #endif */ 11 12 int n,m; 13 string str[105]; 14 15 int main(){ 16 #ifndef ONLINE_JUDGE 17 // freopen("1.txt","r",stdin); 18 #endif 19 cin>>n>>m; 20 int flag=0; 21 map<int,int>mp; 22 for(int i=0;i<n;i++) cin>>str[i]; 23 for(int i=1;i<n;i++){ 24 if(str[i][0]==str[i-1][0]) flag=1; 25 } 26 if(!flag){ 27 if(m==1){ 28 cout<<"YES"<<endl; 29 } 30 else{ 31 for(int k=0;k<n;k++){ 32 for(int i=1;i<m;i++){ 33 if(str[k][i]!=str[k][i-1]) flag=1; 34 } 35 } 36 if(flag) cout<<"NO"<<endl; 37 else cout<<"YES"<<endl; 38 } 39 } 40 else{ 41 cout<<"NO"<<endl; 42 } 43 }
B
贪心
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 maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9 freopen("1.txt","r",stdin); 10 #endif */ 11 12 struct sair{ 13 ll num,val; 14 }a[25]; 15 16 bool cmp(sair a,sair b){ 17 return a.val>b.val; 18 } 19 20 int main(){ 21 #ifndef ONLINE_JUDGE 22 // freopen("1.txt","r",stdin); 23 #endif 24 ll n,m; 25 cin>>n>>m; 26 for(int i=1;i<=m;i++){ 27 cin>>a[i].num>>a[i].val; 28 } 29 sort(a+1,a+m+1,cmp); 30 ll ans=0; 31 for(int i=1;i<=m;i++){ 32 if(n>=a[i].num){ 33 ans+=a[i].num*a[i].val; 34 n-=a[i].num; 35 } 36 else if(n<a[i].num){ 37 ans+=a[i].val*n; 38 n=0; 39 } 40 if(!n) break; 41 } 42 cout<<ans<<endl; 43 }
C
gcd
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 maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9 freopen("1.txt","r",stdin); 10 #endif */ 11 12 ll gcd(ll a,ll b){ 13 if(b==0) return a; 14 return gcd(b,a%b); 15 } 16 int main(){ 17 #ifndef ONLINE_JUDGE 18 // freopen("1.txt","r",stdin); 19 #endif 20 ll a,b,c,d; 21 cin>>a>>b>>c>>d; 22 ll x=gcd(c,d); 23 c/=x; 24 d/=x; 25 x=min(a/c,b/d); 26 cout<<c*x<<" "<<d*x<<endl; 27 }
D
模拟
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 maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9 freopen("1.txt","r",stdin); 10 #endif */ 11 string str; 12 13 int Change(char ch1,char ch2){ 14 return (ch1-‘0‘)*10+(ch2-‘0‘); 15 } 16 17 int main(){ 18 #ifndef ONLINE_JUDGE 19 freopen("1.txt","r",stdin); 20 #endif 21 // std::ios::sync_with_stdio(false); 22 int n; 23 scanf("%d%*c",&n); 24 int pres=25,pref=61; 25 int ans=0; 26 int co=1; 27 for(int i=1;i<=n;i++){ 28 getline(cin,str); 29 // cout<<str<<" "<<i<<endl; 30 char flag=str[7]; 31 int shi=Change(str[1],str[2]); 32 int fen=Change(str[4],str[5]); 33 if(flag==‘p‘) shi+=12; 34 if(shi==12&&flag==‘a‘) shi=0; 35 if(shi==24&&flag==‘p‘) shi=12; 36 // cout<<pres<<" "<<pref<<" "<<shi<<" "<<fen<<endl; 37 if(shi==pres&&fen==pref) co++; 38 else { 39 co=1; 40 } 41 if(co>10) { 42 co=1; 43 ans++; 44 } 45 if(shi<pres||(shi==pres&&fen<pref)){ 46 ans++; 47 } 48 pres=shi; 49 pref=fen; 50 } 51 cout<<ans<<endl; 52 }
E
状压DP
dp(i吃掉j)=dp(i和j同时存在)?p(i战胜j)的概率?prob(i和j相遇)的概率
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 maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9 freopen("1.txt","r",stdin); 10 #endif */ 11 12 int n; 13 double a[25][25]; 14 15 double dp[1<<19]; 16 17 int main(){ 18 #ifndef ONLINE_JUDGE 19 freopen("1.txt","r",stdin); 20 #endif 21 std::ios::sync_with_stdio(false); 22 cin>>n; 23 for(int i=0;i<n;i++){ 24 for(int j=0;j<n;j++){ 25 cin>>a[i][j]; 26 } 27 } 28 dp[(1<<n)-1]=1;///所有鱼都存在的情况 29 for(int i=(1<<n)-1;i;i--){ 30 int num=0; 31 for(int j=0;j<n;j++){ 32 if(i&(1<<j)) num++;///判断存活鱼的个数 33 } 34 for(int j=0;j<n;j++){ 35 if(i&(1<<j)){///j存活 36 for(int k=j+1;k<n;k++){ 37 if(i&(1<<k)){///k存活 38 dp[i-(1<<k)]+=dp[i]*a[j][k]*1.0/(num*(num-1)/2); ///j吃k 39 dp[i-(1<<j)]+=dp[i]*a[k][j]*1.0/(num*(num-1)/2); ///k吃j 40 } 41 } 42 } 43 } 44 } 45 for(int i=0;i<n;i++){ 46 cout<<dp[1<<i]<<" "; 47 } 48 cout<<endl; 49 }
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