Codeforces Round #525 (Div. 2)-A/B/C/E

http://codeforces.com/contest/1088/problem/A

暴力一波就好了。

//题解有O(1)做法是 (n-n%2,2)

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<map>
 #include<set>
 #include<stack>
 #include<deque>
 #include<bitset>
 #include<unordered_map>
 #include<unordered_set>
 #include<queue>
 #include<cstdlib>
 #include<ctype.h>
 #include<ctime>
 #include<functional>
 #include<algorithm>
 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define pii pair<int,int>
 #define mp make_pair
 #define pb push_back
 #define fi first
 #define se second
 #define inf 0x3f3f3f3f
 #define debug puts("debug")
 #define mid ((L+R)>>1)
 #define lc (id<<1)
 #define rc (id<<1|1)
 const int maxn=;
 const int maxm=;
 const double PI=acos(-1.0);
 const double eps=1e-;
 const LL mod=1e9+;
 LL gcd(LL a,LL b){return b==?a:gcd(b,a%b);}
 LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
 LL qpow(LL a,LL b,LL c){LL r=; for(;b;b>>=,a=a*a%c)if(b&)r=r*a%c;return r;}
 template<class T>
 void prt(T v){for(auto x:v)cout<<x<<' ';cout<<endl;}
 struct Edge{int u,v,w,next;};
 int main(){
     int t,n,m,i,j,k,u,v;
     int ans=-;
     cin>>n;
     for(i=;i<=n;++i){
         for(j=;j<=n;++j){
             if(i%j==&&i*j>n&&i/j<n){
                 cout<<i<<' '<<j<<endl;
                 return ;
             }
         }
     }
     cout<<ans<<endl;
     return ;
 }

http://codeforces.com/contest/1088/problem/B

排下序然后遍历一遍。容易发现被减数总等于前一个a[i]。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<map>
 #include<set>
 #include<stack>
 #include<deque>
 #include<bitset>
 #include<unordered_map>
 #include<unordered_set>
 #include<queue>
 #include<cstdlib>
 #include<ctype.h>
 #include<ctime>
 #include<functional>
 #include<algorithm>
 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define pii pair<int,int>
 #define mp make_pair
 #define pb push_back
 #define fi first
 #define se second
 #define inf 0x3f3f3f3f
 #define debug puts("debug")
 #define mid ((L+R)>>1)
 #define lc (id<<1)
 #define rc (id<<1|1)
 const int maxn=;
 const int maxm=;
 const double PI=acos(-1.0);
 const double eps=1e-;
 const LL mod=1e9+;
 LL gcd(LL a,LL b){return b==?a:gcd(b,a%b);}
 LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
 LL qpow(LL a,LL b,LL c){LL r=; for(;b;b>>=,a=a*a%c)if(b&)r=r*a%c;return r;}
 template<class T>
 void prt(T v){for(auto x:v)cout<<x<<' ';cout<<endl;}
 struct Edge{int u,v,w,next;};
 int a[maxn];
 int main(){
     int t,n,m,i,j,k,u,v;
     cin>>n>>k;
     for(i=;i<=n;++i)scanf("%d",a+i);
     sort(a+,a++n);
     int d=;
     for(i=,j=;i<=n&&k;i++){
         if(a[i]-d==)continue;
         printf("%d\n",a[i]-d);
         k--;
         d=a[i];
     }
     while(k--)puts("");
     return ;
 }

http://codeforces.com/contest/1088/problem/C

构造，题目里的n+1就是提示了，从n-1开始倒着遍历，进行n次加法每次都让当前的数%n等于i,最后摸一次n就好。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<map>
 #include<set>
 #include<stack>
 #include<deque>
 #include<bitset>
 #include<unordered_map>
 #include<unordered_set>
 #include<queue>
 #include<cstdlib>
 #include<ctype.h>
 #include<ctime>
 #include<functional>
 #include<algorithm>
 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define pii pair<int,int>
 #define mp make_pair
 #define pb push_back
 #define fi first
 #define se second
 #define inf 0x3f3f3f3f
 #define debug puts("debug")
 #define mid ((L+R)>>1)
 #define lc (id<<1)
 #define rc (id<<1|1)
 const int maxn=;
 const int maxm=;
 const double PI=acos(-1.0);
 const double eps=1e-;
 const LL mod=1e9+;
 LL gcd(LL a,LL b){return b==?a:gcd(b,a%b);}
 LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
 LL qpow(LL a,LL b,LL c){LL r=; for(;b;b>>=,a=a*a%c)if(b&)r=r*a%c;return r;}
 template<class T>
 void prt(T v){for(auto x:v)cout<<x<<' ';cout<<endl;}
 struct Edge{int u,v,w,next;};
 LL a[maxn];
 LL op[maxn],p1[maxn],p2[maxn],tot=;
 int main(){
     int t,n,m,i,j,k,u,v;
     cin>>n;
     for(i=;i<n;++i)scanf("%lld",a+i);
     LL d=;
     for(i=n-;i>=;--i){
         if((a[i]+d)%n==i)continue;
         else{
             LL x=(a[i]+d)%n;
             LL delta=(i-x+n)%n;
             d+=delta;
             op[tot]=;
             p1[tot]=i+;
             p2[tot]=delta;
             tot++;
         }
     }
     op[tot]=;
     p1[tot]=p2[tot]=n;
     tot++;
     cout<<tot<<endl;
     for(i=;i<tot;++i){
         printf("%lld %lld %lld\n",op[i],p1[i],p2[i]);
     }
     return ;
 }

http://codeforces.com/contest/1088/problem/E

给出一棵树，每个节点有权值a[i]，找出k个互不重复覆盖的联通分量，使得SUM{a[i] | i in s} / k的值最大化，如果有多种方案选择k最大的方案。

假设最优解是{b1,b2...bk} 那么有公式  max{b}>=ave{b}  ,也就是说ave{b}的最大值就是max{b},问题转化为求最大权值和的连通分量，然后看有多少个等于ans的联通分量，就是k。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<map>
 #include<set>
 #include<stack>
 #include<deque>
 #include<bitset>
 #include<unordered_map>
 #include<unordered_set>
 #include<queue>
 #include<cstdlib>
 #include<ctype.h>
 #include<ctime>
 #include<functional>
 #include<algorithm>
 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define pii pair<int,int>
 #define mp make_pair
 #define pb push_back
 #define fi first
 #define se second
 #define inf 0x3f3f3f3f
 #define debug puts("debug")
 #define mid ((L+R)>>1)
 #define lc (id<<1)
 #define rc (id<<1|1)
 const int maxn=;
 const int maxm=;
 const double PI=acos(-1.0);
 const double eps=1e-;
 const LL mod=1e9+;
 LL gcd(LL a,LL b){return b==?a:gcd(b,a%b);}
 LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
 LL qpow(LL a,LL b,LL c){LL r=; for(;b;b>>=,a=a*a%c)if(b&)r=r*a%c;return r;}
 template<class T>
 void prt(T v){for(auto x:v)cout<<x<<' ';cout<<endl;}
 struct Edge{int u,v,w,next;};
 LL ans=-inf,f[maxn],a[maxn],k=;
 vector<int>g[maxn];
 void dfs(int u,int fa,bool o){
     f[u]=a[u];
     for(int v:g[u]){
         if(v!=fa){
             dfs(v,u,o);
             if(f[v]>) f[u]+=f[v];
         }
     }
     if(o)ans=max(ans,f[u]);
     if(!o){
         if(f[u]==ans){
             k++;
             f[u]=;
         }
     }
 }
 int main(){
     int t,n=,m,i=,j,u,v;
     scanf("%d",&n);
     for(i=;i<=n;++i)scanf("%lld",a+i);
     for(i=;i<n;++i){
         scanf("%d%d",&u,&v);
         g[u].pb(v),g[v].pb(u);
     }
     dfs(,,);
     dfs(,,);
     printf("%lld %lld\n",ans*k,k);
     return ;
 }