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=200020;
const int maxm=200020;
const double PI=acos(-1.0);
const double eps=1e-6;
const LL mod=1e9+7;
LL gcd(LL a,LL b){return b==0?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=1; for(;b;b>>=1,a=a*a%c)if(b&1)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=-1;
    cin>>n;
    for(i=1;i<=n;++i){
        for(j=1;j<=n;++j){
            if(i%j==0&&i*j>n&&i/j<n){
                cout<<i<<' '<<j<<endl;
                return 0;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}

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=200020;
const int maxm=200020;
const double PI=acos(-1.0);
const double eps=1e-6;
const LL mod=1e9+7;
LL gcd(LL a,LL b){return b==0?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=1; for(;b;b>>=1,a=a*a%c)if(b&1)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=1;i<=n;++i)scanf("%d",a+i);
    sort(a+1,a+1+n);
    int d=0;
    for(i=1,j=1;i<=n&&k;i++){
        if(a[i]-d==0)continue;
        printf("%d
",a[i]-d);
        k--;
        d=a[i];
    }
    while(k--)puts("0");
    return 0;
}

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=200020;
const int maxm=200020;
const double PI=acos(-1.0);
const double eps=1e-6;
const LL mod=1e9+7;
LL gcd(LL a,LL b){return b==0?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=1; for(;b;b>>=1,a=a*a%c)if(b&1)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=0;
int main(){
    int t,n,m,i,j,k,u,v;
    cin>>n;
    for(i=0;i<n;++i)scanf("%lld",a+i);
    LL d=0;
    for(i=n-1;i>=0;--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]=1;
            p1[tot]=i+1;
            p2[tot]=delta;
            tot++;
        }
    }
    op[tot]=2;
    p1[tot]=p2[tot]=n;
    tot++;
    cout<<tot<<endl;
    for(i=0;i<tot;++i){
        printf("%lld %lld %lld
",op[i],p1[i],p2[i]);
    }
    return 0;
}

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=300020;
const int maxm=200020;
const double PI=acos(-1.0);
const double eps=1e-6;
const LL mod=1e9+7;
LL gcd(LL a,LL b){return b==0?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=1; for(;b;b>>=1,a=a*a%c)if(b&1)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=0;
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]>0) f[u]+=f[v];
        }
    }
    if(o)ans=max(ans,f[u]);
    if(!o){
        if(f[u]==ans){
            k++;
            f[u]=0;
        }
    }
}
int main(){
    int t,n=0,m,i=0,j,u,v;
    scanf("%d",&n);
    for(i=1;i<=n;++i)scanf("%lld",a+i);
    for(i=1;i<n;++i){
        scanf("%d%d",&u,&v);
        g[u].pb(v),g[v].pb(u);
    }
    dfs(1,0,1);
    dfs(1,0,0);
    printf("%lld %lld
",ans*k,k);
    return 0;
}