2020 CCPC Wannafly Winter Camp Day7 part of the problem-solving report

K

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Each considered independently clearance, takes most value. For combinations for a title can be converted into the required minimum n, so the answer to half.

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#include <bits/stdc++.h>

using namespace std;

#define ll long long
ll input(){
    ll x=0,f=0;char ch=getchar();
    while(ch<'0'||ch>'9') f|=ch=='-',ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return f? -x:x;
}

const int N=1007;

#define PII pair<ll,ll>
#define fr first
#define sc second
#define mp make_pair

PII a[N];
ll ans[N];

ll get(ll a,ll x){
    ll l=0,r=1e9,res=0;
    while(l<=r){
        ll mid=(l+r)/2;
        if((2*a+mid+1)*mid/2<x) l=mid+1;
        else res=mid,r=mid-1;
    }
    return res;
}

int main(){
    ll a1=input(),a2=input();
    int n=input();
    ll Ans=0x3f3f3f3f3f3f3f;
    for(int i=1;i<=n;i++){
        a[i].fr=input(),a[i].sc=input();
        ans[i]=get(a1,a[i].fr);
        ans[i]=max(ans[i],get(a2,a[i].sc));
        Ans=min(Ans,ans[i]);
    }
    printf("%lld\n",Ans);
}

---

H

---

Each number is deleted situation is equally likely, as long as Hu count how many sets of numbers on the list.

---

#include <bits/stdc++.h>

using namespace std;

#define ll long long
ll input(){
    ll x=0,f=0;char ch=getchar();
    while(ch<'0'||ch>'9') f|=ch=='-',ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return f? -x:x;
}

int a=0,b;

int main(){
    int n=input();
    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++){
            a+=__gcd(i,j)==1;
        }
    }
    // cout<<a<<endl;
    b=(n-1)/2*2+1;
    if(a==0) b=1;
    else{
        int tmp=__gcd(a,b);
        a/=tmp,b/=tmp;
    }
    printf("%d/%d\n",a,b);
}

---

G

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For \ (k \ le nm \) case burst search line.

For \ (k> nm \) case the answer is clearly just find a Hamilton circuit to go, then the answer is clearly: \ (\ SUM A_ {I, J} + \ FRAC {((nm-. 1) nm)} {2} + (k-nm ) nm \)

---

#include <bits/stdc++.h>

using namespace std;

#define ll long long
ll input(){
    ll x=0,f=0;char ch=getchar();
    while(ch<'0'||ch>'9') f|=ch=='-',ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return f? -x:x;
}

ll n,m,x,y,k;
ll a[15][15];
ll vis[15][15];

ll dfs(ll dep,ll tx,ll ty){
    if(dep>=k) return 0;
    if(tx<1||tx>n||ty<1||ty>m) return 0;
    ll res=0,t=vis[tx][ty];
    vis[tx][ty]=-(a[tx][ty]+dep);
    res=max(dfs(dep+1,tx+1,ty),res);
    res=max(dfs(dep+1,tx-1,ty),res);
    res=max(dfs(dep+1,tx,ty+1),res);
    res=max(dfs(dep+1,tx,ty-1),res);
    
    vis[tx][ty]=t;
    
    res+=a[tx][ty]+vis[tx][ty]+dep;

    return res;
}

int main(){
    n=input(),m=input();
    x=input(),y=input();
    k=input();

    ll sum=0,Ans=0;

    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            a[i][j]=input();
            vis[i][j]=0;
            sum+=a[i][j];
        }
    }

    if(k<n*m){
        Ans=dfs(0,x,y);
    }else{
        Ans=sum+((n*m-1)*n*m)/2+(k-(n*m))*(n*m);
    }
    printf("%lld\n",Ans);
}