9/21 harder the more fortunate - thinking game (4.0)

F.CodeForces - 1118D1

Meaning of the questions: There are n cup of coffee, you need to write m-page job; he can drink Any cup of coffee a day (1 or multi-cup); each cup of coffee you can let him write ai-page job; in the same day, he drank coffee can let him write w pages;

w = (first + second cup cup -1 + -2 third cup ......); seeking to drink at least a few days to finish the job (which seems to describe the meaning of the questions are not many, improvise ~ bad language)

Solution: Sorting + binary answer, the lower limit of one day, the upper limit of n days, then directly see Check code;

O (nlogn)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=1e5+10;
int a[MAXN];
int n,m;
bool check(int mid){
    int ans=0;
    int i=1;
    int cnt=0;
    int k=mid;
    while(i<=n){
        for(;i<=min(k,n);i++){
            ans+=max(a[i]-cnt,0);
        }
        if(ans>=m)return true;
        k=k+=mid;
        cnt++;
    }
    return false;
}
bool cmp(int x,int y){
    return x>y;
}
int main(){
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    sort(a+1,a+1+n,cmp);
    int ans=-1;
    int l=1,r=n,mid;
    while(l<=r){
        mid=(l+r)>>1;
        if(check(mid)){
            r=mid-1;
            ans=mid;
        }
        else l=mid+1;
    }
    cout<<ans<<endl;


    return 0;
}

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E.CodeForces - 1221C

Meaning of the questions: acm team, there is a code hands, b the number on the king, c idler;

Seeking up to set the number of teams, each team must have at least one hand and the code number theory a king;

Solution: Analog? ; Easy

#include <bits/stdc++.h>
using namespace std;
const int MAXN=1e5+10;
typedef long long ll;
int main(){
    ios::sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--){
        ll a,b,c,minn;
        cin>>a>>b>>c;
        minn=min(a,min(b,c));
        if(minn!=c){
            cout<<minn<<endl;
        }
        else{
            a-=c;b-=c;
            cout<<minn+min(a,min(b,(a+b)/3))<<endl;
        }
    }



    return 0;
}

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D.CodeForces - 1221B

Question is intended: to a n, the output of a graph (containing only W, B), such that each W or B or B to attack the maximum number of W;

Each character can only attack on the daily diagonal; W attack B, B to attack W;

Solution: BW CROSS output on the line;

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=2e5+10;
int main(){
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(i%2==j%2)cout<<'B';
            else cout<<'W';
        }
        cout<<endl;
    }


    return 0;
}

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C.CodeForces - 1118B 

The meaning of problems: the number n, seeking to remove a number, such that the remaining entries in the array equal to the odd number and even-numbered;

Solution: obtaining first and odd-numbered and even-numbered and not removed;

Then through the array; i representative of the i-th then removed

Odd-numbered (even-numbered after and before i + i and odd-numbered) and =

Even term (before and after the odd-numbered even-numbered and i + i and) and =

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=2e5+10;
int a[MAXN];
int main(){
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    int sum1=0;
    int sum2=0;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        if(i%2)sum1+=a[i];
        else sum2+=a[i];
    }
    //cout<<"sum1="<<sum1<<' '<<"sum2="<<sum2<<endl;
    int ans=0;
    int newsum1=0;
    int newsum2=0;
    for(int i=1;i<=n;i++){
        if(i%2){
            sum1-=a[i];
            if(sum1+newsum2==sum2+newsum1)ans++;
            newsum1+=a[i];
        }
        else {
            sum2-=a[i];
            if(sum1+newsum2==sum2+newsum1)ans++;
            newsum2+=a[i];
        }
    }
    cout<<ans<<endl;


    return 0;
}

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B.CodeForces - 1221A

Meaning of the questions: n give you the number, the same number of each of the two can be combined to double this number, asked whether the synthesis of 2048;

2048 Games ~;

Solution: bucket sort; less than or equal to the number of sort barrel 2048 is, greater than 2048 can not synthesized;

Then traversing from 1 to 2048 (1024 to), [i] num> = 2 num [i * 2] + = num [i] / 2: continue;?

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=1e3+10;
int a[MAXN];
int num[3000];
int main(){
    ios::sync_with_stdio(false);
    int T,n;
    cin>>T;
    while(T--){
        cin>>n;
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++){
            cin>>a[i];
            if(a[i]<=2048){
                num[a[i]]++;
            }
        }
        for(int i=1;i<=2048;i++){
            if(num[i]>=2){
                num[i*2]+=num[i]/2;
            }
        }
        if(num[2048])cout<<"Yes"<<endl;
        else cout<<"No"<<endl;
    }


    return 0;
}

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A.CodeForces - 1118A

Meaning of the questions: two kinds of water bottles, 1L a dollar, 2L b yuan, need to ask a minimum of n L of water, how much;

Solution: 1L and 2L direct comparison price, buy then follow;

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,a,b;

int main(){
    ios::sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--){
        cin>>n>>a>>b;
        if(a*2<=b){
            cout<<n*a<<endl;
        }
        else{
            ll ans=0;
            ans+=n/2*b;
            if(n%2){
                ans+=a;
            }
            cout<<ans<<endl;
        }
    }


    return 0;
}

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