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AtCoder 182

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A - twiblr

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#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
#include <string>
#include <unordered_set>
using namespace std;

int main(){
    int A, B;
    cin>>A>>B;
    cout<< 2 * A + 100 - B <<endl;
    return 0;
}

B - Almost GCD
GCD-ness of k is the number of elements among A_1 to A_n divisible k.
Print the most GCD-ness number greater than or equal to 2;

almost-gcd.cppview raw
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#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
#include <string>
#include <unordered_set>
#include <unordered_map>
using namespace std;

const int INF = 1e9;

int main(){
    unordered_map<int,int> mp;
    int n;
    cin>>n;
    vector<int> a(n);
    int k = 0;
    for(int i=0;i<n;i++) {
        scanf("%d", &a[i]);
        k = max(k, a[i]);
    }

    int res = 0, num = 0;
    for(int i=2;i<=k;i++){
        for(int j=0;j<n;j++) {
            if(a[j] % i==0) mp[i]++;
        }
        if(mp[i] > num){
            num = mp[i];
            res = i;
        }
    }
    cout<<res<<endl;
    return 0;
}

C - To 3
let k be the number of digits in N. Want to make N a multiple of 3 by erasing 0~k-1 digits.
Return the minimum number of digits moved to achieve that.

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#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
#include <string>
#include <unordered_set>
#include <unordered_map>
#include <cstring>
using namespace std;


int main(){
    long long  n;
    cin>>n;
    unordered_map<int,int> mp;
    int k =0, s = 0;
    while(n){
        int c = n%10;
        mp[c%3] ++;
        s += c;
        n/=10;
        k++;
    }
    if(s%3==0) cout<<0<<endl;
    else if(s%3==1) {
        int res = k+1;
        if(mp[1]>=1 && k>1) res = 1;
        if(mp[2]>=2 && k>2) res = min(res, 2);
        if(res==k+1) res = -1;
        cout<<res<<endl;
    }else{
        int res = k+1;
        if(mp[2]>=1 && k>1) res = 1;
        if(mp[1]>=2 && k>2) res = min(res, 2);
        if(res==k+1) res = -1;
        cout<<res<<endl;
    }
    return 0;

}

D - Wandering
Given a sequence A_1 to A_n;
Move A1;
Move A1, then move A2;
…
Move A1, then move A2, … move A_n;
Find the most positive coordinate in the process.

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#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
#include <string>
#include <unordered_set>
#include <unordered_map>
using namespace std;

typedef long long LL;
const int INF = 0x3f3f3f3f;

int main(){
    int n;
    cin>>n;
    LL sum = 0, x = 0, mx = 0;
    LL res = 0;
    for(int i=0;i<n;i++){
        LL a;
        scanf("%lld", &a);
        mx = max(mx, x + a);
        x += a;
        res = max(res, sum + mx);
        sum += x;
    }
    cout<<res<<endl;
    return 0;
}

E - Akari
Given a 2D-grid, there are N bulbs and M blocks on this grid.
Every bulb emits beams of in 4 directions, which can illuminate grids on the way, until reaching a block.
Find the number of grids illuminated.
Solution: The number of bulbs is too large.
Should enumerate all the grids directly, and mark those which are illuminated.

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#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
#include <string>
#include <unordered_set>
#include <unordered_map>
#include <cstring>
using namespace std;

typedef long long LL;
typedef pair<int,int> PII;
int h, w;
int n, m;
const int N = 1510;
int g[N][N];


int main(){
    cin>>h>>w>>n>>m;
    for(int i=0;i<n;i++){
        int a, b;
        scanf("%d%d", &a,&b);
        g[a][b] = 1;
    }
    for(int i=0;i<m;i++){
        int a, b;
        scanf("%d%d", &a, &b);
        g[a][b] = -1;
    }
    int res = 0;
    for(int i=1;i<=h;i++){
        bool flag = false;
        for(int j=1;j<=w;j++){
            if(g[i][j]==-1) flag = false;
            else if(g[i][j]==1) flag = true;
            else if(flag) g[i][j] = 2;
        }
        flag = false;
        for(int j=w;j>=1;j--){
            if(g[i][j]==-1) flag = false;
            else if(g[i][j]==1) flag = true;
            else if(flag) g[i][j] = 2;
        }
    }
    for(int j=1;j<=w;j++){
        bool flag = false;
        for(int i=1;i<=h;i++){
            if(g[i][j]==-1) flag = false;
            if(g[i][j]==1) flag = true;
            else if(flag) g[i][j] = 2;
        }
        flag = false;
        for(int i=h;i>=1;i--){
            if(g[i][j]==-1) flag = false;
            if(g[i][j]==1) flag = true;
            else if(flag) g[i][j] = 2;
        }
    }
    for(int i=1;i<=h;i++)
        for(int j=1;j<=w;j++)
            res += g[i][j] > 0;
    cout<<res<<endl;
    return 0;
}

F - Valid payments