cf593div2

https://codeforces.com/contest/1236/problem/A

#include<bits/stdc++.h>

using namespace std;

int main(){
    int t,a,b,c;
    cin >> t;
    while(t--) {
        cin >> a>> b>>c;
        int x = c/2, sum = 0;
        sum += 3*min(x, b);
        b -= min(x,b);
        int y = b/2;
        sum += min(a, y) *3;
        cout << sum << endl;
    }
    return 0;
}

https://codeforces.com/contest/1236/problem/B

First, we can think about putting one present in m boxs, each box has two conditions（put or not） ,so the res is 2^m, and we should subtract 1(all boxs are empty).(the answer one present in m boxs is 2^m-1)

Then we have n presents, so the amnswer is (2^m-1)^n,

#include<bits/stdc++.h>

using namespace std;
#define ll long long
const ll mod = 1e9+7;
ll fpow(ll a, ll b) {
    ll ret = 1;
    while(b) {
        if(b & 1) ret = ret*a%mod;
        a = a*a%mod;
        b >>= 1;
    }
    return ret;
}
int main(){
    int t;
    t = 1; //cin >> t;
    while(t--) {
        ll a,b;
        cin >> a >> b;
        if(b == 1) cout << "1\n";

        else cout << fpow(fpow(2, b)-1, a) << endl;

    }
    return 0;
}

https://codeforces.com/contest/1236/problem/C

The maximum number is ⌊n22⌋⌊n22⌋.  具体证明不会 emmmmmmmm
可以蛇形填数(如下)
 1 6 7
 2 5 8
 3 4 9

#include<bits/stdc++.h>

using namespace std;
#define ll long long
#define _for(i,a,b) for(int i = (a); i < (b); i++)
#define _rep(i,a,b) for(int i = (a); i <= (b); i++)
const ll mod = 1e9+7;

int main(){
    int t,a,b,c;
    t = 1; //cin >> t;
    while(t--) {
        int dp[303][303];
        cin >> a;
        _rep(j,1,a)
            {
                if(j&1) _for(i,0,a) dp[i][j] = 1+(j-1)*a+i;
                else    _for(i,0,a) dp[i][j] = j*a-i;
            }

        _rep(i,0,a-1) _for(j,1,a+1)
             cout << dp[i][j] <<(j==a?"\n":" ");
    }
    return 0;
}
/*
2 8 5
9 3 4
7 6 1
*/

https://codeforces.com/contest/1236/problem/D

https://codeforces.com/contest/1236/problem/E

1.