Codeforces Round #665 (Div. 2) B. Ternary Sequence

Title

CodeForces 1401 B. Ternary Sequence

Time Limit

2 seconds

Memory Limit

256 megabytes

Problem Description

You are given two sequences $a_1,a_2,\dots ,a_n$ and $b_1,b_2,\dots ,b_n$. Each element of both sequences is either $0$, $1$ or $2$. The number of elements $0$, $1$, $2$ in the sequence $a$ is $x_1$, $y_1$, $z_1$ respectively, and the number of elements $0$, $1$, $2$ in the sequence $b$ is $x_2$, $y_2$, $z_2$ respectively.

You can rearrange the elements in both sequences $a$ and $b$ however you like. After that, let’s define a sequence $c$ as follows:

$c_i=\{\begin{array}{ll}{a}_i{b}_i& \text{ if }{a}_i>b_i\\ 0& \text{ if }{a}_i=b_i\\ -a_i{b}_i& \text{ if }{a}_i<b_i\end{array}$

You’d like to make ${\sum }_{i=1}^n{c}_i$ (the sum of all elements of the sequence $c$) as large as possible. What is the maximum possible sum?

Input

The first line contains one integer $t$ ($1\le t\le 10^4$) — the number of test cases.

Each test case consists of two lines. The first line of each test case contains three integers $x_1$, $y_1$, $z_1$ ($0\le x_1,y_1,z_1\le 10^8$) — the number of $0$-s, $1$-s and $2$-s in the sequence $a$.

The second line of each test case also contains three integers $x_2$, $y_2$, $z_2$ ($0\le x_2,y_2,z_2\le 10^8$; $x_1+y_1+z_1=x_2+y_2+z_2>0$) — the number of $0$-s, $1$-s and $2$-s in the sequence $b$.

Output

For each test case, print the maximum possible sum of the sequence $c$.

Sample Input

3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1

Sample Onput

4
2
0

Note

In the first sample, one of the optimal solutions is:

$a=\{2,0,1,1,0,2,1\}$

$b=\{1,0,1,0,2,1,0\}$

$c=\{2,0,0,0,0,2,0\}$

In the second sample, one of the optimal solutions is:

$a=\{0,2,0,0,0\}$

$b=\{1,1,0,1,0\}$

$c=\{0,2,0,0,0\}$

In the third sample, the only possible solution is:

$a=\{2\}$

$b=\{2\}$

$c=\{0\}$

Source

CodeForces 1401 B. Ternary Sequence

题意

给出ab两个数组0,1,2的数量

$c_i=\{\begin{array}{ll}{a}_i{b}_i& \text{ if }{a}_i>b_i\\ 0& \text{ if }{a}_i=b_i\\ -a_i{b}_i& \text{ if }{a}_i<b_i\end{array}$

随意组合求C的最大值

思路

贪心的想

只有$a_2配b_1$是$2$ $a_1配b_2$是$-2$ 其余都是$0$

于是先把$b_1$分配给$a_2$ 再用其余的去填充$b_2$ 最后无可奈何再减去$a_1配b_2$

AC代码:

#include 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int maxn = 3e5 + 5;
const ll inf = 0x3f3f3f3f;
const ll mod = 1e9 + 7;
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        vector<ll> a(3), b(3);
        for (auto &it : a) cin >> it;
        for (auto &it : b) cin >> it;

        ll mini = min(a[2], b[1]);
        ll ans = 0;
        ans += mini * 2;
        a[2] -= mini, b[1] -= mini;

        mini = min(a[2], b[2]);
        a[2] -= mini, b[2] -= mini;

        mini = min(a[0], b[2]);
        a[0] -= mini, b[2] -= mini;

        mini = min(a[1], b[2]);

        cout << ans - 2 * mini << endl;
    }
    return 0;
}