Justice

题目描述

Put simply, the Justice card represents justice, fairness, truth and the law. You are being called to account for your actions and will be judged accordingly. If you have acted in a way that is in alignment with your Higher Self and for the greater good of others, you have nothing to worry about. However, if you have acted in a way that is out of alignment， you will be called out and required to own up to your actions. If this has you shaking in your boots, know that the Justice card isn’t as black and white as you may think.

On the table there are n weights. On the body of the i-th weight carved a positive integer ki, indicating that its weight is gram. Is it possible to divide then weights into two groups and make sure that the sum of the weights in each group is greater or equal to 1/2 gram? That’s on your call. And please tell us how if possible.

输入

In the first line of the input there is a positive integer T (1≤T≤2000), indicating there are T testcases.
In the first line of each of the T testcases, there is a positive integer n (1≤n≤105 , Σn≤7 × 105 ),indicating there areηweights on the table.
In the next line， there are n integers ki (1≤ki≤109), indicating the number carved on each weight.

输出

For each testcase, first print Case i : ANSWER in one line, i indicating the case number starting from 1 and ANSWER should be either YES or NO, indicating whether or not it is possible to divide the weights. Pay attention to the space between : and ANSWER.
If it’s possible, you should continue to output the dividing solution by print a 0/1 string of length n in the next line. The i-th character in the string indicating whether you choose to put the i-th weight in group 0 or group 1.

样例输入

3
3
2 2 2
3
2 2 1
2
1 1

样例输出

Case 1: NO
Case 2: YES
001
Case 3: YES
10

思路：

当k = 1时，此时是1/2，那么分成两部分，两个组都需要一个1/2，我们设cnt1为第一组组成1/2需要的k的数量，cnt2为第二组组成1/2需要的k的数量，那么当k = 1时，cnt1 = 1, cnt2 = 1。循环着走，我们知道当k = 2时，cnt1 = 2，cnt2 = 2，依次循环，每次我们找在剩余的数中能否满足条件。
AC代码：

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;

int ans[N];

struct node
{
    int val, idx;
    bool operator < (const node &t) const
    {
        return val < t.val;
    }
}a[N];

int main()
{
    int t; cin >> t;
    for (int k = 1; k <= t; k ++)
    {
        int n; cin >> n;
        for (int i = 1; i <= n; i ++) { cin >> a[i].val; a[i].idx = i; }

        sort(a + 1, a + n + 1);

        memset(ans, 0, sizeof ans);

        bool flag = 0;
        int cnt1 = 1, cnt2 = 1, p = 1;
        for (int i = 1; i <= n; i ++)
        {
        ///n - i + 1表示剩余的数，因为我们是排了序的。循环完了的时候p = a[i].val。
            while (cnt1 + cnt2 <= n - i + 1 && p < a[i].val)
            {
                cnt1 *= 2;
                cnt2 *= 2;
                p ++;
            }

			///因为现在的数字只有n - i + 1个数字没有被使用过，所以如果需要的总数超过了这个，说明是不能够成立的。
            if (cnt1 + cnt2 > n - i + 1) { flag = 1; break; }

			///这里是进行分组。如果是第一组需要，我们分给第一组（这里也可以分给第二组，顺序没有影响）。
            if (cnt1) { cnt1 --; ans[a[i].idx] = 1; }
            else cnt2 --;

			///这里是条件成立的情况，就是恰好给两组分完。
            if (!cnt1 && !cnt2) break;
        }

        if (flag || cnt1 || cnt2) printf("Case %d: NO\n", k);
        else
        {
            printf("Case %d: YES\n", k);
            for (int i = 1; i < n; i ++) cout << ans[i];
            cout << ans[n] << endl;
        }
    }

    return 0;
}