Justice(HDU6557+2018年吉林站+二进制）

题目链接

传送门

题意

给你\(n\)个数，每个数表示\(\frac{1}{2^{a_i}}\)，要你把这\(n\)个数分为两堆，使得每堆的和都大于等于\(\frac{1}{2}\)。

思路

首先我们假设第一堆的下标为\(x_1,x_2\dots,x_n\)，且\(2^{a_{x_1}}\leq 2^{a_{x_2}}\leq\dots\leq 2^{a_{x_n}}\)，则：

\[\begin{aligned}
&\sum\limits_{i=1}^{n}\frac{1}{2^{a_{x_i}}}&\\
=&\sum\limits_{i=1}^{n}\frac{2^{a_{x_n}-a_{x_i}}}{2^{a_{x_n}}}\text{（通分）}&
\end{aligned}
\]

我们知道\(2\)个\(2^x\)构成一个\(2^{x+1}\)，因此我们确定每一堆分母的最大值后按照\(2^{a_{x_i}}\)从小到大来凑\(2^{a_{x_n}-1}\)。

注意，一旦相邻两个数的大小差值大于等于\(17\)就直接不用取输出\(NO\)，因为\(2^{17}\geq 10^5\)，后面不管怎么凑都是不可能凑出我们所需要的数的。

代码实现如下

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int t, n;
int vis[maxn];

struct node {
    int val, id;
    bool operator < (const node& x) const {
        return val < x.val;
    }
}a[maxn];

int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
    scanf("%d", &t);
    int icase = 0;
    while(t--) {
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &a[i].val);
            a[i].id = i;
            vis[i] = 0;
        }
        printf("Case %d: ", ++icase);
        sort(a + 1, a + n + 1);
        LL nw;
        if(a[1].val == 1) {
            vis[a[1].id] = 1;
        } else {
            nw = a[1].val - 1;
            if(nw >= 17) {
                printf("NO\n");
                continue;
            }
            nw = (1<<nw) - 1;
            vis[a[1].id] = 1;
            for(int i = 2; i <= n; ++i) {
                if(nw == 0) break;
                if(a[i].val == a[i-1].val) --nw, vis[a[i].id] = 1;
                else {
                    if(a[i].val - a[i-1].val >= 17) {
                        break;
                    }
                    nw = nw * (1<<(a[i].val - a[i-1].val)) - 1;
                    if(nw > n - i) break;
                    vis[a[i].id] = 1;
                }
            }
            if(nw != 0) {
                printf("NO\n");
                continue;
            }
        }
        int idx = -1;
        for(int i = 1; i <= n; ++i) {
            if(!vis[a[i].id]) {
                idx = i;
                break;
            }
        }
        if(idx == -1) {
            printf("NO\n");
            continue;
        }
        if(a[idx].val == 1) {
            printf("YES\n");
            for(int i = 1; i <= n; ++i) {
                printf("%d", vis[i]);
            }
            printf("\n");
        } else {
            nw = a[idx].val - 1;
            if(nw >= 17) {
                printf("NO\n");
                continue;
            }
            nw = (1<<nw) - 1;
            for(int i = idx + 1; i <= n; ++i) {
                if(nw == 0) break;
                if(vis[a[i].id]) continue;
                if(a[i].val == a[i-1].val) --nw;
                else {
                    if(a[i].val - a[i-1].val >= 17) {
                        break;
                    }
                    nw = nw * (1<<(a[i].val - a[i-1].val)) - 1;
                    if(nw > n - i) break;
                }
            }
            if(nw != 0) {
                printf("NO\n");
                continue;
            }
            printf("YES\n");
            for(int i = 1; i <= n; ++i) {
                printf("%d", vis[i]);
            }
            printf("\n");
        }
    }
    return 0;
}