CodeForces_1360H Binary Median(数论+暴力)
Binary Median
Problem Description
Consider all binary strings of length m m m ( 1 ≤ m ≤ 60 1≤m≤60 1≤m≤60). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly 2 m 2^m 2m such strings in total.
The string s is lexicographically smaller than the string t (both have the same length m) if in the first position i from the left in which they differ, we have s [ i ] < t [ i ] s[i]
We remove from this set n n n ( 1 ≤ n ≤ m i n ( 2 m − 1 , 100 ) 1≤n≤min(2^m−1,100) 1≤n≤min(2m−1,100)) distinct binary strings a 1 a_1 a1, a 2 a_2 a2,…, a n a_n an, each of length m. Thus, the set will have k = 2 m − n k=2^m−n k=2m−n strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).
We number all the strings after sorting from 0 0 0 to k − 1 k−1 k−1. Print the string whose index is ⌊ k − 1 ⌋ 2 \frac{⌊k−1⌋}{2} 2⌊k−1⌋ (such an element is called median), where ⌊ x ⌋ ⌊x⌋ ⌊x⌋ is the rounding of the number down to the nearest integer.
For example, if n = 3 n=3 n=3, m = 3 m=3 m=3 and a = [ 010 , 111 , 001 ] a=[010, 111, 001] a=[010,111,001], then after removing the strings ai and sorting, the result will take the form: [ 000 , 011 , 100 , 101 , 110 ] [000, 011, 100, 101, 110] [000,011,100,101,110]. Thus, the desired median is 100 100 100.
Input
The first line contains an integer t t t ( 1 ≤ t ≤ 1000 1≤t≤1000 1≤t≤1000) — the number of test cases. Then, t test cases follow.
The first line of each test case contains integers n n n ( 1 ≤ n ≤ m i n ( 2 m − 1 , 100 ) 1≤n≤min(2^m−1,100) 1≤n≤min(2m−1,100)) and m ( 1 ≤ m ≤ 60 1≤m≤60 1≤m≤60), where n is the number of strings to remove, and m is the length of binary strings. The next n lines contain a 1 a_1 a1, a 2 a_2 a2,…, a n a_n an — distinct binary strings of length m.
The total length of all given binary strings in all test cases in one test does not exceed 1 0 5 10^5 105.
Output
Print t answers to the test cases. For each test case, print a string of length m — the median of the sorted sequence of remaining strings in the corresponding test case.
Sample Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10
Sample Output
100
010
0
1
11
题意
有 2 m 2^m 2m个数，依次为 0 , 1 , 2 , . . . . . , 2 m − 1 0,1,2,.....,2^m-1 0,1,2,.....,2m−1。从中删除n个数，则还剩 k = 2 m − n k=2^m-n k=2m−n个数，求剩下的序列中下标(从0开始)为 ⌊ k − 1 ⌋ 2 \frac{⌊k−1⌋}{2} 2⌊k−1⌋的数是多少。
题解
只考虑原序列，则所求为 x = ⌊ 2 m − 1 ⌋ 2 x = \frac{⌊2^m−1⌋}{2} x=2⌊2m−1⌋。从中删除n个数后，与原先比，结果最多偏移n个。所以可以从 m a x ( 0 , x − n ) max(0, x-n) max(0,x−n)开始考虑，找到符合条件的数即可。
设当前枚举的数为 p o s pos pos，则需要求出序列中剩下的比 p o s pos pos小的数的数量num。num等于pos-删除的序列中小于pos的数的数量。若 n u m = = ⌊ k − 1 ⌋ 2 num==\frac{⌊k−1⌋}{2} num==2⌊k−1⌋， p o s pos pos即为所求。
(若pos被删除，则跳过)。
#include
#define INF 0x3f3f3f3f
#define eps 1e-6
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 120;
const int mod = 1000000007;
LL a[maxn];
char str[maxn];
bool isok(LL x, int n);
int main()
{
int t, n, m, i, j;
LL num, pos, k;
scanf("%d", &t);
while(t--)
{
scanf("%d %d", &n, &m);
for(i=0;i<n;i++){
scanf(" %s", str);
a[i] = 0;
for(j=0;j<m;j++)
a[i] = a[i]*2+str[j]-'0';
}
sort(a, a+n);
pos = max(0LL, (1LL<<m)/2-120);
while(isok(pos, n))
pos++;
k = ((1LL<<m)-n-1)/2;
while(1){
num = pos - (lower_bound(a, a+n, pos)-a);
if(num == k)break;
pos++;
while(isok(pos, n))
pos++;
}
for(i=m-1;i>=0;i--)
if(pos&(1LL<<i))printf("1");
else printf("0");
printf("\n");
}
return 0;
}
bool isok(LL x, int n)
{
int pos = lower_bound(a, a+n, x)-a;
if(pos == n || a[pos] != x)return false;
else return true;
}