The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online
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Live Love
Time Limit: 1 Second Memory Limit: 65536 KB
DreamGrid is playing the music game Live Love. He has just finished a song consisting of notes and got a result sequence ( {PERFECT, NON-PERFECT}). The score of the song is equal to the \textit{max-combo} of the result sequence, which is defined as the maximum number of continuous PERFECTs in the sequence.
Formally speaking, { | is an integer and there exists an integer () such that PERFECT}. For completeness, we define max() = 0.
As DreamGrid is forgetful, he forgets the result sequence immediately after finishing the song. All he knows is the sequence length and the total number of PERFECTs in the sequence, indicated by . Any possible score he may get must satisfy that there exists a sequence of length containing exactly PERFECTs and NON-PERFECTs and . Now he needs your help to find the maximum and minimum among all possible scores.
Input
There are multiple test cases. The first line of the input contains an integer (), indicating the number of test cases. For each test case:
The only line contains two integers and (, , ), indicating the sequence length and the number of PERFECTs DreamGrid gets.
Output
For each test case output one line containing two integers and , indicating the maximum and minimum possible score.
Sample Input
5
5 4
100 50
252 52
3 0
10 10
Sample Output
4 2
50 1
52 1
0 0
10 10
Hint
Let's indicate a PERFECT as and a NON-PERFECT as .
For the first sample test case, the sequence leads to the maximum score and the sequence leads to the minimum score.
思维
#include
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m;
scanf("%d%d",&n,&m);
int maxx=m,minn;
if(n==m)minn=maxx;
else if(m==0)minn=0;
else if(m-1<=n-m)minn=1;
else minn=n/(n-m+1);
printf("%d %d\n",maxx,minn);
}
return 0;
}
XOR Clique
Time Limit: 1 Second Memory Limit: 65536 KB
BaoBao has a sequence . He would like to find a subset of such that , and is maximum, where means bitwise exclusive or.
Input
There are multiple test cases. The first line of input contains an integer , indicating the number of test cases. For each test case:
The first line contains an integer (), indicating the length of the sequence.
The second line contains n integers: (), indicating the sequence.
It is guaranteed that the sum of in all cases does not exceed .
Output
For each test case, output an integer denoting the maximum size of .
Sample Input
3
3
1 2 3
3
1 1 1
5
1 2323 534 534 5
Sample Output
2
3
2/*
题目大意：给定一个数组，在数组中找出最大的子序列，
子序列需要满足的条件是任意两个元素进行异或运算的结果小于两个数中的最小数
解题思路：多找几组数据就会发现若想异或结果小于最小值，只需要两个数二进制形式位数相同，
即二进制前面第一位都为1，进行异或运算后就会第一位变为0，自然比原来两个数都小
所以只需要将序列中所有数的二进制形式位数进行统计即可
看位数即可，把每一个数字都化为二进制，然后把位数相同的就加1，因为只有位数相同，并且两个数字的最高位都1的时候，以后之后的数字才有可能比这两个数字都小，所以只要判断位数相同的有多少个就可以了，这样算出来的数据就是要找的最大值。
*/
#include
#include
#include
#include
using namespace std;
int main()
{
int t, n, x;
scanf("%d", &t);
while(t--)
{
int a[30] = {0};
scanf("%d", &n);
while(n--)
{
scanf("%d", &x);
int y = 0;//用来记录该数转换为二进制后有多少位
while(x > 1)
{
y++;
x /= 2;
}
a[y]++;
}
int maxn = *(max_element(a, a+30));
printf("%d\n", maxn);
}
return 0;
}