2014 ACM-ICPC Beijing Invitational Programming Contest

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Happy Reversal

Time Limit: 1000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

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Elfness is studying in an operation "NOT".

For a binary number A, if we do operation "NOT A", after that, all digits of A will be reversed. (e.g. A=1001101, after operation "NOT A", A will be 0110010).

Now Elfness has N binary numbers of length K, now he can do operations "NOT" for some of his numbers. 

Let's assume after his operations, the maximum number is M, the minimum number is P. He wants to know what's the maximum M - P he can get. Can you help him?

 

Input

The first line of input is an integer T (T ≤ 60), indicating the number of cases.

For each case, the first line contains 2 integers N (1 ≤ N ≤ 10000) and K (1 ≤ K ≤ 60), the next N lines contains N binary numbers, one number per line, indicating the numbers that Elfness has. The length of each binary number is K.

 

Output

For each case, first output the case number as "Case #x: ", and x is the case number. Then you should output an integer, indicating the maximum result that Elfness can get.

Sample Input

2
5 6
100100
001100
010001
010001
111111
5 7
0001101
0001011
0010011
0111000
1001011

Sample Output

Case #1: 51
Case #2: 103

Source

2014 ACM-ICPC Beijing Invitational Programming Contest

给你n组由k个0或者1组成的二进制数。每一个数能够翻转。求两个数的最大差值。

//152 ms 1788 KB
#include<stdio.h>
#include<algorithm>
using namespace std;
char s[107];
long long ans[20007];
int n,m;
long long getnum()
{
    long long res=0,j=1;
    for(int i=m-1;i>=0;i--,j*=2)
        if(s[i]=='1')res+=j;
    return res;
}
int main()
{
    int t,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        int k=0;
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%s",s);
            ans[k++]=getnum();
            for(int j=0;j<m;j++)
                if(s[j]=='1')s[j]='0';
                else s[j]='1';
            ans[k++]=getnum();
        }
        sort(ans,ans+k);
        long long ans1=ans[k-1]-ans[1];
        long long ans2=ans[k-2]-ans[0];
        long long p=ans[0],q=ans[k-1];
        long long e=1<<(m-1);
        if((p^q)&e==e)
            printf("Case #%d: %lld
",cas++,max(ans1,ans2));
        else printf("Case #%d: %lld
",q-p);
    }
    return 0;
}