H. Happy Reversal
Time Limit: 1000ms
Memory Limit: 65536KB
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
题目意思是有多个01串，可以选择反转或者不反转，求出最大值与最小值的最大差
看完题目后，直接暴力，提交后毫无悬念的超时了，这个题的关键是怎么找到最大值与最小值，因为
一个串你只能选择一个值，但又不能保证选择一个值得情况下，就是最好的结果，想了好久也没
想明白，看了别人的代码后，才明白，最大值或最小值是由二进制得来的，应该用异或来解决
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,m,h;
char str[100];
long long a[20010];
long long pow(int k){
return 1LL<<k;
}
long long solve(){
if(n==1)
return 0;
if(a[0]^a[h-1])
return max(a[h-2]-a[0],a[h-1]-a[1]);
else
return a[h-1]-a[0];
}
int main(){
long long Max,p,q;
int i,j,k,t,l=1;
scanf("%d",&t);
while(t--){
h=0;
scanf("%d%d",&n,&m);
for(k=0;k<n;k++){
scanf("%s",str);
p=0;
q=0;
for(i=m-1,j=0;i>=0;i--,j++){
if(str[i]=='0') p+=pow(j);
else q+=pow(j);
}
a[h++]=p;
a[h++]=q;
}
sort(a,a+h);
printf("Case #%d: %lld\n",l++,solve());
}
return 0;
}