ACM-ICPC 2014 Beijing Invitational H Happy Reverse [simulation]

Test instructions: Given n binary strings, you can invert some of these 0 and 1 (i.e. 0 to 0) to find the difference between the maximum and minimum values in n strings after conversion.

Analysis: The idea is to put all the strings and invert the existence of an array, and then sort, find the difference between the maximum and minimum values, (if the same string is reversed, find the second largest and smallest or second small and largest in the maximum value). Note that if there is only one string, the result is 0.

DEBUG:

It's been inscribed for a long time

1. The first time with vim, very cool, but not yet skilled

2. Ignoring the scope of the problem, it is obvious to use Longlong

3. After using the Longlong also WA, with a script run out of data found in Longlong, Min value to become (1<<63)-1 but can not write, should write a longlong is 1<< 30, and then shift on this longlong basis.

Finally, I have to say that Bnuoj really looks a lot better than POJ.

    #include <iostream> #include <cmath> #include <cstring> #include <cstdio> using NA    Mespace std;    Char str[111111][222];    Long Long tmp=1<<30;    Const Long Long max_long= (tmp<<33)-1;    Long Long num[111111];    Long Long m,n; void Getrev (char* to,char* source,long long len) {for (Long long i=0;i<len;i++) to[i]=source[i]== ' 0 '? '        1 ': ' 0 ';    to[len]= ' + ';        A long long getnum (char* s,long long len) {long long sum=0;            For (long long i=0;i<len;i++) {sum*=2;        sum+=s[i]-' 0 ';    } return sum;        } int main () {Long long test;        cin>>test;            For (long long time=1;time<=test;time++) {cin>>m>>n;            For (long long i=1;i<=m;i++) cin>>str[i];                /*if (m==1) {printf ("Case #%d:0\n", time);            Continue   }*/         For (long long i=m+1;i<=2*m;i++) Getrev (str[i],str[i-m],n);            Long long Minn=max_long;long long maxn=0;            For (long long i=1;i<=2*m;i++) num[i]=getnum (str[i],n);            Long Long Max1=0,max1index,min1=max_long,min1index;            Long Long Max2=0,max2index,min2=max_long,min2index; For (long long i=1;i<=2*m;i++) {if (num[i]>max1) {MA                    X1=num[i];                Max1index=i;                    } if (num[i]<min1) {min1=num[i];                Min1index=i; }} for (Long long i=1;i<=2*m;i++) {if (I!=max1index&&num[i]                    &GT;MAX2) {Max2=num[i];                Max2index=i; } if (i!=min1index&&num[i]<min2) {min2=nUm[i];                Min2index=i;                }} if (ABS (Min1index-max1index)!=m) {maxn=max1;            Minn=min1; } else {if (max1-min2>max2-min1) {//cout<&lt ;"                    1 "<<endl;                maxn=max1;minn=min2;                    } else {//cout<< "2" <<endl;                Maxn=max2;minn=min1;            }}//cout<<maxn<< ' <<minn<<endl;        printf ("Case #%lld:%lld\n", Time,maxn-minn);    } return 0; }