12545-bits equalizer (greedy?) )

This problem I am my own idea, once a, 0.000ms. Look at the online most of the solution is similar, I would like to talk about my ideas.

Assume that the string a, b

All we have to do is sweep it backwards. Jump over, encounter a[i] = ' 0 ' &&a[i]!=b[i]; then look back for a a[j]== ' 1 ' &&a[i]!=b[j], that is to say, the first exchange of this skill to compare the provincial steps. If there's nothing to swap, then sweep again? What if a[i]==? && B[i]==0 So obviously we can start with? becomes 1 again and A[i] exchange, need two steps, if still not satisfied, that can only directly put A[i] into 1

A[i]==1 && A[i]!=b[i] The situation is similar to the above, but if the first two can not be satisfied, then this sequence can not become a B sequence

So it's good to start thinking clearly and then knock the code down, sorting it out.

The code is as follows:

#include <bits/stdc++.h>using namespace Std;int T,MAXN = 0;char a[105],b[105];int main () {scanf ("%d", &t);        while (t--) {scanf ("%s%s", A, b);        printf ("Case%d:", ++MAXN);        int n = strlen (a);        bool OK = true;        int cnt = 0;            for (int i=0;i<n;i++) {if (a[i]== '? ') continue;                else if (a[i]!=b[i]&&a[i]== ' 0 ') {bool flage = false; for (int j=i+1;j<n;j++) {if (a[j]!=b[j]&&a[j]== ' 1 ') {flage = true; a[j] = ' 0 '; a[i] = ' 1 '; CN t++; Break  }} if (!flage) {for (int j=0;j<n;j++) {if (A[j] == '?'  &&b[j]== ' 0 ') {flage = true; a[j] = ' 0 '; a[i] = ' 1 '; cnt+=2; Break                }} if (!flage) {a[i] = ' 1 '; cnt++;}                }} else if (a[i]!=b[i]&&a[i]== ' 1 ') {bool flage = false; for (int j=i+1;j<n; j + +) {if (a[j]!=b[j]&&a[j]== ' 0 ') {flage = true; A[j] = ' 1 '; a[i] = ' 0 '; cnt++; break;} } if (!flage) {for (int j=0;j<n;j++) {if (a[j]== '? ')                    &&b[j]== ' 1 ') {flage = true; A[j] = ' 1 '; a[i] = ' 0 '; cnt+=2; break;}            if (!flage) {ok = false; break;}//Cannot Be transformed}}} if (OK) { for (int i=0;i<n;i++) if (a[i] = = '? ') cnt++;        Don't forget here printf ("%d\n", CNT);    } else printf (" -1\n"); } return 0;}

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12545-bits equalizer (greedy?) )