Codeforces 550E Brackets in Implications (construction)

" topic link ": Click here~~

" The main idea" given a logical operation symbol A->B: Currently only if A is 1b for 0 value is 0, the remainder is 1, construct parentheses, change the priority of the operation to make the final result is 0

" problem-solving ideas ":

todo~~

/*
Ideas:
1. If the last digit is 1, the result will not be 0.puts ("no");
2. In the case of a solution, the last one must be 0.
2.1. Further discovery, in fact, the second-to-last must be 1, only 1 of the preceding results and the bit 1 combined to equal 1, further 1->0=0;
2.2. If the 1 front is 0, then merge the two digits, make up 1, recursion 2.1
*/

Code:

#include <bits/stdc++.h>using namespace Std;const int n=1e5+10;typedef long long ll;typedef unsigned long long  Llu;int num[n];int n,m,l,r,ans,cnt,top;int zero,one;int Main () {    while (cin>>n)    {        zero=one=0;        for (int i=0; i<n; ++i)        {            scanf ("%d", &num[i]);            if (num[i]==0) ++zero;            else ++one;        }        int a3=num[n-1];        int a2=num[n-2];        int a1=num[n-3];        if (a3==0&& (a2==1| | zero!=2)//Ensure that the last one is 0, and the penultimate digit is 1        {            puts ("yes");            printf ("(");            for (int i=0; i<=n-3; ++i)            {                printf ("(%d->", Num[i]);            }            printf ("%d", a2);            for (int i=0; i<=n-3; ++i)                printf (")");            if (n-1) printf ("->0");            printf (")");            Puts ("");        }        Else puts ("NO");    }    return 0;}

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Codeforces 550E Brackets in Implications (construction)