Hdoj 5375 Gray Code

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5375

Encoding Rules: TMP = XOR (gr[i],gr[i-1]);

Look for a regular problem, think about it? Before and after characters and? The number of parity can be, a small trick is when the need to reduce the time to reduce the question mark in the interval of the smallest,

Then there is the problem of debugging:

The string is stored from the second bit of the array, scanf ("%s", s+1);

When using Min and Max, remember to empty before each use.

1#include <stdio.h>2 Const intMAXN =200010;3 intXOR (CharACharb) {4     intAA, BB;5AA = A-'0';6BB = B-'0';7     returnaa^BB;8 }9 intMain () {Ten     CharGR[MAXN]; One     intNUM[MAXN]; A     int_min;intT, N; -     inttmpintcnt, sum; -     intLFT, RGT;intTT =0; thescanf"%d",&T); -      while(t--){ -scanf"%s", gr+1); -gr[0] ='0'; +CNT =0; sum =0; N =0; _min =1000000; -          while(Gr[n]! =' /') +n++; A          for(inti =1; i < N; ++i) { atscanf"%d",&num[i]); -         } -gr[n]='0'; -Num[n] =0; -n++; -          for(inti =1; i < N; ++i) { in             if(Gr[i]! ='?'){ -TMP = XOR (gr[i],gr[i-1]); to                 if(TMP) { +Sum + =Num[i]; -                 } the             } *             Else{ $CNT =0;Panax NotoginsengLFT = gr[i-1]; -                  for(; i<n;++i) { the                     if(Gr[i]! ='?'){ +                         if(Num[i] <_min) A_min =Num[i]; theSum + =Num[i]; +                          Break; -                     } $                     Else{ $cnt++; -Sum + =Num[i]; -                         if(Num[i] <_min) the_min =Num[i]; -                     }Wuyi                 } theRGT =Gr[i]; -                 if((LfT! = RGT) && (cnt%2==1))|| (LfT = = RGT) && (cnt%2==0) )){ Wusum = Sum-_min; -                 } About_min =1000000; $             } -         } -printf"Case #%d:%d\n",++tt,sum); -     } A}

Hdoj 5375 Gray Code