Multi-university Training contest-team 1 1002&&hdu 6034 balala power! "string, greedy + sort"

Balala power!

Time limit:4000/2000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 2668 Accepted Submission (s): 562

Problem Description

Sample Input

1a2aabb3abaabc

Sample Output

Case #1:25Case #2:1323Case #3:18221

Source multi-university Training contest-team 1 topics Link: http://acm.hdu.edu.cn/showproblem.php?pid=6034

Test instructions: give you n a string of lowercase letters, let you assign 26 letters 0-25, each string form a 26 binary number, ask how to divide

The weighted value of this n number and maximum. (cannot have a leading 0, but a single 0 can)

Solution: Each character's contribution to the answer can be viewed as a 26-digit number, and the question is equivalent to adding a 0 to 25 weight to these contributions.

Make the answer the most. The maximum number matches 25, the second largest number matches 24, and so on. After ordering this greedy in turn, the only thing to note is not to appear

Leading 0. The key is the sort, in fact the structure of the order can be sorted by the array dictionary order!

The AC code is given below:

1#include <bits/stdc++.h>2 using namespacestd;3typedefLong Longll;4 Constll mod=1e9+7;5 Constll n=1e5+Ten;6ll fac[n]={1};7ll temp[ -];8ll hash[ -];9 BOOLleap[ -];Ten CharStr[n]; One structNode A { - ll Cnt[n]; - ll ID; the     BOOL operator< (ConstNode &p)Const -     { -          for(LL i=n-1; i>=0; i--) -         { +             if(cnt[i]>P.cnt[i]) -                 return true; +             Else if(cnt[i]<P.cnt[i]) A                 return false; at             Else; -         } -     } -}p[ -]; - intMain () - { inll n,tcase=1; -      for(LL i=1; i<n;i++) tofac[i]=fac[i-1]* -%MoD; +      while(SCANF ("%lld", &n)! =EOF) -     { theMemset (P,0,sizeof(P)); *memset (hash,-1,sizeof(Hash)); $memset (Leap,false,sizeof(Leap));Panax Notoginsengll r=0; -          for(LL i=1; i<=n;i++) the         { +scanf"%s", str); All len=strlen (str); theR=max (r,len-1); +             if(len!=1) -leap[str[0]-'a']=1; $              for(LL i=0; i<len;i++) $p[str[i]-'a'].cnt[len-(i+1)]++; -         } -          for(LL i=0;i< -; i++) the         { -              for(LL j=0; j<n;j++)Wuyi             { the                 if(p[i].cnt[j]>= -) -                 { Wup[i].cnt[j+1]+=p[i].cnt[j]/ -; -p[i].cnt[j]%= -; About                 } $             } -P[i].id=i; -         } -Sort (p,p+ -); A          for(LL i=0;i< -; i++) +hash[p[i].id]= --(i+1); the          for(LL i=0;i< -; i++) -         { $             if(leap[p[i].id]&&hash[p[i].id]==0) the             { the                  for(LL j= -; j>=0; j--) the                 { the                     if(!Leap[p[j].id]) -                     { in                          for(LL k= -; k>=j+1; k--) the                         { thehash[p[k].id]=hash[p[k-1].id]; About                         } thehash[p[j].id]=0; the                          Break; the                     } +                 } -                  Break; the             }Bayi         } thell ans=0; the          for(LL i=0;i< -; i++) -         { -              for(LL j=0; j<n;j++) the             { theAns= (ans+fac[j]*p[i].cnt[j]*hash[p[i].id]%MoD); the             } the         } -printf"Case #%lld:%lld\n", tcase++,ans%MoD); the     } the     return 0; the}

Multi-university Training contest-team 1 1002&&hdu 6034 balala power! "string, greedy + sort"