2017杭电多校联赛-Balala Power!_杭电高效联赛-优快云博客

本文链接：https://blog.youkuaiyun.com/wang_heng199/article/details/76115761

Balala Power!

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4809 Accepted Submission(s): 387

Problem Description

Talented Mr.Tang has

n strings consisting of only lower case characters. He wants to charge them with Balala Power (he could change each character ranged from a to z into each number ranged from 0 to 25, but each two different characters should not be changed into the same number) so that he could calculate the sum of these strings as integers in base

26 hilariously.

Mr.Tang wants you to maximize the summation. Notice that no string in this problem could have leading zeros except for string "0". It is guaranteed that at least one character does not appear at the beginning of any string.

The summation may be quite large, so you should output it in modulo

109+7 .

Input

The input contains multiple test cases.

For each test case, the first line contains one positive integers

n , the number of strings.

(1≤n≤100000)

Each of the next

n lines contains a string

si consisting of only lower case letters.

(1≤|si|≤100000,∑|si|≤106)

Output

For each test case, output " Case #

x :

y " in one line (without quotes), where

x indicates the case number starting from

1 and

y denotes the answer of corresponding case.

Sample Input

  
  
   
   1
a
2
aa
bb
3
a
ba
abc

Sample Output

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

  
  
  
  
   
   题目大意：给你一些字符串，这些字符串中的每一个字符都可以表示为0-25中的任何一个数字，但是不能有两个字符表示表示同一个数字，然后
  
  
  
  
   
   要求得到最大的可能值。
  
  
  
  

  
  
  
  
   
   解题思路：题目给你的字母，我们想要得到最大的和值，则需要将字母表示为尽可能大的数，在这里我们看到的这些字符串，实际上是26进制数，
  
  
  
  
   
   所以我们可以先将这些26进制数转换为10进制，然后相加求和。具体看代码。
  
  
  
  

  
  
  
  
   
   ac代码：
  
  
  
  
   
   #include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long 
const int N=105005;
const int mod=1000000007;
ll p[N];
int f[26],q[26],cnt[26][N]; 
char s[N];
int mysort(){
    int i,j,k;
    for(i=0;i<26;i++){
        for(j=i+1;j<26;j++){
            for(k=N-1;k>=0;k--){
                if(cnt[q[i]][k]<cnt[q[j]][k]){
                    int t=q[i];
                    q[i]=q[j];
                    q[j]=t;
                    break;
                }
                if(cnt[q[i]][k]>cnt[q[j]][k])
                    break;
            } 
        } 
    }
    if(f[q[25]]==1)
    {
        for(i=24;i>=0;i--){
            if(!f[q[i]]){
                for(j=i;j<25;j++)
                    q[j]=q[j+1];
                break;
            }
        }
    }
    
}
void init(){
    p[0]=1;
    for(int i=1;i<N;i++)
        p[i]=p[i-1]*26%mod;
}
int main(){
    ll ans;
    int i,j,n,m,x,t=1;
    init(); 
    while(~scanf("%d",&n)){ 
        memset(cnt,0,sizeof(cnt));
        memset(f,0,sizeof(f));
        for(i=0;i<26;i++)
            q[i]=i;
        for(i=0;i<n;i++){
            scanf("%s",s);
            m=strlen(s);
            for(j=0;j<m;j++){
                cnt[s[j]-'a'][m-j-1]++;
            }
            f[s[0]-'a']=1;
        }
        for(i=0;i<26;i++){
            for(j=0;j<N-1;j++){
                cnt[i][j+1]+=cnt[i][j]/26;
                cnt[i][j]%=26;    
            }
        }
        mysort();
        for(ans=i=0;i<25;i++){
            for(j=N-1;j>=0;j--)
            {
                if(cnt[q[i]][j]){
                    ans+=p[j]*cnt[q[i]][j]*(25-i),ans%=mod; 
                }
            }
        }
        printf("Case #%d: %lld\n",t++,ans);
    } 
    return 0;
}
   
   
    
    

   
   题目链接：
   
   点击打开链接http://acm.hdu.edu.cn/contests/contest_showproblem.php?cid=759&pid=1002