hdu 6034 Balala Power!

Balala Power!

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3555    Accepted Submission(s): 844

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

 

Source

2017 Multi-University Training Contest - Team 1

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#include <iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<bits/stdc++.h>

using namespace std;
const long long mod=1e9+7;
struct node
{
    int id;
    double k;
}a[30];
int b[30];
long long num[30][100005];
bool vis[30];
char ch[100005];
int n,maxlen;

bool cmp(node a,node b)
{
    return a.k>b.k;
}
int main()
{
    int cas=0;
    while(~scanf("%d",&n))
    {
         maxlen=0;
         memset(num,0,sizeof(num));
         memset(vis,0,sizeof(vis));
        for(;n>0;n--)
        {
            scanf("%s",&ch);
            int l=strlen(ch);
            maxlen=max(maxlen,l);
            if (l>1) vis[ch[0]-'a']=1;
            for(int i=0;i<l;i++)
                num[ch[i]-'a'][l-i-1]++;
        }

        for(int i=0;i<26;i++)
        {
            a[i].id=i; a[i].k=0;
            for(int j=0;j<maxlen;j++)
                a[i].k=a[i].k/26+num[i][j];
        }

        sort(a,a+26,cmp);

        for(int i=25;i>=0;i--)
        {
            if (vis[a[i].id]) continue;
              else
              {
                  int cnt=25;
                  for(int j=0;j<26;j++)
                    if (j!=i) b[a[j].id]=cnt--;
                        else b[a[j].id]=0;
                  break;
              }
        }

        long long ans=0;
        for(int i=0;i<26;i++)
        {
            long long ret=1;
           for(int j=0;j<maxlen;j++)
           {
               ans=(ans+num[i][j]*ret*b[i])%mod;
               ret=(ret*26)%mod;
           }
        }
        printf("Case #%d: %lld\n",++cas,ans);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/stepping/p/7240175.html