hdu 6034 Balala Power! （2017 Multi-University Training Contest - Team 1）

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

 

/*
题意：给出只含有小写字母的字符串，分别与0-25相匹配，然后把字符串看成是26进制，求所有字符串转化后的总和
解析：记录字母出现在每个字符串的位置以及出现的总次数，按照其权值从大到小分配25，24，23，...
官方题解:每个字符对答案的贡献都可以看作一个26进制的数字，问题相当于要给这些贡献加一个0到25的权重使得答案最大。
最大的数匹配25，次大的数匹配24依次类推。
排序后这样依次贪心即可，唯一注意的是不能出现前导 0。
*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <map>
#include <cmath>
#include <cstring>
typedef long long LL;
const LL N = 100000 + 100;
const LL mod = 1000000007;
using namespace std;
struct ss {
    char sc;
    //str用于记录某个字符在个十百千...位置上分别出现的次数
    int str[N];
    int id;
    bool operator < (const ss &a)const
    {
        for(int i = 0; i < N; i++)
            if(str[i] != a.str[i])
                return str[i] > a.str[i];
        return str[N - 1] > a.str[N - 1];
    }
} num[30];
LL k[N] = {1};
void init()
{
    for(LL i = 1; i < N; i++)
        k[i] = (k[i - 1] * 26) % mod;
}
int main()
{
    int n, tim = 1;
    init();
    while(~scanf("%d%*c", &n)) {
        char s[N];
        int ma = 0;
        int tt[30];
        for(int i = 0; i < 26; i++) {
            tt[i] = 0;
            memset(num[i].str, 0, sizeof(num[i].str));
        }
        for(int i = 0; i < n; i++) {
            scanf("%s", s);
            int len = strlen(s);
            //记录不能为前导0的字母，若是只有一个字母，不算是前导0
            if(len > 1)
                tt[s[0] - 'a']++;
            for(int j = 0; j < len; j++) {
                num[s[j] - 'a'].str[len - 1 - j]++;
            }
        }
        int flag = 0;
        for(int i = 0; i < 26; i++) {
            num[i].sc = i + 'a';
            for(int j = 0; j < N; j++) {
                //26进制，所以满26进一
                num[i].str[j + 1] += num[i].str[j] / 26;
                num[i].str[j] = num[i].str[j] % 26 ;
            }
            int c;
            //交换前后位置，权值按照从大到小的顺序
            for(int j = 0; j < N / 2; j++) {
                c = num[i].str[j];
                num[i].str[j] = num[i].str[N - j - 1];
                num[i].str[N - 1 - j] = c;
            }
        }
        sort(num, num + 26);
        int i;
        //排除前导0
        for(i = 25; i >= 0; i--) {
            if(tt[num[i].sc - 'a'] == 0) {
                break;
            }
        }
        ss t;
        for(int j = 25; j >= 0 && j > i; j--) {
            t = num[i];
            num[i] = num[j];
            num[j] = t;
        }
        LL sum = 0;
        for(int i = 0; i < 26; i++) {
            for(int j = 0; j < N; j++) {
                if(num[i].str[j]) {
                    sum = (sum + ((25 - i) * num[i].str[j] % mod) * k[N - 1 - j]) % mod;
                }
            }
        }
        printf("Case #%d: %lld\n", tim++, sum);
    }
    return 0;
}