HDU 6034 Balala Power! —— Multi-University Training 1

　　Talented Mr.Tang has nn 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 2626 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+7109+7.

Input

　　The input contains multiple test cases. 

Output

　　For each test case, the first line contains one positive integers nn, the number of strings. (1≤n≤100000)

　　Each of the next nn lines contains a string sisi consisting of only lower case letters. (1≤|si|≤100000,∑|si|≤1e6)e
OutputFor each test case, output " Case #xx: yy" in one line (without quotes), where xxindicates the case number starting from 11 and yy 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：用26个字母表示26进制数中的1~26，并且给你n个字符串，以26进制从字符串转化为数字，问这些数字之和最大能有多少（以十进制输出）。为保证数据正确，其中必存在至少一个字母是代表了0。

思路：模拟。首先每个字母代表的数未知，但由于每个字符在字符串的位置是已知的，我们就可以求出每个字母从26进制转化十进制时的“系数”（即像26^a1+26^a2+26^a3……的数），用数组存储26进制数，在求和时考虑进位，去掉必为0的字母，然后对按26进制数的大小对数组排序，最后就能够求和解出答案了。。。比赛的时候是用的字符串存26进制，理论上也是可以做的，但是因为进位的姿势不对（一边加一边进位，先全部加完再进位才是正确的）导致TLE/(ㄒoㄒ)/

 

代码

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<iomanip> 
 6 #include<cmath>
 7 #define MOD 1000000007  
 8 #define MAXN 100010  
 9 using namespace std;
10 long long num[30][MAXN],w[26];
11 long long sum[30];
12 long long a[MAXN];
13 int u;    
14 char ss[MAXN];
15 bool vis[1010];
16 bool cmp(int s, int t)
17 {
18     for(int i=u-1;i>=0;i--){
19         if(num[s][i]!=num[t][i])
20             return num[s][i]<num[t][i];
21     }
22     return 0;
23 }
24 int main()
25 {
26     a[0]=1;
27     for(int i=1;i<MAXN-1;i++)
28         a[i]=26*a[i-1]%MOD;
29         
30     int n;
31     int z=0;
32     while(~scanf("%d", &n))
33     {       
34         u=0;
35         memset(num, 0, sizeof(num));
36         memset(sum, 0, sizeof(sum)); 
37         memset(vis,false,sizeof(vis));
38         int wei,p;
39         for(int i=0;i<n;i++){
40             scanf("%s", ss);
41             int l=strlen(ss);
42             if(l>1)
43                 vis[ss[0]-'a']=true;
44             
45             for(int j=0;j<l;j++){
46                 p=ss[j]-'a';
47                 wei=l-1-j;
48                 num[p][l-1-j]++;
49                 sum[p]+=a[wei];
50                 sum[p]%=MOD;
51             }
52             u=max(u, l);
53         }
54         for(int i=0;i<26;i++){
55             for(int j=0;j<u;j++){ 
56                 num[i][j+1]=num[i][j+1]+num[i][j]/26;
57                 num[i][j]=num[i][j]%26;    
58             }
59             while(num[i][u])
60             {
61                 num[i][u+1]+=num[i][u]/26;
62                 num[i][u++]%=26;
63                 u++;
64             }
65             w[i]=i; 
66         }
67         int cnt=-1;
68         sort(w, w+26, cmp);
69         for(int i=0;i<26;i++){
70             if(!vis[w[i]]){
71                    cnt=w[i];
72                    break;
73             }
74         }
75         //cout<<cnt<<endl;
76         int res=0,x=25;
77         for(int i=25;i>=0;i--){
78             if(cnt!=w[i]){
79                 res+=(long long)(x--)*sum[w[i]]%MOD;
80                 res%=MOD;
81             }
82                 
83         }
84         z=z+1;
85         printf("Case #%d: %lld\n",z,res);
86     }
87 } 

 

转载于:https://www.cnblogs.com/MasterSpark/p/7265712.html