Balala Power!(大数+思维)
Balala Power!
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 714 Accepted Submission(s): 117

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 froma 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.
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)
- #include <iostream>
- #include <string.h>
- #include <stdio.h>
- #include <algorithm>
- using namespace std;
- #define LL long long
- #define MOD 1000000007
- #define MX 100100
- int n;
- int al_n[];
- LL num[][MX];
- LL quan[];
- char temp[MX];
- bool ok[];
- bool cmp(int a,int b)
- {
- if (al_n[a]!=al_n[b])
- return al_n[a]>al_n[b];
- for (int i=al_n[a];i>=;i--)
- if (num[a][i]!=num[b][i])
- return num[a][i]>num[b][i];
- return ;
- }
- int main()
- {
- int cnt=;
- while (scanf("%d",&n)!=EOF)
- {
- memset(quan,,sizeof(quan));
- memset(num,,sizeof(num));
- memset(ok,,sizeof(ok));
- for (int i=;i<n;i++)
- {
- scanf("%s",temp);
- int len = strlen(temp);
- LL k=;
- for (int j=len-;j>=;j--)
- {
- num[temp[j]-'a'][k]++;
- k++;
- }
- if (len!=)
- ok[temp[]-'a']=;
- }
- for (int i=;i<;i++)//字母
- {
- al_n[i]=;
- for (int j=;j<MX;j++) //长度
- {
- if (num[i][j]) al_n[i]=j;
- if (num[i][j]>=)
- {
- int jin = num[i][j]/;
- num[i][j+]+=jin;
- num[i][j]%=;
- }
- }
- }
- int alpa[];
- for (int i=;i<;i++) alpa[i]=i;
- sort(alpa,alpa+,cmp);
- if (al_n[alpa[]]!=&&ok[alpa[]]==)
- {
- for (int i=;i>=;i--)
- {
- if (ok[alpa[i]]==)
- {
- int sbsb=alpa[i];
- for (int j=i+;j<=;j++)
- alpa[j-]=alpa[j];
- alpa[]=sbsb;
- break;
- }
- }
- }
- LL ans = ;
- LL qqq = ;
- for (int i=;i<;i++)
- {
- int zimu = alpa[i];
- if (al_n[zimu]==) continue;
- LL tp=qqq;
- for (int j=;j<=al_n[zimu];j++)
- {
- ans = (ans + (tp * num[zimu][j])%MOD)%MOD;
- tp=(tp*)%MOD;
- }
- qqq--;
- }
- printf("Case #%d: %lld\n",cnt++,ans);
- }
- return ;
- }
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