hdu 6034 Balala Power!（DP）

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Balala Power!

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 998 Accepted Submission(s): 165

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 ato 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 #

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

1a2aabb3abaabc

Sample Output

Case #1: 25Case #2: 1323Case #3: 18221
#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<vector>#include<map>//#include <bits/stdc++.h>using namespace std;const int N = 1e6+10;;typedef long long LL;const LL mod = 1e9+7;map<int,LL>q[30];char str[N];int vis[30], power[30];double dp[100000+7][28];vector<string>p;int a[N];struct node{    int id;    LL cnt;    double x;    bool operator> (const node &A)    {        return x>A.x;    }}p1[30];int cmp(node A,node B){    return A.x>B.x;}int main(){    int ncase=1, n;    while(scanf("%d", &n)!=EOF)    {        memset(vis,0,sizeof(vis));        memset(dp,0,sizeof(dp));        memset(power,0,sizeof(power));        p.clear();        int l=0;        for(int t=0;t<n;t++)        {            scanf("%s",str);            vis[str[0]-'a']=1;            int len=strlen(str);            a[t]=len;            l=max(l,len);            //reverse(str,str+len);            //cout<<str<<endl;            p.push_back(str);            for(int i=0;i<len;i++)  dp[len-i-1][str[i]-'a']++;        }        for(int i=1;i<=l;i++)            for(int j=0;j<26;j++) dp[i][j]+=(dp[i-1][j]/26.0);        for(int i=0;i<26;i++)        {            p1[i].id=i,p1[i].x=dp[l][i];        }        sort(p1,p1+26,cmp);        for(int i=0;i<26;i++)  power[p1[i].id]=25-i;        int k=25;        while(vis[p1[k].id])        {            swap(power[p1[k].id],power[p1[k-1].id]);            k--;        }        LL sum=0;        int m=p.size();        for(int i=0;i<m;i++)        {            LL ans=0;            for(int j=0;j<a[i];j++)            {                ans=(ans*26+power[p[i][j]-'a'])%mod;            }            sum=(sum+ans)%mod;        }        printf("Case #%d: %lld\n",ncase++,sum);    }    return 0;}