HDU 6034 Balala Power!（贪心）

Balala Power!

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

1a2aabb3abaabc

 

Sample Output

Case #1: 25Case #2: 1323Case #3: 18221

 

Source

2017 Multi-University Training Contest - Team 1

题解：将每个字母的个数用26进制统计，从大到小排序（字符串排序），贪心分配数值

如果出现前导0的情况，找出最小的可以赋为0的字母，将后面的字母往前递推

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<ctype.h>#include<iostream>#include<algorithm>#include<string>#include<sstream>#include<queue>#include<stack>#include<math.h>#define N 100005#define pi 3.1415926#define FLAG 9999999#define mod 1000000007#define min(a,b) (a<b ? a:b)using namespace std;typedef long long ll;char str[N];ll map[30][N];ll res[30];ll l[N];ll mm;struct ww{    ll data;    ll cur;    char a[N];}tmp[26];bool cmp(ww x,ww y){    if(strcmp(y.a,x.a)>0)        return 1;    else return 0;}int main(){    ll i,j,k;    ll n;    ll y=0;    while(cin>>n)    {        y++;        k = 0;        memset(map,0,sizeof(map));        memset(l,0,sizeof(l));        mm=0;        for(i = 0; i < n; i++)//统计字母个数        {            cin>>str;            ll len = strlen(str);            if(len>1) l[str[0]-'a']=1;;            k = max(k,len);            ll w=0;            for(j = len-1; j >= 0; j--)            {                map[str[j]-'a'][w] += 1;                w++;            }        }        for(i=0;i<100005;i++)//进位        {            int flag=0;            for(j=0;j<26;j++)            {                if(map[j][i]!=0) flag=1;                if(map[j][i]>=26)                {                    map[j][i+1]+=map[j][i]/26;                    map[j][i]%=26;                }            }            if(flag==1)                mm=max(mm,i);//记录最高位        }        memset(res,0,sizeof(res));        for(int j=0;j<26;j++)//转换为字符串        {            memset(tmp[j].a,0,sizeof(tmp[j].a));            tmp[j].cur=j;            ll w=0;            for(ll i=mm;i>=0;i--)               {                   tmp[j].a[w]=map[j][i]+'0';                   w++;               }        }        sort(tmp,tmp+26,cmp);//排序        ll x=0;        for(j = 0; j < 26; j++)//从小到大分配数值        {            res[tmp[j].cur] = x;            x++;        }        ll f[26];        for(i=0;i<26;i++)            f[res[i]]=i;        for(i=0;i<26;i++)//处理前导0的情况        {            if(l[i]&&res[i]==0)            {                ll ss=26;                ll flag=0;                for(j=0;j<26;j++)                {                    if(l[j]) continue;                    if(res[j]<ss)                    {                        ss=res[j];                        flag=j;                    }                }                for(j=1;j<ss;j++)                {                    res[f[j]]++;                }                res[i]=1;                res[flag]=0;            }        }        ll sum = 0;        ll t=1;        for(j =0; j <=mm; j++)//计算结果        {            for(i = 0; i < 26; i++)            {                if(map[i][j]==0) continue;                if(res[i]==0) continue;                //cout<<res[i]<<endl;                ll tm=t;                //cout<<tm<<" ";                tm = (tm * res[i]) % mod;                tm=(tm * map[i][j])%mod;                //cout<<i<<" "<<res[i]<<endl;                sum = (sum+tm) % mod;            }            t=(t*26)%mod;        }        cout<<"Case #"<<y<<": ";        printf("%lld\n",sum);       }return 0;}