2017多校1 b hdu 6034

来源：互联网发布：写作起名软件编辑：程序博客网时间：2024/05/21 22:56

记录每个字母在每个位数出现的次数，直接用结构体记录并排序可能会超时，再建一个新的数组来排序。找出一个没出现在首位（单字母除外）的字母，其它按排序赋值

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 100000+10;const int mod = 1000000007;int alp[26][N];int fir[26];int vis[26];int num[26];int cnt = 0;bool cmp(int a,int b){    for(int i = N - 1; i >= 0; i--)    {        if(alp[a][i] > alp[b][i]) return true;        else if(alp[a][i] < alp[b][i]) return false;    }    return false;}int main(){     int n;     int tim = 1;     while(~scanf("%d",&n))     {         char op[N];         memset(alp,0,sizeof(alp));         memset(fir,0,sizeof(fir));         memset(vis,0,sizeof(vis));         memset(num,-1,sizeof(num));         for(int i = 0; i < n; i++)         {           scanf("%s",op);           int len = strlen(op) - 1;            for(int j = 0; j <= len; j++)            {               int d = op[j] - 'a';               if(j == 0 && len > 0)                   fir[d] = 1;               alp[d][len - j] ++;               vis[d] = 1;           //    cout << d <<"   " <<len - j <<"     "<<alp[d][len- j]<< endl;            }         }         for(int i = 0; i <= 25; i++)         {             for(int j = 0; j <= N-1; j++)             {                 if(alp[i][j] >= 26)                 {                     alp[i][j+1] += alp[i][j]/26;                     alp[i][j] %= 26;                 }             }         }         int aa[26];          for(int i = 0; i < 26; i++)              aa[i] = i;              sort(aa,aa+26,cmp);             for(int i = 25; i >= 0; i--)             {                 int te = aa[i];                 if(!fir[te])                 {                     num[te] = 0;                     break;                 }             }         int biap = 25;         for(int i = 0; i <= 25; i++)         {             int te = aa[i];             if(num[te] == -1)             {                 num[te] = biap--;             }         }         long long sum = 0;         long long mul = 1;         for(int i = 0; i <= N -1; i++)         {             for(int j = 0; j <= 25; j++)             {                   sum = (sum+(alp[j][i]%mod)*(mul*num[j])%mod)%mod;             }             mul *= 26;             mul %= mod;         }        printf("Case #%d: %lld\n",tim++,sum);     }}

阅读全文

0 0