tjut 4899

来源：互联网发布：linux 挂载硬盘编辑：程序博客网时间：2024/06/07 12:11
#include<stdio.h>  #include<iostream>  #include<algorithm>  #include<string>  #include<string.h>  #include<math.h>  #include<limits.h>  #include<time.h>  #include<stdlib.h>  #include<map>  #include<queue>  #include<set>  #include<stack>  #include<vector>  #define LL long long  using namespace std;  const int mod = 1e9 + 7;  char dna[5] = "ATGC";  LL dp[2][70000], prt[20];  int res[20], res1[20], ans[70000][5];  int main()  {      int T;      while(~scanf("%d", &T))          while(T--)          {              memset(res, 0, sizeof res);              memset(ans, 0, sizeof ans);              memset(dp, 0, sizeof dp);              string s;              cin >> s;              int n;              cin >> n;              int len = s.size();              int maxx = 1 << len;              for(int i = 0; i < maxx; i++)              {                  res[0] = 0;                  int tmp = i;                  int cnt = 1;                  for(int j = 0; j < len; j++)//转化当前状态为正常dp                  {                      if(tmp & 1)                          res[cnt] = res[cnt - 1] + 1;                      else                          res[cnt] = res[cnt - 1];                      cnt++;                      tmp >>= 1;                  }                  for(int j = 0; j < 4; j++)//计算分别加4个字符后状态变化                  {                      res1[0] = 0;                      for(int k = 1; k <= len; k++)                      {                          if(s[k - 1] == dna[j])                              res1[k] = res[k - 1] + 1;                          else                              res1[k] = max(res1[k - 1], res[k]);                      }                      for(int k = len; k > 0; k--)//记录i状态后添加字符j后的状态                      {                          ans[i][j] <<= 1;                          if(res1[k] > res1[k - 1])                              ans[i][j] += 1;                      }                  }              }              dp[0][0] = 1;              for(int i = 0; i < n; i++)              {                  memset(dp[(i + 1) & 1], 0, sizeof dp[(i + 1) & 1]);                  for(int j = 0; j < maxx; j++)                  {                      for(int k = 0; k < 4; k++)                      {                          dp[(i + 1) & 1][ans[j][k]] += dp[i & 1][j];                          if(dp[(i + 1) & 1][ans[j][k]] > mod)                              dp[(i + 1) & 1][ans[j][k]] %= mod;                      }                  }              }              memset(prt, 0, sizeof prt);              for(int i = 0; i < maxx; i++)              {                  int cnt = 0;                  int tmp = i;                  while(tmp)                  {                      if(tmp & 1)                          cnt++;                      tmp >>= 1;                  }                  prt[cnt] += dp[n & 1][i];                  if(prt[cnt] > mod)                      prt[cnt] %= mod;              }              for(int i = 0; i <= len; i++)                  printf("%lld\n", prt[i]);          }      return 0;  }
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