A - Por Costel and Azerah-dp求和

• A - Por Costel and Azerah

•  Gym - 100923A 

• #include <iostream>
  #include <stdio.h>
  #include <cmath>
  using namespace std;
  #define maxn 1000052
  #define ll long long
  #define md 1000000007
  ll t,n,x,dp[maxn][5];
  int main()
  {
      freopen("azerah.in","r",stdin);
      freopen("azerah.out","w",stdout);
      scanf("%lld",&t);
      while(t--)
      {
          scanf("%lld",&n);
          dp[0][1]=0;
          dp[0][0]=1;
          for(ll i=1; i<=n; i++)
          {
              scanf("%lld",&x);
              if(x%2==0)
              {
                  dp[i][0]=(2*dp[i-1][0])%md;
                  dp[i][1]=2*dp[i-1][1]%md;
                  //重组替换两种操作。
              }
              else
                  dp[i][1]=dp[i][0]=(dp[i-1][1]+dp[i-1][0])%md;
          }
          printf("%lld\n",dp[n][0]-1);
      }
      return 0;
  }