CSP 202112-3 登机牌条码 100分题解

分析：

一步步按照它的要求来就是了。
多项式除法的过程要手动模拟（有点像高精度）。

一开始我50分，是只对答案取了模没对过程取模。
然后80分，发现3^k其实会爆long long。

注意每一处的细节（要不要取模）

代码

#include <bits/stdc++.h>

using namespace std;

#define File(x) freopen("(x)","r",stdin)
#define pf prlong longf
#define ull unsigned long long
#define db double
#define ll long long
#define MAXN  
long long w,s,f;
string x;
vector<long long>v;

int main(){
   cin>>w>>s>>x;

   long long tp=0;

   for(long long i=0;i<x.length();i++){
      if(x[i]>='A'&&x[i]<='Z'){
            if(tp==1){
            v.push_back(28);
            v.push_back(28);
         }
         else if(tp==2){
            v.push_back(28);                     
         }
         tp=0;
         v.push_back(x[i]-'A');
      }
      else if(x[i]>='a'&&x[i]<='z'){
         if(tp==0||tp==2)v.push_back(27);
         v.push_back(x[i]-'a');
         tp=1;
      }
      else {
         if(tp!=2)v.push_back(28);
         tp=2;
         v.push_back(x[i]-'0');
      }
   }

   if(v.size()%2){
      v.push_back(29);
   }
   vector<long long>v2;
   for(long long i=0;i<v.size();i+=2){
      v2.push_back(v[i]*30+v[i+1]);
   }
   
   long long len=v2.size()+1;
   if(s==-1){
      while(len%w){
         v2.push_back(900);
         ++len;
      }
      cout<<len<<endl;
      for(auto x:v2)cout<<x<<endl;
      return 0;
   }
   while((len+(1<<(s+1)))%w){
      ++len;
      v2.push_back(900);
   } 
 
   vector<long long>v3;
   v3.push_back(len);
   for(auto X:v2){
      v3.push_back(X);
   }
     for(auto x:v3)printf("%d\n",x);

   vector<long long>v4;
   v4.push_back(1);
   v4.push_back(-3);
   
   long long mul=-3;

   for(long long i=2;i<=(1<<(s+1));i++){
      v4.push_back(0);
      mul=mul*3%929; 
      for(long long j=v4.size()-1;j>=1;j--)v4[j]+=v4[j-1]*mul%929,v4[j]%=929;
   }  
      long long w4=1<<(s+1);
      for(long long i=1;i<=w4;i++) v3.push_back(0);
      for(long long i=0;i<v3.size();i++){
          long long num=v3[i]; 
          for(long long j=i+1;j<=i+w4;j++)v3[j]-=num*v4[j-i]%929,v3[j]%=929;
          if(i==v3.size()-w4-1){ f=i+1; break; }    
      }
      while(f<v3.size()){ 
      long long ans1=(-v3[f])%929;  
      if(ans1<0)ans1+=929; 
      cout<<ans1<<endl;
      ++f;
      } 
      return 0;
}

 

标签：题解,back,long,202112,v3,v4,登机牌,push,define
来源： https://www.cnblogs.com/GUOGaby/p/16061917.html