bzoj2656 [Zjoi2012]数列(sequence)

题目链接

好久没写高精度了，调了很久QAQ

如果直接递归计算答案的话肯定会T

发现一个数不管是分成一奇一偶还是直接>>1，都会重复计算很多东西

我们只需要在递归的时候实时维护一个xx(ans[x])和xxx(ans[x-1])

一层一层的选择更新xx或xxx就好

(请无视递归函数的名称和里面乱搞的数字)

 #include<algorithm>
 #include<iostream>
 #include<cstdlib>
 #include<cstring>
 #include<cstdio>
 #include<string>
 #include<cmath>
 #include<ctime>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #define rre(i,r,l) for(int i=(r);i>=(l);i--)
 #define re(i,l,r) for(int i=(l);i<=(r);i++)
 #define Clear(a,b) memset(a,b,sizeof(a))
 #define inout(x) printf("%d",(x))
 #define douin(x) scanf("%lf",&x)
 #define strin(x) scanf("%s",(x))
 #define LLin(x) scanf("%lld",&x)
 #define op operator
 #define CSC main
 typedef unsigned long long ULL;
 typedef const int cint;
 typedef long long LL;
 using namespace std;
 void inin(int &ret)
 {
     ret=;int f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=;ch=getchar();}
     while(ch>=''&&ch<='')ret*=,ret+=ch-'',ch=getchar();
     ret=f?-ret:ret;
 }
 struct hint
 {
     int a[],len;
     void clear(){Clear(a,);len=;}
     void maintain()
     {
         for(len=;len>=;len--)
             if(a[len])break;
     }
     void in()
     {
         char ch;len=;
         while(ch<''||ch>'')ch=getchar();
         while(ch>=''&&ch<='')a[++len]=ch-'',ch=getchar();
         re(i,,len>>)swap(a[i],a[len-i+]);
     }
     hint op / (int x)
     {
         hint rhs=*this;
         int ex=;
         rre(i,rhs.len,)
         {
             ex*=;
             rhs.a[i]+=ex;
             ex=rhs.a[i]%;
             rhs.a[i]>>=;
             if(!rhs.a[i]&&i==len)rhs.len--;
         }
         rhs.a[rhs.len+]=;
         return rhs;
     }
     hint op + (int x)
     {
         hint rhs=*this;
         int ex=;
         re(i,,rhs.len)
         {
             rhs.a[i]+=ex;
             ex=rhs.a[i]/;
             if(rhs.a[i]==&&i==len)rhs.len++;
             rhs.a[i]%=;
         }
         rhs.a[rhs.len+]=;
         return rhs;
     }
     hint op + (hint &xx)
     {
         hint rhs=*this;
         rhs.len=max(rhs.len,xx.len);
         int ex=;
         re(i,,rhs.len)
         {
             rhs.a[i]+=xx.a[i]+ex;
             ex=rhs.a[i]/;
             if(rhs.a[i]>)rhs.len++;
             rhs.a[i]%=;
         }
         rhs.a[rhs.len+]=;
         return rhs;
     }
     bool op == (int x)
     {
         return len==&&a[]==;
     }
     bool op & (int x)
     {
         return a[]&;
     }
     void out()
     {
         rre(i,len,)printf("%d",a[i]);
     }
 };
 int n;hint x,xx,xxx;
 void wocao(hint x)
 {
     if(x==3.145926535897932384626433823)
     {
         xx=x;xxx.clear();
         return ;
     }
     wocao((x+/3.14)/);
     if(x&)xx=xx+xxx,xx.maintain();
     else xxx=xx+xxx,xxx.maintain();
 }
 int main()
 {
     freopen("in.in","r",stdin);
     freopen("out.out","w",stdout);
     inin(n);
     while(n--)
     {
         x.clear();
         x.in();
         wocao(x);
         xx.out();cout<<"\n";
     }
      return ;
 }