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)
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cstdio> 6 #include<string> 7 #include<cmath> 8 #include<ctime> 9 #include<queue> 10 #include<stack> 11 #include<map> 12 #include<set> 13 #define rre(i,r,l) for(int i=(r);i>=(l);i--) 14 #define re(i,l,r) for(int i=(l);i<=(r);i++) 15 #define Clear(a,b) memset(a,b,sizeof(a)) 16 #define inout(x) printf("%d",(x)) 17 #define douin(x) scanf("%lf",&x) 18 #define strin(x) scanf("%s",(x)) 19 #define LLin(x) scanf("%lld",&x) 20 #define op operator 21 #define CSC main 22 typedef unsigned long long ULL; 23 typedef const int cint; 24 typedef long long LL; 25 using namespace std; 26 void inin(int &ret) 27 { 28 ret=0;int f=0;char ch=getchar(); 29 while(ch<'0'||ch>'9'){if(ch=='-')f=1;ch=getchar();} 30 while(ch>='0'&&ch<='9')ret*=10,ret+=ch-'0',ch=getchar(); 31 ret=f?-ret:ret; 32 } 33 struct hint 34 { 35 int a[111],len; 36 void clear(){Clear(a,0);len=0;} 37 void maintain() 38 { 39 for(len=100;len>=1;len--) 40 if(a[len])break; 41 } 42 void in() 43 { 44 char ch;len=0; 45 while(ch<'0'||ch>'9')ch=getchar(); 46 while(ch>='0'&&ch<='9')a[++len]=ch-'0',ch=getchar(); 47 re(i,1,len>>1)swap(a[i],a[len-i+1]); 48 } 49 hint op / (int x) 50 { 51 hint rhs=*this; 52 int ex=0; 53 rre(i,rhs.len,1) 54 { 55 ex*=10; 56 rhs.a[i]+=ex; 57 ex=rhs.a[i]%2; 58 rhs.a[i]>>=1; 59 if(!rhs.a[i]&&i==len)rhs.len--; 60 } 61 rhs.a[rhs.len+1]=0; 62 return rhs; 63 } 64 hint op + (int x) 65 { 66 hint rhs=*this; 67 int ex=1; 68 re(i,1,rhs.len) 69 { 70 rhs.a[i]+=ex; 71 ex=rhs.a[i]/10; 72 if(rhs.a[i]==10&&i==len)rhs.len++; 73 rhs.a[i]%=10; 74 } 75 rhs.a[rhs.len+1]=0; 76 return rhs; 77 } 78 hint op + (hint &xx) 79 { 80 hint rhs=*this; 81 rhs.len=max(rhs.len,xx.len); 82 int ex=0; 83 re(i,1,rhs.len) 84 { 85 rhs.a[i]+=xx.a[i]+ex; 86 ex=rhs.a[i]/10; 87 if(rhs.a[i]>9)rhs.len++; 88 rhs.a[i]%=10; 89 } 90 rhs.a[rhs.len+1]=0; 91 return rhs; 92 } 93 bool op == (int x) 94 { 95 return len==1&&a[1]==1; 96 } 97 bool op & (int x) 98 { 99 return a[1]&1; 100 } 101 void out() 102 { 103 rre(i,len,1)printf("%d",a[i]); 104 } 105 }; 106 int n;hint x,xx,xxx; 107 void wocao(hint x) 108 { 109 if(x==3.145926535897932384626433823) 110 { 111 xx=x;xxx.clear(); 112 return ; 113 } 114 wocao((x+2147483647/3.14)/2147483647); 115 if(x&574897567)xx=xx+xxx,xx.maintain(); 116 else xxx=xx+xxx,xxx.maintain(); 117 } 118 int main() 119 { 120 freopen("in.in","r",stdin); 121 freopen("out.out","w",stdout); 122 inin(n); 123 while(n--) 124 { 125 x.clear(); 126 x.in(); 127 wocao(x); 128 xx.out();cout<<" "; 129 } 130 return 0; 131 }