UVA 11809 Floating-Point Numbers

题目链接：https://vjudge.net/problem/UVA-11809

题目翻译摘自《算法禁赛入门经典》

题目大意

　　计算机常用阶码-尾数的方法保存浮点数。如图所示，如果阶码有 6 位，尾数有 8 位， 可以表达的最大浮点数为$0.111111111_2 * 2^{111111_2}$。注意小数点后第一位必须为 1，所以一共有 9 位小数。 

　　

　　这个数换算成十进制之后就是$0.998046875 * 2^{63} = 9.205357638345294 * 10^{18}$。你的任务是根据这个最大浮点数，求出阶码的位数 E 和尾数的位数 M。输入格式为 AeB，表示最大浮点数为$A * 10^B$。$0 < A < 10$，并且恰好包含 15 位有效数字。输入结束标志为 0e0。对于每组数据，输出 M 和 E。输入保证有唯一解，且$0 \leq M \leq 9，1 \leq E \leq 30$。在本题中，M + E + 2 不必为 8 的整数倍。

分析

　　设最大值为 v，则$v = (1 - \frac{1}{2^{M + 1}}) * 2^{2^E - 1} = A * 10^B$。

　　两边求以 10 为底的对数可得$\lg v = \lg (2^{M + 1} - 1) - (M + 1) * \lg 2 + (2^E - 1) * \lg 2 = \lg A + B$。

　　于是可以暴力枚举所有 M，算出 E 后再带回去检验误差。

　　此题的 EPS 选取有讲究，不能太大，可不能太小，10e-6 就差不多了，不然会 Wa。

代码如下

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
  9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 12  
 13 #define pr(x) cout << #x << " = " << x << "  "
 14 #define prln(x) cout << #x << " = " << x << endl
 15  
 16 #define LOWBIT(x) ((x)&(-x))
 17  
 18 #define ALL(x) x.begin(),x.end()
 19 #define INS(x) inserter(x,x.begin())
 20  
 21 #define ms0(a) memset(a,0,sizeof(a))
 22 #define msI(a) memset(a,inf,sizeof(a))
 23 #define msM(a) memset(a,-1,sizeof(a))
 24 
 25 #define MP make_pair
 26 #define PB push_back
 27 #define ft first
 28 #define sd second
 29  
 30 template<typename T1, typename T2>
 31 istream &operator>>(istream &in, pair<T1, T2> &p) {
 32     in >> p.first >> p.second;
 33     return in;
 34 }
 35  
 36 template<typename T>
 37 istream &operator>>(istream &in, vector<T> &v) {
 38     for (auto &x: v)
 39         in >> x;
 40     return in;
 41 }
 42  
 43 template<typename T1, typename T2>
 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 45     out << "[" << p.first << ", " << p.second << "]" << "\n";
 46     return out;
 47 }
 48 
 49 inline int gc(){
 50     static const int BUF = 1e7;
 51     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 52     
 53     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 54     return *bg++;
 55 } 
 56 
 57 inline int ri(){
 58     int x = 0, f = 1, c = gc();
 59     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
 60     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 61     return x*f;
 62 }
 63  
 64 typedef long long LL;
 65 typedef unsigned long long uLL;
 66 typedef pair< double, double > PDD;
 67 typedef pair< int, int > PII;
 68 typedef pair< int, PII > PIPII;
 69 typedef pair< string, int > PSI;
 70 typedef pair< int, PSI > PIPSI;
 71 typedef set< int > SI;
 72 typedef set< PII > SPII;
 73 typedef vector< int > VI;
 74 typedef vector< VI > VVI;
 75 typedef vector< PII > VPII;
 76 typedef map< int, int > MII;
 77 typedef map< int, PII > MIPII;
 78 typedef map< PII, int > MPIII;
 79 typedef map< string, int > MSI;
 80 typedef multimap< int, int > MMII;
 81 //typedef unordered_map< int, int > uMII;
 82 typedef pair< LL, LL > PLL;
 83 typedef vector< LL > VL;
 84 typedef vector< VL > VVL;
 85 typedef priority_queue< int > PQIMax;
 86 typedef priority_queue< int, VI, greater< int > > PQIMin;
 87 const double EPS = 1e-6;
 88 const LL inf = 0x7fffffff;
 89 const LL infLL = 0x7fffffffffffffffLL;
 90 const LL mod = 1e9 + 7;
 91 const int maxN = 1e4 + 7;
 92 const LL ONE = 1;
 93 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 94 const LL oddBits = 0x5555555555555555;
 95 
 96 int B, M, E;
 97 double A;
 98 string str;
 99 
100 int main(){
101     //freopen("MyOutput.txt","w",stdout);
102     //freopen("input.txt","r",stdin);
103     INIT();
104     while(getline(cin, str)) {
105         if(str == "0e0") break;
106         str[str.find('e')] = ' ';
107         stringstream sin(str);
108         sin >> A >> B;
109         double v = log10(A) + B;
110         
111         For(x, 1, 10) {
112             double tmp = log10(pow(2, x) - 1) - (x + 1) * log10(2);
113             E = log2((v - tmp) / log10(2)) + EPS;
114             if(fabs(v - tmp - pow(2, E) * log10(2)) < EPS) {
115                 M = x - 1;
116                 break;
117             }
118         }
119         cout << M << " " << E << endl;
120     }
121     return 0;
122 }

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