这题目好难理解，看了一下午效率好低，看着题解做出来的...不过也算把刘汝佳老师第三章的题全做完了...
题意：将一浮点数转换成标准储存需要多少位尾数和指数
思路：因输入为响应最大值，为对应关系，故可打表，枚举m位与e位，存储相应的科学记数法表示，输入时再查表。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <cmath> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <string> 10 #include <queue> 11 #include <stack> 12 #include <vector> 13 #include <fstream> 14 #include <cctype> 15 #include <ctime> 16 #include <iomanip> 17 #include <sstream> 18 19 #define fi for (int i = 1; i <= n; ++i) 20 #define fj for (int j = 1; j <= m; ++j) 21 #define INF 0x3fffffff 22 #define pau system("pause") 23 #define printclock printf("Time used = %.2fs ", (double) clock() / CLOCKS_PER_SEC); 24 #define eps 1e-5 25 #define cin std::cin 26 #define cout std::cout 27 #define endl std::endl 28 #define pair std::pair 29 30 const double pi = acos(-1.0); 31 const int max_n = 1e5 + 5; 32 typedef long long ll; 33 typedef pair<int, int> pii; 34 35 template<class Q> void getint(Q &res, char ch = getchar()) { 36 while (!isdigit(ch) && ch != '-') ch = getchar(); 37 bool flag = true; 38 if ('-' == ch) flag = false, ch = getchar(); 39 for (res = 0; isdigit(ch); ch = getchar()) res = res * 10 + ch - 48; 40 if (!flag) res = -res; 41 } 42 void putint(int x) { 43 int i = 0, a[10]; 44 bool flag = true; 45 if (x < 0) { 46 flag = false; 47 x = -x; 48 } else if (x == 0) { 49 a[i++] = 0; 50 } 51 while (x > 0) { 52 a[i++] = x % 10; 53 x /= 10; 54 } 55 if (flag == false) putchar('-'); 56 for (int j = i - 1; j >= 0; --j) putchar(a[j] + 48); 57 } 58 int mmax(int a, int b) {return a > b ? a : b;} 59 int mmin(int a, int b) {return a < b ? a : b;} 60 int gcd(int a, int b) {while (b) b ^= a ^= b ^= a %= b; return a;} 61 62 /*int parent[max_n], rank[max_n]; 63 void ini() { 64 for (int i = 1; i < 10001; ++i) { 65 parent[i] = i; 66 rank[i] = 0; 67 } 68 } 69 int find(int x) { 70 return x == parent[x] ? x : parent[x] = find(parent[x]); 71 } 72 void unite(int x, int y) { 73 x = find(x); 74 y = find(y); 75 if (x == y) return; 76 77 if (rank[x] < rank[y]) { 78 parent[x] = y; 79 } else { 80 parent[y] = x; 81 if (rank[x] == rank[y]) rank[x]++; 82 } 83 }*/ 84 85 //#define LOCAL 86 87 int main() { 88 #ifdef LOCAL 89 freopen("data.in", "r", stdin); 90 freopen("data.out", "w", stdout); 91 #endif // LOCAL 92 double A[15][35]; 93 int B[15][35]; 94 for (int i = 0; i <= 9; ++i) { 95 for (int j = 1; j <= 30; ++j) { 96 double t1 = 1 - pow(2, -i - 1); 97 int t2 = pow(2, j) - 1; 98 double t3 = log10(t1) + t2 * log10(2); 99 B[i][j] = t3; 100 A[i][j] = pow(10, t3 - B[i][j]); 101 } 102 } 103 std::string s; 104 while (cin >> s, s != "0e0") { 105 for (std::string::iterator i = s.begin(); i != s.end(); ++i) { 106 if ('e' == *i) { 107 *i = ' '; 108 break; 109 } 110 } 111 std::istringstream ss(s); 112 double a, b; 113 ss >> a >> b; 114 for (int i = 0; i < 10; ++i) { 115 for (int j = 1; j <= 30; ++j) { 116 if (b == B[i][j] && fabs(a - A[i][j]) < 1e-5) { 117 cout << i << ' ' << j << endl; 118 } 119 } 120 } 121 } 122 return 0; 123 }