1.链接地址：
2.题目：
Exponentiation
Time Limit: 500MS Memory Limit: 10000K Total Submissions: 127573 Accepted: 31149 Description
Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems. This problem requires that you write a program to compute the exact value of R n where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.Input
The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.Output
The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don't print the decimal point if the result is an integer.Sample Input
95.123 120.4321 205.1234 156.7592 998.999 101.0100 12Sample Output
548815620517731830194541.899025343415715973535967221869852721.0000000514855464107695612199451127676715483848176020072635120383542976301346240143992025569.92857370126648804114665499331870370751166629547672049395302429448126.76412102161816443020690903717327667290429072743629540498.1075960194566517745610440100011.126825030131969720661201Hint
If you don't know how to determine wheather encounted the end of input: s is a string and n is an integerC++ while(cin>>s>>n) { ... } c while(scanf("%s%d",s,&n)==2) //to see if the scanf read in as many items as you want /*while(scanf(%s%d",s,&n)!=EOF) //this also work */ { ... }Source
3.思路：
高精度浮点数乘法，利用大整数乘法的基础计算
（1）判断是否为小数，不为小数则转为小数
（2）获取小数点位置，计算出相乘后的小数点的位置。并去除原小数的小数点
（3）利用大整数乘法求结果
（4）补全小数点，并补零
tip：注意一下例子
0.0000 1 结果为 0
1.0000 0 结果为 1
0.0000 0 结果为 1
并注意小于1的浮点数要去除前导的0
4.代码：
1 #include2 #include 3 #include 4 #include 5 #include 6 7 using namespace std; 8 9 string mul(string str1,string str2) 10 { 11 vector v_res(str1.size()+str2.size(),0); 12 string::size_type i,j; 13 vector ::size_type k,p; 14 15 reverse(str1.begin(),str1.end()); 16 reverse(str2.begin(),str2.end()); 17 for(i = 0; i != str1.size(); ++i) 18 { 19 for(j = 0; j != str2.size(); ++j) 20 { 21 v_res[i+j] += (str1[i]-'0') * (str2[j] - '0'); 22 } 23 } 24 for(k = 0; k != v_res.size() - 1; ++k) 25 { 26 v_res[k+1] += v_res[k] / 10; 27 v_res[k] = v_res[k] % 10; 28 } 29 30 for(p = v_res.size() - 1; p != -1; --p) 31 { 32 if(v_res[p] != 0) break; 33 } 34 if(p == -1) p = 0; 35 36 string s_res(p+1,'0'); 37 for(k = p; k != -1; --k) s_res[p-k] = char(v_res[k] + '0'); 38 39 40 return s_res; 41 42 } 43 44 string real_mul(string str1,string str2) 45 { 46 string::size_type idx_str1_point = str1.find("."); 47 if(idx_str1_point == string::npos) 48 { 49 str1 += ".0"; 50 idx_str1_point = str1.find("."); 51 } 52 str1.erase(idx_str1_point,1); 53 54 string::size_type idx_str2_point = str2.find("."); 55 if(idx_str2_point == string::npos) 56 { 57 str2 += ".0"; 58 idx_str2_point = str2.find("."); 59 } 60 str2.erase(idx_str2_point,1); 61 62 string::size_type dec_res_len = (str1.size() - idx_str1_point) + (str2.size() - idx_str2_point); 63 64 string res = mul(str1,str2); 65 66 if(res.size() < dec_res_len + 1) 67 { 68 res = string(dec_res_len + 1 - res.size(),'0') + res; 69 } 70 71 res.insert(res.size() - dec_res_len,"."); 72 73 string::size_type idx_res_tail = res.find_last_not_of("0"); 74 res = res.substr(0,idx_res_tail+1); 75 76 if(res[res.size() - 1] == '.') res.erase(res.size()-1); 77 78 return res; 79 } 80 81 int main() 82 { 83 string r; 84 int n; 85 while(cin>>r>>n) 86 { 87 string res = "1"; 88 while(n--) 89 { 90 res = real_mul(res,r); 91 } 92 string::size_type idx_res_pre = res.find_first_not_of("0"); 93 if(idx_res_pre != string::npos) res.erase(0,idx_res_pre); 94 95 96 cout< <