TOJ 2861 ZOJ 1086 Octal Fractions / 大数 高精度
Octal Fractions
时间限制(普通/Java):1000MS/3000MS???? 运行内存限制:65536KByte
描述 Fractions in octal (base 8) notation can be expressed exactly in decimal notation. For example,0.75 in octal is 0.953125 (7/8 + 5/64) in decimal. All octal numbers of n digits to the right of the octal point can be expressed in no more than 3n decimal digits to the right of the decimal point. 输入 The input to your program will consist of octal numbers,one per line,to be converted. Each input number has the form 0.d1d2d3 ... dk,where the di are octal digits (0..7). There is no limit on k. 输出 Your output will consist of a sequence of lines of the form where the left side is the input (in octal),and the right hand side the decimal (base 10) equivalent. There must be no trailing zeros,i.e. Dm is not equal to 0. 样例输入 0.75
0.0001
0.01234567
样例输出 0.75 [8] = 0.953125 [10]
0.0001 [8] = 0.000244140625 [10]
0.01234567 [8] = 0.020408093929290771484375 [10]
可以怎么做 0.75 = (5/8+7)/8 0.01234567 = (((7/8+6)/8)+5)/8.... 模拟高精度就行 代码高队长提供 #include <stdio.h> #include <string.h> #include <string> #include <iostream> #include <algorithm> using namespace std; string s; char res[30000]; int reslen; void js() { int i,j,k,num=0; for(i=0;;i++) { if(res[i]!=0) num = num * 10 + res[i]-'0'; else num = num * 10; res[i]=num/8+'0'; num=num%8; if(res[i+1]==0&&num==0) break; } } int main() { while(cin >> s) { reverse(s.begin(),s.end()); memset(res,sizeof(res)); int i,k; for(i=0;s[i]!='.';i++) { res[0]=s[i]; js(); } reverse(s.begin(),s.end()); cout << s <<" [8] = 0."<< res + 1 << " [10]"<<endl; } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |