poj1001:1001:Exponentiation解题报告

1001:Exponentiation

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总时间限制: 

500ms 

内存限制: 

65536kB

描述

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 Rⁿwhere R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.

输入

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.

输出

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.

样例输入

95.123 120.4321 205.1234 156.7592  998.999 101.0100 12

样例输出

548815620517731830194541.899025343415715973535967221869852721.0000000514855464107695612199451127676715483848176020072635120383542976301346240143992025569.92857370126648804114665499331870370751166629547672049395302429448126.76412102161816443020690903717327667290429072743629540498.1075960194566517745610440100011.126825030131969720661201

提示

If you don't know how to determine wheather encounted the end of input:
s is a string and n is an integer

C++while(cin>>s>>n){...}cwhile(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    */{...}

来源

East Central North America 1988

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• 代码；

  #include<iostream>#include<string>#include <algorithm>using namespace std;string multi(string str1, string str2){    reverse(str1.begin(), str1.end());    reverse(str2.begin(), str2.end());    size_t npos1 = str1.find('.');    size_t npos2 = str2.find('.');    if(npos1 == string::npos) npos1 = 0;    else str1.erase(str1.begin() + npos1);    if(npos2 == string::npos) npos2 = 0;    else str2.erase(str2.begin() + npos2);    string ans(str1.length() + str2.length(), '0');    for(size_t i = 0; i < str1.length(); ++i){        int carry = 0;        for(size_t j = 0; j < str2.length(); ++j){            int val = ((str1[i] - '0') * (str2[j] - '0') + carry + ans[i + j] - '0');            ans[i + j] = val % 10 + '0';            carry = val / 10;        }        if(carry){            ans[i + str2.length()] = ans[i + str2.length()] + carry;        }    }    size_t npos = npos1 + npos2;    bool hasPoint = false;    if(npos > 0){        ans.insert(npos, 1 ,'.');        hasPoint = true;    }    npos = ans.length() - 1;    while(ans[npos--] == '0') ans.pop_back();    reverse(ans.begin(),ans.end());       if(hasPoint){        npos = ans.length() - 1;        while(ans[npos--] == '0') ans.pop_back();        if(ans[npos + 1] == '.') ans.pop_back();    }    return ans;}int main(){    string s,ans;    int n;     while(cin>>s>>n){            ans.assign(s);     if(n == 1){                while(*(ans.begin()) == '0'){                    ans.erase(ans.begin());                }                size_t pos = ans.find('.');                if(pos != string ::npos){                    pos = ans.length() - 1;                    while (ans[pos--] == '0') {                        ans.pop_back();                    }                    if(ans[pos + 1] == '.') ans.pop_back();                }                          }else{                int bit = 1;     for(int i = 2; i <= n; i<<=1){                    bit = i;     ans = multi(ans, ans);     }                for(int i = bit; i < n; ++i){                    ans = multi(s,ans);                }     }         cout<<ans<<endl;     }        return 0;}