• 每日一题2014/7/30


    Exponentiation
    Time Limit: 500MS   Memory Limit: 10000K
    Total Submissions: 134035   Accepted: 32776

    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 Rn 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 12
    0.4321 20
    5.1234 15
    6.7592  9
    98.999 10

    1.0100 12

    Sample Output

    548815620517731830194541.899025343415715973535967221869852721
    .00000005148554641076956121994511276767154838481760200726351203835429763013462401
    43992025569.928573701266488041146654993318703707511666295476720493953024
    29448126.764121021618164430206909037173276672
    90429072743629540498.107596019456651774561044010001
    1.126825030131969720661201

    没做出来,抄的别人的代码,一开始思路差不多,但是我代码水平太差了

    #include<iostream>
    using namespace std;
    
    #define LENGTH 200
    class BigNumber 
    {
    private:
        int* value; 
        int decimal_bits;
        int exp;
    
        int previousBit(int bit) {
            if (bit < 10)
                return 0;
            else
                return bit / 10;
        }
    public:
        BigNumber(char* str, int e) {
            value = new int[LENGTH];
            //decimal = new int[LENGTH];
            for(int i = 0; i < LENGTH; i++)
            {
                value[i] = 0;
                //decimal[i] = 0;
            }
            int index = -1;
            int str_len = strlen(str);
            for(int i = 0; i < str_len; i++)
                if(str[i] == '.')
                    index = i;
            decimal_bits = (index == -1 ? 0 : str_len - index - 1);
            //将输入的字符串转化为相应的数值
            if (index == -1)
            {
                for (int i = 0; i < str_len; i++)
                    value[LENGTH - str_len + i] = str[i] - '0';
            }
            else
            {
                for (int i = 0; i < index; i++)
                    value[LENGTH - str_len + i + 1] = str[i] - '0';
                for (int i = index + 1; i < str_len; i++)
                    value[LENGTH - str_len + i] = str[i] - '0';
            }
            exp = e;
        }
    
        ~BigNumber()
        {
            delete[] value;
        }
        void pow() {
            if (exp == 0) {
                for (int i = 0; i < LENGTH - 1; i++)
                    value[i] = 0;
                value[LENGTH - 1] = 1;
                return;
            }
    
            int* tmp_value = new int[LENGTH];
            //int* tempDecimal = new int[LENGTH];
            for(int i = 0; i < LENGTH; i++)
                tmp_value[i] = value[i];
    
            for (int i = 0; i < exp - 1; i++) 
            {
                //乘一次保存一次值
                int* result_value = new int[LENGTH];
    
                for(int j = 0; j < LENGTH; j++)
                    result_value[j] = 0;
                
                for (int j = 0; j < LENGTH; j++)
                {
                    if (tmp_value[j] != 0)
                    {
                        for (int k = 0; k < LENGTH; k++)
                        {
                            if (value[k] != 0)
                                result_value[j + k - LENGTH + 1] += tmp_value[j]*value[k];
                        }
                    }
                }
                for (int j = 0; j < LENGTH; j++)
                    value[j] = result_value[j];
    
                //进行进位,以防在上述相乘的过程中溢出
                for (int j = LENGTH - 1; j >=0; j--)
                {
                    value[j - 1] += previousBit(value[j]);
                    value[j] %=10;
                }
    
                delete result_value;
            }
    
            delete[] tmp_value;
    
        }
    
        void print()
        {
            if (exp == 0)
                printf("%d
    ", 1);
            else
            {
                if (decimal_bits == 0)
                {
                    bool flag = false;
                    for (int i = 0; i < LENGTH; i++)
                    {
                        if (value[i] == 0 && !flag)
                            continue;
                        else
                        {
                            flag = true;
                            printf("%d", value[i]);
                        }
                    }
                    printf("
    ");
                }
                else
                {
                    int bits = decimal_bits*exp;
                    bool is_zero = true;
                    for (int i = 0; i < LENGTH - bits; i++)
                    {
                        if (value[i] == 0 && is_zero)
                            continue;
                        is_zero = false;
                        printf("%d", value[i]);
                    }
                    bool output_dot = false;
                    int output_pos = -1;
                    for (int i = LENGTH - 1; i >= LENGTH - bits; i--)
                    {
                        if (value[i] == 0 && !output_dot)
                            continue;
                        output_dot = true;
                        output_pos = i;
                        break;
                    }
                    if (output_dot)
                    {
                        printf(".");
                        for (int i = LENGTH - bits; i <=output_pos; i++)
                            printf("%d", value[i]);
                    }
                    printf("
    ");
                }
            }
        }
    };
    
    int main() {
        char* str=new char[6];
        int n;
    
        while(cin>>str>>n)
        {
            BigNumber* bn = new BigNumber(str, n);
            bn->pow();
            bn->print();
            delete bn;
        }
        return 0;
    }
    因为弱小,所以要变强,因为不想灭亡,所以选择战斗
  • 相关阅读:
    全球疫情实时监控——约翰斯&#183;霍普金斯大学数据大屏实现方案
    少儿编程崛起?2020年4月编程语言排名发布——Java,C,Python分列前三,Scratch挤进前20
    干货来了!阿里发布近300页Flink实战电子书
    Druid 0.17 入门(3)—— 数据接入指南
    Druid 0.17 入门(2)—— 安装与部署
    Flink 1.10 正式发布!——与Blink集成完成,集成Hive,K8S
    Druid入门(1)—— 快速入门实时分析利器-Druid_0.17
    程序员需要了解依赖冲突的原因以及解决方案
    每日一技|活锁,也许你需要了解一下
    Dubbo 服务 IP 注册错误踩坑经历
  • 原文地址:https://www.cnblogs.com/cmjason/p/3877190.html
Copyright © 2020-2023  润新知