zoukankan      html  css  js  c++  java
  • POJ 1001 Exponentiation

    Time Limit: 500MS   Memory Limit: 10000K
    Total Submissions: 179524   Accepted: 43274

    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

    Hint

    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)
    {
    ...
    }
    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

    思路:先化成整数,然后跑高静。
    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n,sum,poi;
    char s[601];
    int num[6001],ans[6001],bns[6001];
    void pre1(){
        for(int i=5;i>=0;i--)
            if(s[i]=='.')    break;
            else sum++;
        sum*=n;int flag=0;num[0]=5;
        for(int i=0;i<=4;i++)
        if(s[i]=='.')    flag=1,s[i]=s[i+1];
        else if(flag)    s[i]=s[i+1];    
        for(int i=4;i>=0;i--)    num[5-i]=s[i]-'0';
        for(int i=5;i>=1;i--)    if(num[i]==0)num[0]--;else break;
        //for(int i=num[0];i>=1;i--)    cout<<num[i];cout<<endl;
    }
    void work(){
        memset(bns,0,sizeof(bns));
        for(int i=1;i<=num[0];i++){
            for(int j=1;j<=ans[0];j++){
                bns[j+i-1]+=num[i]*ans[j];
                bns[0]=j+i-1;
                if(bns[j+i-1]>=10){
                    if(j+i>ans[0])    bns[0]=j+i;
                    bns[j+i]+=bns[j+i-1]/10;
                    bns[j+i-1]%=10;
                }
            }
        }
        
        for(int i=0;i<=bns[0];i++)    ans[i]=bns[i];
        for(int i=ans[0];i>=1;i--)    if(ans[i]==0)ans[0]--;else break;
    }
    void work1(){
        for(int i=ans[0];i>=poi;i--){
            if(i==sum)    cout<<".";
            cout<<ans[i];
        }
        cout<<endl;
    }
    void work2(){
        cout<<".";
        for(int i=ans[0];i>=poi;i--)
        cout<<ans[i];
        cout<<endl;
    }
    void work3(){
        cout<<".";
        for(int i=1;i<=sum-ans[0];i++)
        cout<<"0";
        for(int i=ans[0];i>=poi;i--)    cout<<ans[i];cout<<endl;
    }
    int main(){
        while(scanf("%s%d",s,&n)!=EOF){
            memset(ans,0,sizeof(ans));sum=0;
            memset(num,0,sizeof(num));pre1();
            for(int i=0;i<=5;i++)    ans[i]=num[i];
            for(int i=1;i<n;i++)    work();
            //for(int i=ans[0];i>=1;i--)    cout<<ans[i];cout<<endl;
            poi=1;
            for(int i=1;i<=ans[0];i++)
                if(ans[i]!=0||i>sum){ poi=i;break; }        
            if(ans[0]>sum)    work1();
            else if(ans[0]=sum) work2();
            else if(ans[0]<sum)    work3();
        }
    }
    /*
    95.123 12
    0.4321 20
    5.1234 15
    6.7592  9
    98.999 10
    1.0100 12
    */
    细雨斜风作晓寒。淡烟疏柳媚晴滩。入淮清洛渐漫漫。 雪沫乳花浮午盏,蓼茸蒿笋试春盘。人间有味是清欢。
  • 相关阅读:
    安装VCSA6.5(vCenter Server Appliance 6.5)
    VMware文章总结
    kbmmw 的远程桌面功能2-android手机端
    kbmmw 的远程桌面功能
    delphi 中的win32 以外到平台的字符串处理一定慢吗?(转载)
    delphi 中如何访问另一个类中到私有方法?(转载)
    使用 kbmmw 的ORM开发纯REST数据库访问服务
    kbmmw 5.05.00 发布
    使用delphi-cross-socket 开发kbmmw smart http service
    利用Delphi-cross-socket 库提升kbmmw 跨平台开发
  • 原文地址:https://www.cnblogs.com/cangT-Tlan/p/9192105.html
Copyright © 2011-2022 走看看