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  • POJ1001(C++处理大数)

    Exponentiation
    Time Limit: 500MS   Memory Limit: 10000K
    Total Submissions: 158025   Accepted: 38470

    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

    #include <cstdio>
    #include <cstring>
    using namespace std;
    const int MAXN=105;
    struct BigInt{
        int e[MAXN],len,pp;
        BigInt()
        {
            memset(e,0,sizeof(e));
            len=0;
            pp=0;
        }
        void set(const char s[])
        {
            int l=strlen(s);
            for(int i=0;i<l;i++)
            {
                if(s[i]=='.')
                {
                    pp=l-i-1;
                    continue;
                }
                e[len++]=s[i]-'0';
            }
            for(int i=0;i<len/2;i++)
            {
                int t=e[i];
                e[i]=e[len-i-1];
                e[len-i-1]=t;
            }
            while(len>1&&e[len-1]==0)    len--;
        }
        BigInt operator*(const BigInt &b)
        {
            BigInt ret;
            for(int i=0;i<len;i++)
            {
                int up=0;
                for(int j=0;j<b.len;j++)
                {
                    int z=e[i]*b.e[j]+up+ret.e[i+j];
                    ret.e[i+j]=z%10;
                    up=z/10;
                }
                if(up!=0)
                {
                    ret.e[i+b.len]=up;
                }
            }
            ret.len=len+b.len;
            ret.pp=pp+b.pp;
            while(ret.len>1&&ret.e[ret.len-1]==0)    ret.len--;
            return    ret; 
        }
        void print()
        {
            int limit=0;
            while(e[limit]==0&&limit<pp)    limit++;//去后导0 
            if(pp>=len)
            {
                printf(".");
                for(int i=pp;i>len;i--)
                {
                    printf("0");
                }
                for(int i=len-1;i>=limit;i--)
                {
                    printf("%d",e[i]);
                }
                printf("
    ");
            }
            else
            {
                for(int i=len-1;i>=pp;i--)
                {
                    printf("%d",e[i]);
                }
                if(limit<pp)
                {
                    printf(".");
                    for(int i=pp-1;i>=limit;i--)
                    {
                        printf("%d",e[i]);
                    }
                }
                printf("
    ");
            }
        }
    };
    char buf[10];
    int n;
    int main()
    {
        while(scanf("%s%d",buf,&n)!=EOF)
        {
            BigInt t;
            t.set(buf);
            BigInt res;
            res.set("1.0");
            for(int i=0;i<n;i++)
            {
                res=res*t;
            }
            res.print();
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/program-ccc/p/5650682.html
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