POJ 1001 Exponentiation 求高精度幂

最新推荐文章于 2019-08-19 17:54:23 发布

蒟蒻的ACMer

最新推荐文章于 2019-08-19 17:54:23 发布

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分类专栏：模拟只会猜题意文章标签： c++ iostream poj 编程 numbers

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本文介绍了一种解决特定大数幂运算问题的方法，通过精确计算实数R的整数次幂R^n（0.0<R<99.999且0<n≤25），避免了传统浮点数运算带来的精度损失。文中详细展示了如何处理输入数据、进行运算及输出结果的具体步骤。

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Exponentiation

Time Limit: 500MS		Memory Limit: 10000K
Total Submissions: 147507		Accepted: 36006

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 R ⁿ 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

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

East Central North America 1988

ACcode：

#include <iostream>
#include <cstring>
#include <cstdio>
#define maxn 999999
using namespace std;
char a[10];
int add[10],ans[maxn],res[maxn],n,doc,len;
void put1(){
    if(doc==-1){
        bool flag=false;
        for(int i=ans[0];i>=1;--i){
            if(ans[i]!=0)flag=true;
            if(flag)printf("%d",ans[i]);
        }
    }
    else {
        doc=5-doc;
        doc*=n;
        int flag1,flag2;
        for(int i=1;i<=ans[0];++i)
            if(ans[i]!=0){
                flag1=i;
                break;
            }
        for(int i=ans[0];i>=1;--i)
            if(ans[i]!=0){
                flag2=i;
                break;
            }
        if(flag2<doc)flag2=doc;
        if(flag1>doc)flag1=doc+1;
        bool flag=false;
        while(flag2>=flag1){
            if(flag2==doc)putchar('.');
            printf("%d",ans[flag2]);
            flag2--;
        }
    }
    putchar('\n');
}
void fun(){
    memset(res,0,sizeof(res));
    for(int i=1;i<=ans[0];i++)  {
        for (int j=1;j<=len;j++)  {
            res[i+j-1]+=ans[i]*add[j];
            if (res[i+j-1]>9)
            {
                res[i+j] += res[i+j-1] / 10;
                res[i+j-1] %= 10;
            }
        }
    }
    if(res[ans[0]+len-1]>9){
        res[ans[0]+len]+=res[ans[0]+len-1]/10;
        res[ans[0]+len-1]%=10;
    }
    ans[0]=ans[0]+len;
    for(int i=1;i<=ans[0];i++)ans[i]=res[i];
}
int main(){
    while(cin>>a>>n){
        if(n==0){
            printf("1\n");
            continue;
        }
        memset(ans,0,sizeof(ans));
        memset(add,0,sizeof(a));
        doc=-1;len=6;
        int i,j,cnt=0;
        for(i=0;i<6;++i){
            if(a[i]=='.'){
                doc=i;
                len=5;
            }
            if(a[i]=='0')cnt++;
        }
        if(cnt==len){
            printf("0\n");
            continue;
        }
        for(i=5,j=1;i>=0;--i)
            if(i!=doc){
                add[j]=(a[i]-'0');
                ans[j]=add[j];
                j++;
            }
            ans[0]=len;
        for(i=1;i<n;++i)
            fun();
            put1();
    }
    return 0;
}