【百炼oj】1001:Exponentiation

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 12
0.4321 20
5.1234 15
6.7592  9
98.999 10
1.0100 12

样例输出

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

#include<stdio.h>
#include<string.h>
#include<math.h>
#define p 7
#define len 120
int main(){
    int mult(int*a,int*b);
    int n,i,j,k,location,flag;
    int an[len],bn[len];
    char sz[p];
    while(scanf("%s %d",sz,&n)==2){
            memset(an,0,sizeof(an));
            memset(bn,0,sizeof(bn));
            an[0]=1;
            location=-1;
        for(i=0;i<p-1;i++){
            if(sz[i]=='.'){
                location=i;//memory the location of the dot
                location=n*(p-2-i);//get the new location
                for(j=i;j<p-1;j++)
                    sz[j]=sz[j+1];//remove the dot
                    sz[p-1]='\0';
            }
        }
        for(i=0;i<strlen(sz);i++)
            bn[i]=sz[strlen(sz)-1-i]-'0';
             for(i=0;i<n;i++)
            mult(an,bn);
        if(location==-1){
        flag=0;
        for(i=len-1;i>=0;i--){
            if(flag)printf("%d",an[i]);
        else if(an[i]){
            printf("%d",an[i]);
            flag=1;
            }
        }
         if(!flag)printf("0");
         printf("\n");
        }//整数情况
        else//小数
        {
            for(i=0;;i++){
                 k=i;
                if(an[i]!=0)break;
            }
           for(i=len-1;i>=k;i--){
               if(an[i])break;
           }
           if(i<location){
            printf(".");
            for(i=location-1;i>=k;i--)printf("%d",an[i]);
            printf("\n");
           }
           else{
                if(k<location){
            flag=0;
            for(i=len-1;i>=k;i--){
            if(flag){
                    printf("%d",an[i]);
                    if(i==location&&i!=k)printf(".");
            }
            else if(an[i]){
            printf("%d",an[i]);
            if(i==location&&i!=k)printf(".");
            flag=1;
            }
        }
                    printf("\n");
           }
           else{
            flag=0;
            for(i=len-1;i>=location;i--){
            if(flag){
                    printf("%d",an[i]);
            }
            else if(an[i]){
            printf("%d",an[i]);
            flag=1;
            }
        }
          if(!flag)printf("0");
          printf("\n");
           }
           }
        }
    }
}
int mult(int*a,int*b){
    int i,j,k,l;
    int c[len];
    memset(c,0,sizeof(c));
    for(i=0;i<len;i++){
        k=0;
        for(j=0;j<len-i;j++){
            l=c[i+j]+a[j]*b[i]+k;
            c[i+j]=l%10;
            k=l/10;
        }
    }
    for(i=0;i<len;i++)a[i]=c[i];
}