| Exponentiation |
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 .
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 Rn. Leading zeros and insignificant trailing zeros should be suppressed in the output.
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
Miguel A. Revilla
2000-02-09
计算高精度幂R^n
就是高精度乘法反复算即可，输出时算一下小数点的位置将其输出即可
一个算是坑的地方就是输出时末尾的0在去掉的时候要注意别删多了，比如10.000 10这组测试数据，别输出了1
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 5 using namespace std; 6 7 int main() 8 { 9 int r; 10 int a[200],b[10],c[200]; 11 char s[20]; 12 13 14 while(gets(s)) 15 { 16 memset(a,0,sizeof(a)); 17 memset(b,0,sizeof(b)); 18 a[0]=1; 19 20 int t=0,pos; 21 for(int i=5;i>=0;i--) 22 if(s[i]!=‘.‘) 23 b[t++]=s[i]-‘0‘; 24 else 25 pos=i; 26 sscanf(&s[7],"%d",&r); 27 28 for(int k=1;k<=r;k++) 29 { 30 memset(c,0,sizeof(c)); 31 for(int i=0;i<t;i++) 32 { 33 int u=0; 34 for(int j=0;i+j<200;j++) 35 { 36 int temp=c[j+i]; 37 c[j+i]=(a[j]*b[i]+c[j+i]+u)%10; 38 u=(a[j]*b[i]+temp+u)/10; 39 } 40 } 41 for(int i=0;i<200;i++) 42 a[i]=c[i]; 43 } 44 45 int Start,End; 46 for(Start=199;a[Start]==0;Start--); 47 for(End=0;a[End]==0;End++); 48 if(Start<(5-pos)*r) 49 { 50 putchar(‘.‘); 51 Start=(5-pos)*r-1; 52 } 53 if(End>(5-pos)*r) 54 End=(5-pos)*r; 55 for(int i=Start;i>=End;i--) 56 if(i==(5-pos)*r) 57 printf("%d.",a[i]); 58 else 59 printf("%d",a[i]); 60 putchar(‘\n‘); 61 } 62 63 return 0; 64 }