Exponentiation
| Time Limit: 500MS | Memory Limit: 10000K | |
| Total Submissions: 127046 | Accepted: 31014 |
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.
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
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
与UVa 748是同一道题(http://www.cnblogs.com/lzj-0218/p/3528164.html)，但是在输出的地方有一点小小的差别，就是这POJ的这道题中，如果算出来是个整数的话，不要输出小数点
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&&i!=End) 57 printf("%d.",a[i]); 58 else 59 printf("%d",a[i]); 60 putchar(‘\n‘); 61 } 62 63 return 0; 64 }