POJ
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Exponentiation
Time Limit: 500MS Memory Limit: 10000KTotal Submissions: 171554 Accepted: 41539
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
95.123 120.4321 205.1234 156.7592 998.999 101.0100 12
Sample Output
548815620517731830194541.899025343415715973535967221869852721.0000000514855464107695612199451127676715483848176020072635120383542976301346240143992025569.92857370126648804114665499331870370751166629547672049395302429448126.76412102161816443020690903717327667290429072743629540498.1075960194566517745610440100011.126825030131969720661201
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){...}cwhile(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
输入n,m,输出n的m次幂。。
func1获取有效数字。(写的有点繁琐)
func2大数乘法。
#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;int func1(char *a){ int point=0,vju1=0; for(int i=0; i<6; i++) if(a[i]=='.') vju1=1; if(vju1) for(int i=5; i>=0; i--) if(!vju1) if(a[i]=='.') break; else point++; else if(a[i]!=48) { vju1=0; if(a[i]!='.') point++; else break; } else a[i]=0; int n=strlen(a),vju2=0,flag=0; char b[187]; for(int i=0; i<n; i++) { if((vju2&&a[i]!='.')||(!vju2&&a[i]!=48&&a[i]!='.')) b[flag++]=a[i]; if(!vju2&&a[i]!=48) vju2=1; } b[flag]=0; strcpy(a,b); return point;}void func2(char *a,char *b){ char s[187]; int n=strlen(a),m=strlen(b),t[187]= {}; reverse(a,a+n); reverse(b,b+m); for(int i=0; i<n; i++) for(int j=0; j<m; j++) t[i+j]+=(a[i]-48)*(b[j]-48); for(int i=0; i<=180; i++) t[i+1]+=t[i]/10,t[i]=t[i]%10; int flag=0,vju=0; for(int i=185; i>=0; i--) if(vju) s[flag++]=t[i]+48; else if(t[i]!=0) vju=1,s[flag++]=t[i]+48; s[flag]=0; strcpy(a,s); reverse(b,b+m);}int main(){ char a[187]; int n; while(cin>>a>>n) { int point=func1(a),t=n-1; char b[187]; strcpy(b,a); while(t--) func2(a,b); int m=strlen(a); if(m-point*n<0) { cout<<'.'; int temp=point*n-m; while(temp--) printf("0"); } for(int i=0; i<m; i++) { if(i==m-point*n) printf("."); printf("%c",a[i]); } cout<<endl; } return 0;}
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