POJ 1001 exponentiation seeking high precision power

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 are a taxing experience for many computer systems.

This problem requires so 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 a integer such that 0 < n <= 25.

Input

The input would consist of a set of pairs of values for R and N. The R value would occupy columns 1 through 6, and the N value would be in columns 8 and 9.

Output

The output would consist of one line for each line of input giving the exact value of r^n. Leading zeros should is suppress Ed in the output. Insignificant trailing zeros must not being printed. Don ' t print the decimal point if the result was an integer.

Sample Input

95.123 120.4321 205.1234 156.7592  998.999 101.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721.00000005148554641076956121994511276767154838481760200726351 20383542976301346240143992025569.92857370126648804114665499331870370751166629547672049395302429448126.7641210216181644302 0690903717327667290429072743629540498.1075960194566517745610440100011.126825030131969720661201

Hint

If you don ' t know what to determine wheather encounted the end of input:
sis a string and Nis an integer

C + +while (cin>>s>>n) {...} cwhile (scanf ("%s%d", S,&n) ==2)//to See  If the scanf read on 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 999999using 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&GT;=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;}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

POJ 1001 exponentiation seeking high precision power