Exponential multiplication of an infinite number of poj 1001 exponentiation

Poj is doing very well. This question requires an infinitely large exponential multiplication result.

Basic requirements: infinitely large digits multiply

Additional requirements: ability to handle special situations-the key is to test this ability.

Therefore, the use cases in this question are particularly important, and even smart people will neglect some use cases.

The test on program robustness has reached the abnormal level.

Someone posted the token trial data address: http://poj.org/showmessage? Message_id = 76017.

With these cases, debugging will be completed in a few times.

Key examples I have missed:

000.10 1
000000 1
000.00 1
. 00000 0
000010 1

# Include <stdio. h> # include <iostream> # include <string> # include <vector> # include <algorithm> using namespace STD; bool standardizenumnodot (string & S) {While (! S. empty () & '0' = s [0]) s. erase (S. begin (); If (S. empty () S = "0"; // when n = 1, 0 bool notdot = true; For (unsigned I = 0; I <S. size () & notdot; I ++) {If ('. '= s [I]) notdot = false;} If (notdot) return true; while (! S. Empty () & '0' = s [S. Size ()-1]) S. Erase (S. End ()-1); If (! S. empty ()&&'. '= s [S. size ()-1]) s. erase (S. end ()-1); If (S. empty () S = "0"; return false;} int handledecimalpoint (string & S) {If (standardizenumnodot (s) return 0; int fraction = 0; int J = 0; For (unsigned I = 0; I <S. size (); I ++) {If (fraction> 0) fraction ++; If (s [I]! = '. ') S [J ++] = s [I]; else fraction ++;} s. erase (S. end ()-1); Return fraction-1;} string mulstr (string a, string B) {If ("0" = A | "0" = B) return "0"; int ap = handledecimalpoint (a); int BP = handledecimalpoint (B); string ans (. size () + B. size (), '0'); For (INT I =. size ()-1; I> = 0; I --) {int carry = 0; int an = A [I]-'0'; For (Int J = B. size ()-1; j> = 0; j --) {int bn = B [J]-'0 '; int sum = An * bn + carry + ans [I + J + 1]-'0'; carry = sum/10; ans [I + J + 1] = sum % 10 + '0';} If (carry) ans [I] + = carry ;} if (AP> 0 | BP> 0) ans. insert (ans. end ()-ap-BP ,'. '); standardizenumnodot (ANS); Return ans;} string spow (string S, int N) {If (S. empty () | "0" = s) Return "0"; // For program robustness, add If (0 = N) Return "1 "; if (1 = N) return s; string dividestr = spow (S, N/2); dividestr = mulstr (dividestr, dividestr); If (N % 2) dividestr = mulstr (dividestr, S); Return dividestr;} void exponentiation () {string s; int N; while (CIN >>s> N) {standardizenumnodot (s ); // cout <spow (S, N) <Endl ;}} int main () {exponentiation (); Return 0;} When n = 1 ;}

This algorithm takes 0 ms, haha.