[Problem] uva11029 leading and trailing

Link: http://vjudge.net/problem/viewProblem.action? Id = 19597

Description: calculates the first three digits and the last three digits of N ^ K.

Idea: the question should solve two problems. The last three digits can be used to evaluate the modulo of the higher power. The following describes how to evaluate the first three digits.

N ^ k = (10 ^ lg n) ^ K = 10 ^ (K * lg n)

For ease of description, make X = n ^ K. The integer part of lg x indicates the number of digits of X. If the integer is Zs and the fractional part is XS, x = (10 ^ ZS) * (10 ^ XS ). (10 ^ ZS) in the form of 100 ......, Like the number of digits in X, (10 ^ XS) is at 100 ...... Then, modify the number on each digit to convert it into X. So (10 ^ XS) * 100 is the first three digits of X (if you don't think clearly, you can draw a picture on the paper ).

 

The following is my implementation.

 1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 using namespace std; 5 #define MOD 1000 6 int T,N,K; 7 int Trail; 8 double Lead; 9 inline void Get_int(int &Ret)10 {11     char ch;12     bool flag=false;13     for(;ch=getchar(),ch<‘0‘||ch>‘9‘;)14         if(ch==‘-‘)15             flag=true;16     for(Ret=ch-‘0‘;ch=getchar(),ch>=‘0‘&&ch<=‘9‘;Ret=Ret*10+ch-‘0‘);17     flag&&(Ret=-Ret);18 }19 int My_Pow(int n,int k)20 {21     if(k==1)22         return n%MOD;23     int p=My_Pow(n,k/2);24     if(k%2)25         return (p*p*(n%MOD))%MOD;26     return (p*p)%MOD;27 }28 int main()29 {30     Get_int(T);31     while(T--)32     {33         Get_int(N);Get_int(K);34         Lead=(double)K*log10(N);35         Lead=Lead-(int)Lead;36         Lead=pow((double)10,2+Lead);37         Trail=My_Pow(N,K);38         printf("%d...%03d\n",(int)Lead,Trail);39     }40     return 0;41 }

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