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  • light_oj 1282 求n^k的前几位数和后几位数

    light_oj 1282 求n^k的前几位数和后几位数

    E - Leading and Trailing
    Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

    Description

    You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of nk.

    Input

    Input starts with an integer T (≤ 1000), denoting the number of test cases.

    Each case starts with a line containing two integers: n (2 ≤ n < 231) and k (1 ≤ k ≤ 107).

    Output

    For each case, print the case number and the three leading digits (most significant) and three trailing digits (least significant). You can assume that the input is given such that nk contains at least six digits.

    Sample Input

    5

    123456 1

    123456 2

    2 31

    2 32

    29 8751919

    Sample Output

    Case 1: 123 456

    Case 2: 152 936

    Case 3: 214 648

    Case 4: 429 296

    Case 5: 665 669

    题意:输出n^k的前三位和后三位。

    思路:后三位直接快速幂取模,前三位化为科学计数法取对数推导:

                n^k=a.bc*10^m ( m为n^k的位数,即m=(int)lg(n^k)=(int)(k*lgn) );

                求对数:  k*lgn=lg(a.bc)+m

                 即 a.bc=10^(k*lgn-m)=10^(k*lgn-(int)(k*lgn));

                  abc=a.bc*100;

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include<vector>
    #include<stack>
    #include<queue>
    #include<set>
    #include<map>
    #include<string>
    #include<math.h>
    #include<cctype>
    
    using namespace std;
    
    typedef long long ll;
    const int maxn=1000100;
    const int INF=(1<<29);
    const double EPS=0.0000000001;
    const double Pi=acos(-1.0);
    const int p=1000;
    
    ll n,k;
    
    ll qpow(ll n,ll k)
    {
        ll res=1;
        while(k){
            if(k&1) res=(res%p)*(n%p)%p;
            n=(n%p)*(n%p)%p;
            k>>=1;
        }
        return res;
    }
    
    ll f(ll n)
    {
        double x=k*log10(n)-(int)(k*log10(n));
        return pow(10,x)*100;
    }
    
    int main()
    {
        int T;cin>>T;
        int tag=1;
        while(T--){
            cin>>n>>k;
            printf("Case %d: %lld %03lld
    ",tag++,f(n),qpow(n,k));
        }
        return 0;
    }
    View Code
    没有AC不了的题,只有不努力的ACMER!
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  • 原文地址:https://www.cnblogs.com/--560/p/4550278.html
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