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  • Leading and Trailing(数论/n^k的前三位)题解

    Leading and Trailing

     

    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

    思路:

    n^k的后三位用快速幂。前三位计算方法:指数级一般用对数解决,令x为a^k整数部分, y为a^k小数部分 ,所以10^(x+y)==n^k,其中10^x是10.....0,那么10^y其实就是各个位数上的值,所以10^y前三位就是n^k前三位

    新学了一个函数 double a=modf(double x,double *i ),返回x的整数部分给i,小数部分给a

    代码:

    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cctype>
    #include<queue>
    #include<cmath>
    #include<string>
    #include<map>
    #include<stack> 
    #include<set>
    #include<vector>
    #include<iostream>
    #include<algorithm>
    #define INF 0x3f3f3f3f
    #define ll long long
    const int N=10000003;	//18
    const int MOD=1000; 
    using namespace std;
    int pow_mod(ll a,ll b){
    	ll ans=1;
    	while(b){
    		if(b&1) ans=ans*a%MOD;
    		a=a*a%MOD;
    		b/=2;
    	} 
    	return (int)ans;
    }
    int main(){
    	int T,res1,res2,num=1;
    	ll a,k;
    	double y;
    	double x;
    	scanf("%d",&T);
    	while(T--){
    		scanf("%lld%lld",&a,&k);
    		res2=pow_mod(a,k);
    		y=modf((double)(k*log10(a)),&x);
    		res1=floor(pow(10,y)*100);
    		printf("Case %d: %d %03d
    ",num++,res1,res2);
    	}
    	return 0;
    }



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  • 原文地址:https://www.cnblogs.com/KirinSB/p/9409120.html
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