题解:求一个数的次幂,然后输出前三位和后三位,后三位注意有前导0的情况. 后三位直接用快速幂取模求解. 前三位求得时候只需要稍微变形一下,可以把乘过的结果拆成用科学计数法,那么小数部分只有由前面决定,所以取前三位利用double来计算就可以了. #include <bits/stdc++.h> using namespace std; typedef long long ll; const int Mod = 1000; ll ppow(ll a, ll k) // 后三位 { ll ans…

[LightOJ1282]Leading and Trailing(数论) 题面 Vjudge 给定两个数n,k 求n^k的前三位和最后三位 题解 这题..真的就是搞笑的 第二问,直接输出快速幂\(mod \ 1000\)的值,要补前导零 第一问...就是搞笑的 依旧是快速幂 但是用double来算 每次中间值只要大于1000 直接除得小于1000就行了 不会就看代码把.. 这题就是搞笑的... #include<iostream> #include<cstdio> #includ…

Pairs Forming LCM (LightOJ - 1236)[简单数论][质因数分解][算术基本定理](未完成) 标签: 入门讲座题解 数论 题目描述 Find the result of the following code: long long pairsFormLCM( int n ) { long long res = 0; for( int i = 1; i <= n; i++ ) for( int j = i; j <= n; j++ ) if( lcm(i, j) ==…

Sigma Function (LightOJ - 1336)[简单数论][算术基本定理][思维] 标签: 入门讲座题解 数论 题目描述 Sigma function is an interesting function in Number Theory. It is denoted by the Greek letter Sigma (σ). This function actually denotes the sum of all divisors of a number. For exam…

Goldbach`s Conjecture(LightOJ - 1259)[简单数论][筛法] 标签: 入门讲座题解 数论 题目描述 Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states: Every even integer, greater than 2, can be expressed as the sum of…

Leading and Trailing https://vjudge.net/contest/288520#problem/E 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), den…

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 st…

题意:求nk的前三位和后三位. 分析: 1.后三位快速幂取模,注意不足三位补前导零. 补前导零:假如nk为1234005,快速幂取模后,得到的数是5,因此输出要补前导零. 2.前三位: 令n=10a,则nk=10ak=10x+y,x为ak的整数部分,y为ak的小数部分. eg:n=19,k=4,则nk=130321, a=log10(n)=1.2787536009528289615363334757569 ak=5.1150144038113158461453339030277, 因此,x=5,…

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:求一个数n的k次方后的前三位与后三位.并且后三位要求控制格式. 思路:这道题后三位可以用快速幂求出来,前三位就要用到log了.先说一下怎么求n^k的前三位. 我先设10^p=n^k,同时取log10…

http://www.lightoj.com/volume_showproblem.php?problem=1013   Yes, you are developing a 'Love calculator'. The software would be quite complex such that nobody could crack the exact behavior of the software. So, given two names your software will gene…