light_oj 1282 求n^k的前几位数和后几位数
2016-08-24 08:48
302 查看
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 TrailingTime Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit Status Practice LightOJ
1282
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\n",tag++,f(n),qpow(n,k)); } return 0; }
相关文章推荐
- Light-oj-1282 Leading and Trailing(数学取位)
- light_oj 1282
- E - Leading and Trailing Light oj 1282
- Light-oj 1282 - Leading and Trailing
- 【Light-oj】-1282 - Leading and Trailing(数论,快速幂,log,好)
- Light-oj-1045 lightoj1045 - Digits of Factorial (N!不同进制的位数)
- light oj 1422 - Halloween Costumes (区间dp)
- 【Light-oj】-1198 - Karate Competition(贪心+双端队列)
- 【Light-oj】-1022 - Circle in Square(水)
- 杭电 OJ 1282 回文数猜想
- OJ上关于阶乘位数的公式
- 【light-oj】-1109 - False Ordering(数学)
- Light OJ Integer Divisibility 【取模】
- Light OJ Basic Math
- 寒假刷oj——折腾五位数
- Crazy Calendar Light OJ 1393 (Nim博弈)
- Light oj Trailing Zeroes (III) (二分查找)
- 【Light-oj】-Integer Divisibility(同余定理)
- POJ 2576 / Light OJ 1147 Tug of War 状态压缩DP
- 杭电OJ(HDOJ)1018题:求n的阶乘位数(数学公式)