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 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).
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.
5 123456 1 123456 2 2 31 2 32 29 8751919
Case 1: 123 456 Case 2: 152 936 Case 3: 214 648 Case 4: 429 296 Case 5: 665 669
解題方法寫在這兒
代碼如下
#include<stdio.h>#include<cmath>typedef long long LL;LL _pow(LL a,int n,LL MOD){ LL ans=1; while(n){ if(n%2) ans=ans*a%MOD; a=a*a%MOD; n>>=1; } return ans;}int main(){ LL n;int T,k; scanf("%d",&T); for(int t=1;t<=T;t++){ scanf("%lld%d",&n,&k); double a=(double)k*log10(n)-floor((double)k*log10(n)); a=pow(10.0,a); a=floor(100*a); LL b=_pow(n,k,1000); printf("Case %d: %0.0lf %0.3lld/n",t,a,b); } return 0;}新聞熱點
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