LightOJ - 1282 - Leading and Trailing(数学技巧,快速幂取余)
链接:
https://vjudge.net/problem/LightOJ-1282
题意:
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.
思路:
后三位快速幂取余,考虑前三位。
\(n^k\)可以表示为\(a*10^m\)即使用科学计数法。
对两边取对数得到\(k*log10(n) = log10(a)+m\)
则x = log10(a)是k*log10(n)的小数部分。
a = pow(10, x).就是科学计数法的前面部分。
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<math.h>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;
const int INF = 1e9;
const int MAXN = 1e6+10;
const int MOD = 1e9+7;
LL n, k;
LL PowMod(LL a, LL b)
{
LL res = 1;
while(b)
{
if (b&1)
res = res*a%1000;
a = a*a%1000;
b >>= 1;
}
return res;
}
int main()
{
int t, cnt = 0;
scanf("%d", &t);
while(t--)
{
printf("Case %d:", ++cnt);
scanf("%lld%lld", &n, &k);
double v = 1.0*k*log10(n);
v -= (LL)v;
LL r1 = (LL)(pow(10, v)*100);
LL r2 = PowMod(n, k);
printf(" %lld %03lld\n", r1, r2);
}
return 0;
}