LightOJ1282 Leading and Trailing

题面

给定两个数n,k 求n^k的前三位和最后三位

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 < 2^31) and k (1 ≤ k ≤ 10^7).

Solution

后面一问是搞笑的

前面一问？

也是搞笑的，用double除成小于1的数算就好了

滑稽题

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

IL ll Read(){
	char c = '%'; ll x = 0, z = 1;
	for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
	return x * z;
}

ll n, k, ansr, ansl;

IL ll Powr(RG ll x, RG ll y, RG ll MOD){
	RG ll cnt = 1;
	for(; y; y >>= 1, x = x * x % MOD) if(y & 1) cnt = cnt * x % MOD;
	return cnt;
}

IL double Powl(RG double x, RG ll y){
	while(x >= 1) x /= 10; RG double cnt = 1.0;
	for(; y; y >>= 1){
		if(y & 1){
			cnt = cnt * x;
			while(cnt < 0.1) cnt *= 10;
		}
		x = x * x;
		while(x < 0.1) x *= 10;
	}
	while(cnt < 100) cnt *= 10;
	return cnt;
}

int main(RG int argc, RG char *argv[]){
	for(RG int T = Read(), i = 1; i <= T; ++i){
		n = Read(); k = Read();
		ansl = Powl(n, k); ansr = Powr(n, k, 1000);
		printf("Case %d: %lld ", i, ansl);
		printf("%lld%lld%lld\n", ansr / 100, (ansr % 100) / 10, ansr % 10);
	}
	return 0;
}