Leading and Trailing（LightOJ - 1282）
Leading and Trailing（LightOJ - 1282）
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 = 1;
while(k)
{
if(k%2)ans *= a;
ans %= Mod;
a *= a;
a %= Mod;
k /= 2;
}
return ans;
}
double Merge(double x)
{
while(x >=1000.0)
{
x /= 10.0;
}
return x;
}
double dopow(double a, int k) // 前三位
{
double ans = 1.0;
while(k)
{
if(k%2)ans *= a;
ans = Merge(ans);
a *= a;
a = Merge(a);
k /= 2;
}
return ans;
}
int main()
{
int T,cas = 0;
ll n, k;
scanf("%d",&T);
while(T--)
{
cas ++;
scanf("%lld %lld", &n, &k);
ll ans2 = ppow(n,k);
double m = n * 1.0;
double ans1 = dopow(m,k);
printf("Case %d: %d %03lld\n",cas, (int)ans1, ans2);
}
return 0;
}Problem
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