LightOJ 1282 Leading and Trailing (数学)

题意：求 n^k 的前三位和后三位。

析：后三位，很简单就是快速幂，然后取模1000，注意要补0不全的话，对于前三位，先取10的对数，然后整数部分就是10000....，不用要，只要小数部分就好，然后取前三位。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const int mod = 1000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}

int fast_pow(int a, int n){
int res = 1;
a %= mod;
while(n){
if(n & 1)  res = res * a % mod;
a = a * a % mod;
n >>= 1;
}
return res;
}

int main(){
int T;  cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%d %d", &n, &m);
double x = m * log10(n) - (int)(m * log10(n));
int ans1 = (int)(pow(10, x) * 100);
int ans2 = fast_pow(n, m);
printf("Case %d: %d %03d\n", kase, ans1, ans2);
}
return 0;
}