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;

}