uva11029 - Leading and Trailing

题目：

　　求n的k次方，然后将答案用前三位和最后三位表示。

Sample Input 2 123456 1 123456 2 Sample Output 123...456 152...936

分析：　　题目中其实有提示，用double来表示n的k次方，double神奇的地方在于能转化为string类型的字符串。用到了sprintf这个函数。代码：　　

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
const int INF = 1000000000;
#define MAX 200

int n, k;

ll power_mod(ll a, ll n, ll mod)
{
    if(n == 0) return 1LL;
    ll ans = power_mod(a, n/2, mod);
    ans = ans*ans%mod;
    if(n%2) ans = ans*a%mod;
    return ans;
}

double pow(double a, int n)
{
    if(n == 0) return 1;
    double ans = pow(a, n/2);
    ans = ans*ans;
    if(n%2) ans = ans*a;
    while( ans > INF ) ans /= INF;
    return ans;
}

int main()
{
//    freopen("input.txt", "r", stdin);
    int caseNum;
    scanf("%d", &caseNum);
    while(caseNum--)
    {
        scanf("%d %d", &n, &k);
        double head = pow( (double)n, k );
        char str[MAX];
        sprintf(str, "%lf", 1000*head);
        str[3] = ‘\0‘;

        ll last = power_mod(n, k, 1000);
        printf("%s...%03lld\n", str, last);
    }

    return 0;
}