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;
}