poj 2109 Power of Cryptography

题目链接：

　　http://poj.org/problem?id=2109

题目描述：

　　给n，p。肯定有k(1<=k<=10⁹)，满足kⁿ=p，1<=n<=200,1<=p<=10¹⁰¹,问k是几？

解题思路：

　　由于p太大，无法对p进行开方，只有枚举k，又因为k太大又知道k的范围，所以很自然就想到了二分枚举，大树乘法加上二分枚举就可以解决。^_^……（奉上代码）！！

代码：　

#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#define INF 1000000000
#define N 110

int a[N], b[N], n, len;
char str[N];
int judge (int x);
int main ()
{
    int i, low, high, mid;
    while (scanf ("%d %s", &n, str) != EOF)
    {
        memset (a, 0, sizeof(a));
        len = strlen (str);
        for (i=0; i<len; i++)
            a[i] = str[i] - '0';

        low = 1;
        high = INF;

        while (low <= high)
        {
            mid = (low + high) / 2;
            int m = judge (mid);
            if (m < 0)
                low = mid + 1;
            else if (m > 0)
                high = mid - 1;
            else
                break;
        }
        printf ("%d
", mid);
    }
    return 0;
}

int judge (int x)
{
    int m , i, j;
    m = n * log10(x) + 1;
    if (m > len)
        return 1;
    else if (m < len)
        return -1;
    else
    {
        memset (b, 0, sizeof(b));
        b[0] = 1;
        for (i=0; i<n; i++)
        {
            int yu = 0;
            for (j=0; j<110; j++)
            {
                int s = x * b[j] + yu;
                b[j] = s % 10;
                yu = s / 10;
            }
        }
        for (i=0; i<len; i++)
            if (b[len-1-i] > a[i])
                return 1;
            else if (b[len-1-i] < a[i])
                return -1;
        return 0;
    }
}