UVA 10061－How many zero’s and how many digits ?

题目大意：给数字N和数字b，问N！转化成b进制后的位数和尾数的0的个数。

解题思路：一个m位的b进制数N，最小是b^(m-1)，最大不超过b^m，b^(m-1)≤N

#include <stdio.h>#include <iostream>#include <math.h>using namespace std;int main() {    double N, B;    while (scanf("%lf%lf", &N, &B) != EOF) {        double s = 0;        for (int i = 1; i <= N; i++) {            s = s + log(i) / (log(B) * 1.0);            }        int m = (int)B;        int y;        y = (int)s + 1;  //位数        int min = -1;        for (int i = 2; i <= m ; i++) {            int count = 0;            for (; m % i == 0; m = m / i) {                count++;            }            int count2 = 0;            if (count != 0) {                for (int j = i; j <= N; j++) {                    for (int k = j; k % i == 0; k = k / i)                        count2++;                }                int o = count2 / count;                if(o < min || min == -1)                    min = o;            }        }        if(min == -1)            min = 0;        printf("%d %d\n", min, y);    }    return 0;}