light oj 1090 - Trailing Zeroes (II)

Find the number of trailing zeroes for the following function:

ⁿC_r * p^q

where n, r, p, q are given. For example, if n = 10, r = 4, p = 1, q = 1, then the number is 210 so, number of trailing zeroes is 1.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains four integers: n, r, p, q (1 ≤ n, r, p, q ≤ 10⁶, r ≤ n).

Output

For each test case, print the case number and the number of trailing zeroes.

Sample Input	Output for Sample Input
2 10 4 1 1 100 5 40 5	Case 1: 1 Case 2: 6

 

题意：求组合数C（n，r）乘p的q次方末尾0个数。

分析：大家都做过n！末尾0的个数这一题吧，这题类似。C(n，r）=n！/（（n-r）！×r！）；

由末尾0，来自最基本因子2×5，分别求出式子中因子2，因子5的个数，取两者最小值即可

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
using namespace std;
int sum5, sum2;
void five(int n)//计算n！中包含因子5的个数
{
    while(n!=0)
        sum5+=n/5,n/=5;

}
void two(int n)//计算n！中包含因子2的个数
{
    while(n!=0)
        sum2+=n/2,n/=2;
}
int main()
{
    int T, t=1, n, r, p, q;
    scanf("%d", &T);
    while(T--)
    {
        sum2=sum5=0;
        scanf("%d%d%d%d", &n, &r, &p, &q);
        int s=p;
        while(s%5==0)//求p中因子5的个数
            sum5++,s/=5;
        while(p%2==0)//求p中因子2的个数
            sum2++,p/=2;
        sum5*=q;sum2*=q;
        five(-r);two(-r);
        five(n);two(n);
        five(r-n);two(r-n);
        printf("Case %d: %d\n", t++, min(sum5, sum2));
    }
}