LightOJ 1090Trailing Zeroes (II) [分解]【数论】

题目链接 ：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=120197#problem/Y

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Trailing Zeroes (II)
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit Status Practice LightOJ 1090

Description

Find the number of trailing zeroes for the following function:

C(n,r) x 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
2
10 4 1 1
100 5 40 5
Sample Output
Case 1: 1
Case 2: 6

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题目大意 ： 就是求C(n,r) x p^q 末尾有几个 0

题解 : 求末尾有几个0 就将其展开成(2^n) * (5^m) * k 然后输出min（n，m） 即可。。

但是数据量特别大 所以需要先预处理

用 five[N],two[N]出 N！中n，m的值(值的含义如上所述) 然后 稍加操作即可求出C(n,m)中n,m的值 (这里并不难,看代码就明白了)

剩下的p^q 就更好办了 p=(2^n) * (5^m) * k
求出的n，m的值在乘上q

最后加上C(n,m)中n,m的值 就是方程里n,m的值

总复杂度 O（1e6*（log（2,1e6）+log（5,1e6））+T *（log（2,p）+log（5,p）））

附本题代码

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#include <stdio.h>#include <stdlib.h>#include <string.h>#include <limits.h>#include <malloc.h>#include <ctype.h>#include <math.h>#include <string>#include <iostream>#include <algorithm>#include <map>#include <vector>#include <set>using namespace std;#define LL long long int#define uLL unsigned long long int#define _LL __int64struct Num{    int five,two;};Num Search( int n){    Num a;    a.five=a.two=0;    while(n%2==0)    {        n/=2;        a.two++;    }    while(n%5==0)    {        n/=5;        a.five++;    }    return a ;}Num num[1010101];Num add(Num a,Num b){    Num tem;    tem.five=a.five+b.five;    tem.two =a.two +b.two;    return tem;}Num jian(Num a,Num b){    Num tem;    tem.five=a.five-b.five;    tem.two =a.two -b.two;    return tem;}void init(){    memset(num,0,sizeof(num));    for(int i=2;i<=1e6;i++)    {        num[i]=add(num[i-1],Search(i));    }    return ;}int solve(Num a){    return min(a.five,a.two);}int main(){    init();    LL t,pp=0;    scanf("%llu",&t);    while(t--)    {        int n,p,q,r;        scanf("%d%d%d%d",&n,&r,&p,&q);        Num m;        m=Search(p);        m.five*=q,m.two*=q;        Num sum = add(jian(jian(num[n] ,num[n-r] ) ,num[r] ), m);        printf("Case %llu: %d\n",++pp,solve(sum));    }    return 0;}