阶乘 0 的个数
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| Time Limit: 1500MS | Memory Limit: 65536K | |
| Total Submissions: 15843 | Accepted: 9775 |
Description
ACM technicians faced a very interesting problem recently. Given a set of BTSes to visit, they needed to find the shortest path to visit all of the given points and return back to the central company building. Programmers have spent several months studying this problem but with no results. They were unable to find the solution fast enough. After a long time, one of the programmers found this problem in a conference article. Unfortunately, he found that the problem is so called "Travelling Salesman Problem" and it is very hard to solve. If we have N BTSes to be visited, we can visit them in any order, giving us N! possibilities to examine. The function expressing that number is called factorial and can be computed as a product 1.2.3.4....N. The number is very high even for a relatively small N.
The programmers understood they had no chance to solve the problem. But because they have already received the research grant from the government, they needed to continue with their studies and produce at least some results. So they started to study behaviour of the factorial function.
For example, they defined the function Z. For any positive integer N, Z(N) is the number of zeros at the end of the decimal form of number N!. They noticed that this function never decreases. If we have two numbers N1 < N2, then Z(N1) <= Z(N2). It is because we can never "lose" any trailing zero by multiplying by any positive number. We can only get new and new zeros. The function Z is very interesting, so we need a computer program that can determine its value efficiently.
Input
Output
Sample Input
6 3 60 100 1024 23456 8735373
Sample Output
0 14 24 253 5861 2183837
题意：求 n! 中末尾 0 的个数，比如 10! = 3628800，末尾有 2 个零。
思路：
1）直接求解 n!
// n = 17 爆 int
// n = 21 爆 long long
int fact1(int n){
LL ans = 1LL;
for(int i = 1; i <= n; i++)
ans *= i;
int count = 0;
while(ans){
int val = ans % 10;
if(!val) count++;
else break;
ans /= 10;
}
return count;
}
2）模拟大数乘法计算 n!，用数组存储结果
// 模拟大数乘法
int fact2(int n){
enum {SIZE = 3000};
int f[SIZE];
memset(f, 0, sizeof(f));
f[0] = 1;
for(int val = 2; val <= n; val++){
int carry = 0;
for(int i = 0; i < SIZE; i++){
f[i] = f[i] * val + carry;
carry = f[i] / 10;
f[i] %= 10;
}
}
int len = SIZE - 1;
while(!f[len]) len--;
int count = 0;
#if 0
for(int i = len; i >= 0; i--)
printf("%d", f[i]);
printf("\n");
#endif
for(int i = 0; i <= len; i++)
if(f[i] == 0) count++;
else break;
return count;
}3）认真分析发现，n! = K * 10^M，所以结果就是求 M。而把 n! 分解成质因数相乘的形式为：n! = 2^x * 3^y * 5^z * ...；发现只有每一对 2 * 5 才能为结果添加一个 0，所以结果变成求 min{x, z}。而根据 n! 特性知道 x 必定大于 z，所以结果又变成求 z。也就是求 n! 的质因数相乘形式中 5 的指数；我们只需要判断 5 的倍数即可。
int fact3(int n){
int count = 0;
for(int val = 5; val <= n; val += 5){
int tmp = val;
while(tmp % 5 == 0){
count++;
tmp /= 5;
}
}
return count;
}非常遗憾，时间复杂度为 O(n/5)，TLE，理由同上。
4）在上面的启发下，可以使用类似的思路，就是
z = n / 5 + n / 5^2 + n / 5^3 ... + 0
可转换为不断用 n / 5，直到 n 为 0。比如 n = 100
100 / 5 = 20
20 / 5 = 4
4 / 5 = 0
那结果就是 20 + 4 + 0 = 24。
为什么可以这样？
其实 n / 5 意思就是取 n 以内的 5 的倍数的第一个 5（比如 50 = 5 * 5 * 2 就有 2 个 5）
n / 5^2 意思就是取 n 以内的 5 的倍数的第二个 5
...
举个例子，比如 n = 100
100 / 5 = 20
表示取 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 这些数的第一个 5（能被 5 整除）。
所有 5 的倍数都符合，故有 20 个。
100 / 5^2 = 4
表示取 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 这些数的第二个 5（能被 5^2 整除）。
只有 25, 50, 75, 100 这 4 个数符合。
100 / 5^3 = 0
表示取 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 这些数的第三个 5（能被 5^3 整除）。
没有符合要求的数字
把结果加起来就是 n! 的质因数相乘形式中 5 的指数。
int fact4(int n){
int count = 0;
while(n){
n /= 5;
count += n;
}
return count;
}结果~ AC