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求N！最后非0位的值。比方2是120的最后一個不是0的值。
輸入N比較大，要大數保存。
注意到最后0的個數是與5的因數的個數相等。設f（n）為n！的最后非0位。
那么f(n)=((n%5)!* f(n/5) *2^(n/5))%10
因數2的個數始終大于5，從1開始每連續5個劃分為1組，當中5的倍數僅僅提取出一個因數5后，
組成一個新的數列1到n/5，我們有1*2*3*4*5=6*7*8*9*5=2（取最后一個非0位），這里就是2^（n/5）。
再乘上剩下來的幾個數字就可以
（比方n是123，那么第一次會剩下121,122,123三個數沒有被分配）。

比如：23 就能夠變為 f(23) = ((3)! * f(4) * 2^(4))%10; f(4) = 4;

故f(23) = 4； 參考http://blog.csdn.net/yihuikang/article/details/7721875

Last non-zero Digit in N!

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5908????Accepted Submission(s): 1471


Problem Description The expression N!, read as "N factorial," denotes the product of the first N positive integers, where N is nonnegative. So, for example,
N N!
0 1
1 1
2 2
3 6
4 24
5 120
10 3628800

For this problem, you are to write a program that can compute the last non-zero digit of the factorial for N. For example, if your program is asked to compute the last nonzero digit of 5!, your program should produce "2" because 5! = 120, and 2 is the last nonzero digit of 120.

Input Input to the program is a series of nonnegative integers, each on its own line with no other letters, digits or spaces. For each integer N, you should read the value and compute the last nonzero digit of N!.

Output For each integer input, the program should print exactly one line of output containing the single last non-zero digit of N!.

Sample Input 1 2 26 125 3125 9999
Sample Output 1 2 4 8 2 8

#include<stdio.h> #include<string.h> const int di[4] = { 6, 2, 4, 8};//這是2的次冪最后一位的循環； const int pre[10] = { 1, 1, 2, 6, 4,2,2,4,2,8};//前十個數的最后一位； int a[200], ls; char s[200]; void tran( int ls )//轉換 將個位放在a[0]處 {for( int i =ls-1; i >= 0; i -- )a[ls-i-1] = s[i]-'0'; } void mult( ) {int i, t=0;//t是借位； for( i = ls-1; i >= 0; i -- ){int q = t*10+a[i];a[i] = q/5;t = q%5;}while( ls > 0&&a[ls-1] == 0 ) --ls;//排除后面的0 細致考慮一下 } int la_no_num( ) {if( ls == 1 ) return pre[a[0]]; //假設僅僅有一位直接輸出或返回int x1 = pre[a[0]%5]; //這是f（n%5）mult( );int x2 = di[(a[0]+a[1]*10)%4];//這是2^(n/5) 為什么僅僅算前兩位（提示：同余定理）int ans = (x1*x2*la_no_num())%10;//f(n)=((n%5)!* f(n/5) *2^(n/5))%10return ans; } int main() {int la, i;while( ~scanf( "%s", s ) ){ls = strlen(s);tran(ls); printf( "%d\n", la_no_num() );}}

轉載于:https://www.cnblogs.com/hrhguanli/p/3822988.html

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