今天在HackerRank上翻到一道高精度题,于是乎就写了个高精度的模板,说是模板其实就只有乘法而已。
Extra long factorials
Problem Statement
You are given an integer N. Print the factorial of this number.
Note: Factorials of N>20 can't be stored even in a 64−bit long long variable. Big integers must be used for such calculations. Languages like Java, Python, Ruby etc. can handle big integers but we need to write additional code in C/C++ to handle such large values.
We recommend solving this challenge using BigIntegers.
Input Format
Input consists of a single integer N.Constraints
1≤N≤100Output Format
Output the factorial of N.Sample Input
25
Sample Output
15511210043330985984000000 ------------------------------------------------------------------------------------- 从前看C++ Primer时,看到Constructor处发现有很多方便的feature,但一般也不写Class,大多忘却了。这回用了一下,还不错。
const int N(10000);int tmp[N];struct BigInt{ int d[N], len; BigInt(int n){ len=!n; for(int tmp=n; tmp; len++, tmp/=10); for(int i=len-1; i>=0; i--) d[i]=n%10, n/=10; } BigInt(){} void multi(const BigInt &n){ memset(tmp, 0, sizeof(tmp)); for(int i=0; i
注意到这里写了两个Constructor,一个带参数的BigInt(int n)以及一个不带参数的BigInt()。如果一个class里没写constructor的话其实还是带有一个default constructor的,这里就是BigInt::BigInt(),但如果定义了Constructor的话default constructor就会被覆盖,所以这里还要将BigInt()显式(explicitly)定义一下,否则 BigInt a;这种声明变量的方式就会报错。
和constructor有关的概念还有type conversion。我们定义了BigInt::BigInt(int n)后求阶乘就可很简洁地写成
BigInt res(1); for(int i=2; i<=n; i++){ res.multi(i); }
我们将int型的变量i传递给函数BigInt::void multi(const BigInt &n)时,int将会自动转换成BigInt,之所以可以这样正是由于我们定义了constructor BigInt(int n) 。