# Solving notes (38) -- big integer factorial calculation

Source: Internet
Author: User

Problem description: Calculate the factorial of an integer N, 0 <= n <= 5000.

For example, if n is 50, the result is 30414093201713378043612608166064768844377641568960512000000000000.

Idea: Starting from the factorial definition, n! = N * (n-1) * (n-2) *... * 2*1. Because N is large, basic data types cannot be stored. However, we can use an integer array to store results (including intermediate results). Each element of the array stores one of the results, and the first element of the array is initialized to 1. The calculation process is relatively simple, N, n-1, N-2, 1 multiplied by the current value, that is, the value of the integer array to save. The final result is printed.

Reference code:

`Void caln_solution1 (int n) {int A [20000] = {1, 0}; // Save the result, including the intermediate result int m = 0; For (; n --) // multiply by N, n-1, 1 {int C = 0; // carry value for (INT I = 0; I <= m; I ++) // multiply the intermediate result by N {C + = A [I] * n; // calculate the result first and then consider carrying a [I] = C % 10; // The intermediate result of the I-th digit C/= 10; // carry-up} For (; C/= 10) // carry a [++ m] = C % 10;} For (; m> = 0; m --) // print the result cout <A [m]; cout <Endl ;}`

In the above program, each element of the integer array represents only one bit of the result, which wastes a lot of space. The improved method is to let each element in the array Save the four digits of the result, which can greatly save space. The code is modified as follows.

`Void caln_solution2 (int n) {int A [5000] = {1, 0}; // Save the result, including the intermediate result int m = 0; For (; n --) // multiply N, n-1, 1 {int C = 0, I = 0; For (; I <= m; I ++) // multiply the intermediate result by N {C + = A [I] * n; // calculate the result first and then consider carrying a [I] = C % 10000; // The intermediate result of the I-th digit. Note that the modulo 10000 is used to indicate the 4-digit C/= 10000; // carry up} A [M + 1] = C; // carry a maximum of 1 bits, so you do not need to use the loop if (C> 0) m ++;} cout <A [M --]; for (; m> = 0; M --) cout <SETW (4) <setfill ('0') <A [m]; cout <Endl ;}`

The Code is as follows.

`#include<iostream>#include<iomanip>using namespace std;int main(){int n;cin>>n;CalN_Solution1(n);CalN_Solution2(n);return 0;} `

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