C ++中前N个阶乘的乘积
给定数字N，任务是找到前N个阶乘以1000000007为模的乘积。阶乘表示当我们找到该数字以下所有数字的乘积（包括该数字）时，表示为！（感叹号），例如-4！=4x3x2x1=24。
因此，我们必须找到n阶乘和1000000007模的乘积。
约束
1 ≤ N ≤ 1e6.
输入值
n = 9
输出结果
27
说明
1! * 2! * 3! * 4! * 5! * 6! * 7! * 8! * 9! Mod (1e9 + 7) = 27
输入值
n = 3
输出结果
12
说明
1! * 2! * 3! mod (1e9 +7) = 12
解决问题的方法如下
从i=1到n递归找到阶乘并乘积所有阶乘
将所有阶乘的乘积乘以1e9+7
返回结果。
算法
In Fucntion long long int mulmod(long long int x, long long int y, long long int mod)
Step 1→ Declare and Initialize result as 0
Step 2→ Set x as x % mod
Step 3→ While y > 0
If y % 2 == 1 then,
Set result as (result + x) % mod
Set x as (x * 2) % mod
Set y as y/ 2
Step 4→ return (result % mod)
In Function long long int nfactprod(long long int num)
Step 1→ Declare and Initialize product with 1 and fact with 1
Step 2→ Declare and Initialize MOD as (1e9 + 7)
Step 3→ For i = 1 and i <= num and i++
Set fact as (call function mulmod(fact, i, MOD))
Set product as (call function mulmod(product, fact, MOD))
If product == 0 then,
Return 0
Step 4→ Return product
In Function int main() Step 1→ Declare and Initialize num = 3
Step 2→ Print the result by calling (nfactprod(num))
Stop示例
#include <stdio.h>
long long int mulmod(long long int x, long long int y, long long int mod){
long long int result = 0;
x = x % mod;
while (y > 0) {
//在y为奇数的地方加x。
if (y % 2 == 1)
result = (result + x) % mod;
//将x乘以2-
x = (x * 2) % mod;
//y除以2-
y /= 2;
}
return result % mod;
}
long long int nfactprod(long long int num){
//初始化产品和事实
long long int product = 1, fact = 1;
long long int MOD = 1e9 + 7;
for (int i = 1; i <= num; i++) {
//查找每次迭代的阶乘
fact = mulmod(fact, i, MOD);
//第一个i阶乘的乘积
product = mulmod(product, fact, MOD);
//当产品可被MOD整除时返回0-
if (product == 0)
return 0;
}
return product;
}
int main(){
long long int num = 3;
printf("%lld \n", (nfactprod(num)));
return 0;
}输出结果
如果运行上面的代码，它将生成以下输出-
12