题目内容

对于一个长度为n的十进制整数N=(b1,b2….bn)(0<=bi<=9,b1!=0)，定义P(N)=b1^1 * b2^2 *….bn^n,当然这个数很大，我们只要你输出P(N)%1000000007的结果 P(123)=(1^1 * 2 ^ 2 * 3^3)mod 1000000007=108

输入说明

多组输入，每一行一个数字1<=N<=10^1000000

输出说明

P(N)%1000000007的结果

输入样例1

123

输出样例1

108

提示

无

解题思路：

模拟即可

AC代码：

#![allow(warnings)]
use std::io;
use std::error::Error;
use std::boxed::Box;
use std::convert::TryInto;
use std::cmp::Ordering;
use std::cmp::min;
use std::cmp::max;
const p: u64 = 1000000007;
fn pow(num: u64, j: u64) -> u64 {
let mut res: u64 = 1;
for i in 1 ..= j {
res *= num % p;
res %= p;
}
res
}
fn getAns(str_num: &String) -> u64 {
let mut res: u64 = 1;
for (index, str_char) in str_num.char_indices() {        
if let Ok(num) = str_char.to_string().parse::<u64>() {
res *= pow(num, (index + 1) as u64) as u64;
res %= p as u64;
}
}
return res;
}
fn main() -> Result<(), Box<dyn Error>> {
let mut cin = String::new();
io::stdin().read_line(&mut cin).unwrap();
print!("{}", getAns(&cin));
Ok(())
}