求长度为 L 的神奇串对的个数
一对弦 s 和 r 如果每个指标 i 的 s 的字符小于 r ，即s【I】<r【I】，则称之为神奇。任务是计算长度为 L 的可能字符串对的数量。由于该值可以很大，所以以 10 9 为模给出答案。
注意:字符串只包含小写英文字母。
示例:
输入: L = 1 输出: 325 因为需要的字符串长度是 1。 如果 s =“a”，那么 r 可以是“b”、“c”、“d”、…“z”(25 种可能性) 如果 s =“b”，那么 r 可以是“c”、“d”、“e”、…“z”(24 种可能性) …。 如果 s =“y”，那么 r 只能是“z”(1 种可能性) s 不能是“z”，因为它是最大小写字符。 因此总的可能性是 1 + 2 + 3 + … + 25 = 325
输入:L = 2 T3】输出: 105625
进场:对于 L = 1 ，总可能性为 325 。对于 L = 2 ，总可能性为 325 2 。任何数值 L 的总可能性为 325 L 。由于该值可能很大，打印答案时取模 10 9 。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Iterative Function to calculate (x^y)%p in O(log y)
int power(int x, unsigned int y, int p)
{
// Initialize result
int res = 1;
// Update x if it is >= p
x = x % p;
while (y > 0) {
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// Y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Driver Code
int main()
{
int L = 2, P = pow(10, 9);
int ans = power(325, L, P);
cout << ans << "\n";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Iterative Function to calculate (x^y)%p in O(log y)
static int power(int x, int y, int p)
{
// Initialize result
int res = 1;
// Update x if it is >= p
x = x % p;
while (y > 0)
{
// If y is odd, multiply x with result
if (y % 2 == 1)
{
res = (res * x) % p;
}
// Y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Driver Code
public static void main(String[] args)
{
int L = 2;
int P = (int) Math.pow(10, 9);
int ans = power(325, L, P);
System.out.println(ans);
}
}
// This code has been contributed by 29AjayKumar
Python 3
# Python implementation of the approach
# Iterative Function to calculate (x^y)%p in O(log y)
def power(x, y, p):
# Initialize result
res = 1;
# Update x if it is >= p
x = x % p;
while (y > 0):
# If y is odd, multiply x with result
if (y %2== 1):
res = (res * x) % p;
# Y must be even now
y = y >> 1; # y = y/2
x = (x * x) % p;
return res;
# Driver Code
L = 2; P = pow(10, 9);
ans = power(325, L, P);
print(ans);
# This code contributed by Rajput-Ji
C
// C# implementation of the approach
using System;
class GFG
{
// Iterative Function to calculate (x^y)%p in O(log y)
static int power(int x, int y, int p)
{
// Initialize result
int res = 1;
// Update x if it is >= p
x = x % p;
while (y > 0)
{
// If y is odd, multiply x with result
if (y % 2 == 1)
{
res = (res * x) % p;
}
// Y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Driver Code
public static void Main()
{
int L = 2;
int P = (int) Math.Pow(10, 9);
int ans = power(325, L, P);
Console.WriteLine(ans);
}
}
// This code is contributed by AnkitRai01
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Iterative Function to calculate (x^y)%p in O(log y)
function power($x, $y, $p)
{
// Initialize result
$res = 1;
// Update x if it is >= p
$x = $x % $p;
while ($y > 0)
{
// If y is odd, multiply x with result
if ($y & 1)
$res = ($res * $x) % $p;
// Y must be even now
$y = $y >> 1; // y = y/2
$x = ($x * $x) % $p;
}
return $res;
}
// Driver Code
$L = 2;
$P = pow(10, 9);
$ans = power(325, $L, $P);
echo $ans , "\n";
// This code is contributed by ajit.
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Iterative Function to calculate (x^y)%p in O(log y)
function power(x, y, p)
{
// Initialize result
let res = 1;
// Update x if it is >= p
x = x % p;
while (y > 0)
{
// If y is odd, multiply x with result
if (y % 2 == 1)
{
res = (res * x) % p;
}
// Y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
let L = 2;
let P = Math.pow(10, 9);
let ans = power(325, L, P);
document.write(ans);
</script>
Output:
105625
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