原文:
给定正整数 k，a 和 b，我们需要打印 a^b ie 的最后 k 位数字..功率(a，b)。
input constraint:
k <= 9, a <= 10^6, b<= 10^6
示例:
input : a = 11, b = 3, k = 2
output : 31
explanation : a^b = 11^3 = 1331, hence
last two digits are 31
input : a = 10, b = 10000, k = 5
output : 00000
explanation : a^b = 1000..........0 (total
zeros = 10000), hence last 5 digits are 00000
天真的pg电子试玩链接的解决方案
首先计算 a^b，然后通过与 10^k.取模取最后 k 位。当 a^b 太大时，上述pg电子试玩链接的解决方案失败，因为我们在 c/c 中最多只能容纳 2^64 -1。
有效解
有效的方法是每次乘法后只保留 k 个数字。这个想法与中讨论的非常相似，在这里我们讨论了一种寻找的一般方法(a^b)%c，这里的 c 是 10^k.
下面是实现。
c
// c code to find last k digits of a^b
#include
using namespace std;
/* iterative function to calculate (x^y)%p in o(log y) */
int power(long long int x, long long int y, long long int p)
{
long long int res = 1; // initialize result
x = x % p; // update x if it is more than or
// equal to 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;
}
// c function to calculate
// number of digits in x
int numberofdigits(int x)
{
int i = 0;
while (x) {
x /= 10;
i ;
}
return i;
}
// c function to print last k digits of a^b
void printlastkdigits(int a, int b, int k)
{
cout << "last " << k;
cout << " digits of " << a;
cout << "^" << b << " = ";
// generating 10^k
int temp = 1;
for (int i = 1; i <= k; i )
temp *= 10;
// calling modular exponentiation
temp = power(a, b, temp);
// printing leftmost zeros. since (a^b)%k
// can have digits less then k. in that
// case we need to print zeros
for (int i = 0; i < k - numberofdigits(temp); i )
cout << 0;
// if temp is not zero then print temp
// if temp is zero then already printed
if (temp)
cout << temp;
}
// driver program to test above functions
int main()
{
int a = 11;
int b = 3;
int k = 2;
printlastkdigits(a, b, k);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java code to find last k digits of a^b
public class gfg
{
/* iterative function to calculate (x^y)%p in o(log y) */
static int power(long x, long y, long p)
{
long res = 1; // initialize result
x = x % p; // update x if it is more than or
// equal to p
while (y > 0) {
// if y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return (int) res;
}
// method to print last k digits of a^b
static void printlastkdigits(int a, int b, int k)
{
system.out.print("last " k " digits of " a
"^" b " = ");
// generating 10^k
int temp = 1;
for (int i = 1; i <= k; i )
temp *= 10;
// calling modular exponentiation
temp = power(a, b, temp);
// printing leftmost zeros. since (a^b)%k
// can have digits less then k. in that
// case we need to print zeros
for (int i = 0; i < k - integer.tostring(temp).length() ; i )
system.out.print(0);
// if temp is not zero then print temp
// if temp is zero then already printed
if (temp != 0)
system.out.print(temp);
}
// driver method
public static void main(string[] args)
{
int a = 11;
int b = 3;
int k = 2;
printlastkdigits(a, b, k);
}
}
python 3
# python 3 code to find last
# k digits of a^b
# iterative function to calculate
# (x^y)%p in o(log y)
def power(x, y, p):
res = 1 # initialize result
x = x % p # update x if it is more
# than or equal to 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
# function to calculate
# number of digits in x
def numberofdigits(x):
i = 0
while (x) :
x //= 10
i = 1
return i
# function to print last k digits of a^b
def printlastkdigits( a, b, k):
print("last " str( k) " digits of "
str(a) "^" str(b), end = " = ")
# generating 10^k
temp = 1
for i in range(1, k 1):
temp *= 10
# calling modular exponentiation
temp = power(a, b, temp)
# printing leftmost zeros. since (a^b)%k
# can have digits less then k. in that
# case we need to print zeros
for i in range( k - numberofdigits(temp)):
print("0")
# if temp is not zero then print temp
# if temp is zero then already printed
if (temp):
print(temp)
# driver code
if __name__ == "__main__":
a = 11
b = 3
k = 2
printlastkdigits(a, b, k)
# this code is contributed
# by chitranayal
c
// c# code to find last k digits of a^b
using system;
class gfg
{
// iterative function to calculate
// (x^y)%p in o(log y)
static int power(long x, long y, long p)
{
long res = 1; // initialize result
x = x % p; // update x if it is more
// than or equal to p
while (y > 0)
{
// if y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return (int) res;
}
// method to print last k digits of a^b
static void printlastkdigits(int a, int b, int k)
{
console.write("last " k " digits of "
a "^" b " = ");
// generating 10^k
int temp = 1;
for (int i = 1; i <= k; i )
temp *= 10;
// calling modular exponentiation
temp = power(a, b, temp);
// printing leftmost zeros. since (a^b)%k
// can have digits less then k. in that
// case we need to print zeros
for (int i = 0;
i < k - temp.tostring().length; i )
console.writeline(0);
// if temp is not zero then print temp
// if temp is zero then already printed
if (temp != 0)
console.write(temp);
}
// driver code
public static void main()
{
int a = 11;
int b = 3;
int k = 2;
printlastkdigits(a, b, k);
}
}
// this code is contributed
// by 29ajaykumar
服务器端编程语言（professional hypertext preprocessor 的缩写）
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 $res;
}
// function to calculate
// number of digits in x
function numberofdigits($x)
{
$i = 0;
while ($x)
{
$x = (int)$x / 10;
$i ;
}
return $i;
}
// function to print last k digits of a^b
function printlastkdigits( $a, $b, $k)
{
echo "last ",$k;
echo " digits of " ,$a;
echo "^" , $b , " = ";
// generating 10^k
$temp = 1;
for ($i = 1; $i <= $k; $i )
$temp *= 10;
// calling modular exponentiation
$temp = power($a, $b, $temp);
// printing leftmost zeros. since
// (a^b)%k can have digits less
// then k. in that case we need
// to print zeros
for ($i = 0;
$i < $k - numberofdigits($temp); $i )
echo 0;
// if temp is not zero then print temp
// if temp is zero then already printed
if ($temp)
echo $temp;
}
// driver code
$a = 11;
$b = 3;
$k = 2;
printlastkdigits($a, $b, $k);
// this code is contributed by ajit
?>
java 描述语言
输出:
last 2 digits of 11^3 = 31
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