快速指数取模运算与用扩展欧几里得算法求解最大公约数和求乘法逆元

实验1.1快速指数取模运算
一、实验1.1源代码：#include来自"stdio.h"
#include "stdlib.h"
#include "iostream"
using namespace std;
void Mode(int a, int b, int n)
{
int c=1;
do{
if(a%2==0)
return 1;
}
q = x3 / y3;
t1 = x1 - q*y1;
t2 = x2 - q*y2;
t3 = x3 - q*y3;
x1 = y1;
x2 = y2;
x3 = y3;
y1 = t1;
y2 = t2;
y3 = t3;
}
}
int main()
{
int x, y, z ;
int a, b;
int extended_Gcd(int a,int b, int &x, int &y) //求最大公约数
{
if (b == 0)
{
x = 1;
y = 0;
return a;
}
else
{
int gcd = extended_Gcd(b, a%b, x, y);
int t = x;
x = y;
y = t - (a / b) * y;
cout<<"请输入指数a："<<endl;
cin>>a;
cout<<"请输入该基数b："<<endl;
cin>>b;
cout<<"请输入被除数n："<<endl;
cin>>n;
Mode(a,b,n);
}
二、实验效果图：
实验1.2用扩展欧几里得算法求解最大公约数和求乘法逆元
一、实验1.2源代码：
#include <stdio.h>
y3 = (f >= d) ? d : f;
while (1)
{
if (y3 == 0)
{
*result = x3; //两个数不互素则result为两个数的最大公约数，此时返回值为零
return 0;
}
if (y3 == 1)
{
*result = y2; //两个数互素则resutl为其乘法逆元，此时返回值为1
return gcd;
}
}
int extended_Ivn(int f, int d, int *result) //求乘法逆元
{
int x1, x2, x3, y1, y2, y3, t1, t2, t3, q;
x1 = y2 = 1;
x2 = y1 = 0;
x3 = (f >= d) ? f : d;
int *p, *q;
p = &x; q = &y;
z = 0;
printf("请输入两个数: ");
scanf("%d%d", &a, &b);
if (extended_Gcd(a,b, *p, *q) == 1)
{
extended_Ivn(a, b, &z);
printf("%d和%d互素，乘法的逆元是：%d\n", a, b, z);
}
else
{
z=extended_Gcd(a,b, *p, *q);
printf("%d和%d不互素，最大公约数为：%d\n", a, b, z);
}
return 0;
}
二、实验效果图：
{
a=a/2;
b=(b*b)%n;
}
else
{
a=a-1;
c=(b*c)%n;
}
}while(a!=0);
cout<<"取余结果为："<<c<<endl;
}
void main ()
{
int a, b, n;
cout<<"输入格式范例b^a mod n"<<endl;
cout<<"........................................................."<<endl;