X^2 Mod P
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
ll num[100005];
int main()
{
ll p, a;
while (cin >> p >> a) {
ll i;
int u = 0;
for (i = 0; i < p; i ++) {
if (i * i % p == a) {
cout << i << " ";
u = 1;
}
}
if(u == 0)
cout << "No Solution";
cout << endl;
}
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
ll num[100005];
int main()
{
ll p, a;
while (cin >> p >> a) {
ll i;
int u = 0;
for (i = 0; i < p; i ++) {
if (i * i % p == a) {
cout << i << " ";
u = 1;
}
}
if(u == 0)
cout << "No Solution";
cout << endl;
}
}
本文介绍了一个简单的C++程序,该程序通过遍历的方法求解形如X^2 ≡ a (mod p)的方程,其中a和p为输入参数。对于每个可能的X值,程序检查其平方是否等于a模p,并输出符合条件的解。
2133

被折叠的 条评论
为什么被折叠?



