标签：des style blog http os io for ar 2014
Description
A square root of a number x is a number
r such that r2 = x. A discrete square root of a non-negative integer
x is a non-negative integer
r such that r2 
x mod N , 0
r <
N , where N is a specific positive integer and mod is the modulo operation.
It is well-known that any positive real number has exactly two square roots, but a non-negative integer may have more than two discrete square roots. For example, for N = 12 , 1 has four discrete square roots 1, 5, 7 and 11.
Your task is to find all discrete square roots of a given non-negative integer x. To make it easier, a known square root r of x is also given to you.
Input
The input consists of multiple test cases. Each test case contains exactly one line, which gives three integers
x, N and
r. (1x <
N, 2N < 1, 000, 000, 000, 1r
< N). It is guaranteed that r is a discrete square root of
x modulo N. The last test case is followed by a line containing three zeros.
Output
For each test case, print a line containing the test case number (beginning with 1) followed by a list of corresponding discrete square roots, in which all numbers are sorted increasingly..
Sample Input
1 12 1 4 15 2 0 0 0
Sample Output
Case 1: 1 5 7 11 Case 2: 2 7 8 13
题意：r^2≡x(mod n)求（0<r<n）r的所有解
思路：已知一个x,n和一个解r1，假设还有一个解是r，那么就可以通过：r^2≡x(mod n) 和 ：r1^2≡x(mod n)
得到：：r1^2 + k1*n=x 和
：r^2 + k2*n=x,连立解得：r^2 - r1^2 = k3*n =>(r+r1)*(r-r1) = k3*n
那么我们假设A*B = n，我们就可以得到：：r+r1≡0(mod A) , ：r-r1≡0(mod B),也就可以得到：
r1+K1*A = r1-K2*B = r,变形得到：K1*A≡2*r1(mod B)，那么这个就是解模方程了，但是我们带入系数的时候，
是带入A和B，这个通过枚举N的因子就可以得到了，因为还有一个K1，这个就留着我们最后放方程都扩大K1倍的时候
解决.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <vector>
typedef long long ll;;
using namespace std;
const int maxn = 11111;
int N, r;
set<ll> ans;
void exgcd(ll a, ll b, ll &d, ll &x, ll &y) {
if (b == 0) {
d = a;
x = 1, y = 0;
}
else {
exgcd(b, a%b, d, y, x);
y -= a / b * x;
}
}
void solve(ll c, ll a, ll b) {
ll x, y, d;
exgcd(a, b, d, x, y);
ll n = 2*r;
if (n % d == 0) {
x *= n / d;
x = (x % (b/d) + (b/d)) % (b/d);
ll m = x * a - r;
while (m < N) {
if (m >= 0 && m * m % N == c)
ans.insert(m);
m += b / d * a;
}
}
}
int main() {
int x, cas = 1;
while (scanf("%d%d%d", &x, &N, &r) != EOF && x) {
ans.clear();
for (int i = 1; i*i <= N; i++)
if (N % i == 0) {
solve(x, i, N/i);
solve(x, N/i, i);
}
set<ll>::iterator it;
printf("Case %d:", cas++);
for (it = ans.begin(); it != ans.end(); it++)
printf(" %lld", *it);
printf("\n");
}
return 0;
}UVA - 1426 Discrete Square Roots (模方程)
标签：des style blog http os io for ar 2014
原文地址：http://blog.csdn.net/u011345136/article/details/38826387
