65 lines
2.0 KiB
C++
65 lines
2.0 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 100009;
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//洛谷P4777 【模板】扩展中国剩余定理（EXCRT）
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LL n; // n组数据
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LL mod[N], yu[N]; //每组测试数据中，模记为mod[i],余数记为yu[i]
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//龟速乘
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LL qmul(LL a, LL k, LL b) {
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LL res = 0;
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while (k) {
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if (k & 1) res = (res + a) % b;
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a = (a + a) % b;
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k >>= 1;
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}
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return res;
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}
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//扩展欧几里得
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LL exgcd(LL a, LL b, LL &x, LL &y) {
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if (!b) {
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x = 1, y = 0;
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return a;
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}
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LL d = exgcd(b, a % b, y, x);
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y -= a / b * x;
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return d;
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}
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//最小公倍数
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LL lcm(LL a, LL b) {
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LL x, y;
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LL d = exgcd(a, b, x, y);
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return (a / d * b);
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}
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//扩展中国剩余定理
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LL excrt() {
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LL M = mod[1], ans = yu[1];
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// M为前i-1个数的最小公倍数,即lcm(mod[1],mod[2],...,mod[n])
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// ans为前i-1个方程的通解
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LL x, y;
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for (int i = 2; i <= n; i++) {
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LL m = mod[i]; //当前要合并方程的模
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LL res = (yu[i] - ans % m + m) % m; // r2-r1，之所以%mi,是为了防止余数差出现负数
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LL d = exgcd(M, m, x, y); // 1、求特解x0; 2、求最大公约数d
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if (res % d) return -1; //无解的情况
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LL t = qmul(x, res / d, m); //根据ax+by=1的特解翻番得到方程ax+by=c的特解
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ans += t * M; //获得前i个方程的通解 r=r + x*lcm(m1,m2,..)
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M = lcm(M, m); //更改M的值,把mi也加到M中去，相当于不断的取所有mi的最小公倍数
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ans = (ans % M + M) % M; //对最小公倍数取模就是最小的合理解
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}
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return ans;
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}
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int main() {
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ios::sync_with_stdio(false); //加速cin和cout
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> mod[i] >> yu[i];
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cout << excrt() << endl;
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return 0;
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}
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