求解 n 个同余方程
x ≡ Ci (mod Mi)
推荐一下讲证明的文章：
1.https://www.cnblogs.com/TheRoadToTheGold/p/8458326.html 很简洁易懂。
2.http://blog.csdn.net/clove_unique/article/details/54571216 排版不太好；这篇文章还讲了扩展卢卡斯 (Lucas) 定理。
可以用滚动数组优化空间，但我的代码里没这么写。
1 void China() 2 { 3 scanf("%lld", &n); 4 for (i = 1; i <= n; ++i) 5 scanf("%lld%lld", &M[i], &C[i]); 6 bool flag = false; 7 for (i = 2; i <= n; ++i) { 8 LL m1 = M[i-1], m2 = M[i], c1 = C[i-1], c2 = C[i]; 9 LL t = gcd(m1, m2); 10 if ((c2-c1) % t) { flag = true; break; } 11 M[i] = m1*m2 / t; 12 C[i] = Inv(m1/t, m2/t) * (c2-c1)/t % (m2/t) * m1 + c1; 13 C[i] = (C[i]%M[i]+M[i]) % M[i]; 14 } 15 if (flag) { printf("-1 "); continue; } 16 printf("%lld ", C[n]); 17 }
模板题 POJ2891
1 #include <stdio.h> 2 3 typedef long long LL; 4 5 const int _N = 1200; 6 7 LL C[_N], M[_N]; 8 9 void exgcd(LL m, LL n, LL &x, LL &y, LL &d) 10 { 11 if (n) { exgcd(n, m%n, y, x, d); y -= m/n*x; return; } 12 x = 1, y = 0, d = 1; 13 return; 14 } 15 16 LL gcd(LL t1, LL t2) { return t2 ? gcd(t2, t1 % t2) : t1; } 17 18 LL Inv(LL v, LL p) 19 { 20 LL x, y, d; 21 exgcd(v, p, x, y, d); 22 x = (x%p+p)%p; 23 if (!x) x = p; 24 return x; 25 } 26 27 int main() 28 { 29 LL n, i; 30 while (~scanf("%lld", &n)) { 31 for (i = 1; i <= n; ++i) 32 scanf("%lld%lld", &M[i], &C[i]); 33 bool flag = false; 34 for (i = 2; i <= n; ++i) { 35 LL m1 = M[i-1], m2 = M[i], c1 = C[i-1], c2 = C[i]; 36 LL t = gcd(m1, m2); 37 if ((c2-c1) % t) { flag = true; break; } 38 M[i] = m1*m2 / t; 39 C[i] = Inv(m1/t, m2/t) * (c2-c1)/t % (m2/t) * m1 + c1; 40 C[i] = (C[i]%M[i]+M[i]) % M[i]; 41 } 42 if (flag) { printf("-1 "); continue; } 43 printf("%lld ", C[n]); 44 } 45 return 0; 46 }