X问题
时限：1000MS
题意很明确，就是让你解一元同余方程组。题目的要求是找出小于等于\(N\)个数。
利用同余方程的性质，可以找到\(X\)的最小值\(x_0\)，同时也知道\(X\equiv t(mod\,m)\),其中\(m\)为\([m_0,m_1,\cdots,m_M]\),根据上面的条件可以得到\(x_0+m*z\leq N\)，然后将z解出来就是要的答案。

#include "iostream"
#include "stdio.h"
using namespace std;
typedef long long LL;
LL m[20], r[20];
void ext_gcd(LL a, LL b, LL &d, LL& x, LL& y) {LL res;if (!b){x = 1; y = 0; d = a;}else {ext_gcd(b, a%b, d, y, x);y -= x*(a/b);}
}
LL mod_line(LL a, LL b, LL m, LL& d) {LL x, y; ext_gcd(a, m, d, x, y);if (b%d) return -1;LL e = x*(b/d)%(m/d) + (m/d);return e%(m/d);
}
int main(int argc, char const *argv[])
{int T;LL N, M;scanf("%d", &T);while (T--) {bool flag = false;scanf("%lld%lld", &N, &M);for (int i = 0; i < M; i++) scanf("%lld", m+i);for (int j = 0; j < M; j++) scanf("%lld", r+j);  for (int i = 1; i < M; i++) {LL d;LL x0 = mod_line(m[i-1],r[i] - r[i-1], m[i], d);if (flag || x0 == -1) {flag = true;  break;}r[i] = x0*m[i-1]+r[i-1];m[i] = m[i-1]*(m[i]/d);r[i] %= m[i];} if (flag) printf("0\n");else {LL t = 0;if (r[M-1] <= N) t =(N-r[M-1])/m[M-1] + 1;if (t && r[M-1] == 0) t--;   //去掉解是0的情况printf("%lld\n", t);}}return 0;
}

Hello Kiki
时限：1000MS
这道题相比上道题就简单多了，直接计算，然后输出最小的。同样注意要是答案是0的时候的输出。

#include "bits/stdc++.h"
using namespace std;
typedef long long LL;
LL m[20], r[20];
void ext_gcd(LL a, LL b, LL &d, LL& x, LL& y) {LL res;if (!b){x = 1; y = 0; d = a;}else {ext_gcd(b, a%b, d, y, x);y -= x*(a/b);}
}
LL mod_line(LL a, LL b, LL m, LL& d) {LL x, y; ext_gcd(a, m, d, x, y);if (b%d) return -1;LL e = x*(b/d)%(m/d) + (m/d);return e%(m/d);
}
int main(int argc, char const *argv[])
{int T; LL N;int Kcase = 0;scanf("%d", &T);while (T--) {bool flag = false;scanf("%lld", &N);for (int i = 0; i < N; i++) scanf("%lld", m+i);for (int j = 0; j < N; j++) scanf("%lld", r+j);  for (int i = 1; i < N; i++) {LL d;LL x0 = mod_line(m[i-1],r[i] - r[i-1], m[i], d);if (flag || x0 == -1) {flag = true;  break;}r[i] = x0*m[i-1]+r[i-1];m[i] = m[i-1]*(m[i]/d);r[i] %= m[i];} printf("Case %d: ", ++Kcase);if (flag) printf("-1\n");else {if (r[N-1] == 0) r[N-1] = m[N-1];printf("%lld\n", r[N-1]);}}return 0;
}