题目链接：https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4078

很容易列出方程组，但因为模数并不一定是质数，可能不存在逆元，因此用辗转相除法来消元

每个齿轮一定不会转动超过 \(n\) 下(鸽巢原理)，\(dfs\) 统计回代即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 25;
const int INF = 1000000007;

int n, K; 
int a[maxn];
int A[maxn][maxn];
int ans;

void Gauss(){
	int i, j, k, r;
	for(i = 0 ; i < K ; ++i){
		r = i;
		for(j = i + 1 ; j < K ; ++j)
			if(A[j][i]) r = j;
		if(A[r][i] == 0) continue;
		if(r != i)
			for(j = 0 ; j <= K ; ++j) swap(A[i][j], A[r][j]);
	
		for(j = i + 1 ; j < K ; ++j){ // 
			if(j != i){
				while(A[j][i]){ // 辗转相除法消元，避免除法（模代数系统） 
					int rate = A[i][i] / A[j][i], tmp;
					for(k = 0 ; k <= K ; ++k){
						tmp = A[j][k];
						A[j][k] = ((A[i][k] - rate * A[j][k]) % n + n) % n;
						A[i][k] = tmp;
					}
				}
			} 
		}
	}
}

int t[maxn][maxn], val[maxn];

void dfs(int pos, int res){
	if(res >= ans) return;
	if(pos == -1){
		ans = res;
		return;
	}
	
	for(int i = 0 ; i < K ; ++i){
		for(int j = 0 ; j <= K ; ++j){
			t[i][j] = A[i][j];
		}
	}
	
	for(int i = 0 ; i < n ; ++i){
		val[pos] = i;
		int now = 0;
		for(int j = pos ; j < K ; ++j){
			now = (now + A[pos][j] * val[j]) % n;
		}
		if(now % n == A[pos][K] % n) dfs(pos - 1, res + i);
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < ‘0‘ || ch > ‘9‘){ if(ch == ‘-‘) f = -1; ch = getchar(); } while(ch >= ‘0‘ && ch <= ‘9‘){ s = s * 10 + ch - ‘0‘; ch = getchar(); } return s * f; }

int main(){
	while(scanf("%d%d", &K, &n) == 2 && K){
		memset(val, 0, sizeof(val));
		memset(A, 0, sizeof(A));
		int x;
		for(int i = 0 ; i < K ; ++i) {
			scanf("%d", &x);
			A[i][K] = n + 1 - x;
		}
		
		int p, u, v;
		for(int i = 1 ; i <= K ; ++i){
			scanf("%d", &p);
			for(int j = 1 ; j <= p ; ++j){
				scanf("%d%d", &u, &v);
				A[u-1][i-1] = v;
			}
		}

		Gauss();
		
		ans = INF;
		dfs(K-1, 0);
		
		if(ans == INF){
			printf("No solution\n");
		}
		else printf("%d\n", ans); 
	}
	return 0;
}

UVa 1332 - Kid's Problem (高斯消元+dfs)