Day7 - E - Strange Way to Express Integers POJ - 2891

Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:

Choose k different positive integers a₁, a₂, …, a_k. For some non-negative m, divide it by every a_i (1 ≤ i ≤ k) to find the remainder r_i. If a₁, a₂, …, a_k are properly chosen, m can be determined, then the pairs (a_i, r_i) can be used to express m.

“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”

Since Elina is new to programming, this problem is too difficult for her. Can you help her?

Input

The input contains multiple test cases. Each test cases consists of some lines.

• Line 1: Contains the integer k.
• Lines 2 ~ k + 1: Each contains a pair of integers a_i, r_i (1 ≤ i ≤ k).

Output

Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.

Sample Input

2
8 7
11 9

Sample Output

31

Hint

All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

思路：解同余方程，中国剩余定理，由于每个余数不一定互质，就需要两两合并方程组，给出代码与参考博客：

typedef long long LL;
typedef pair<LL, LL> PLL;

const int maxm = 1e5+5;

LL r[maxm], a[maxm];

void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d) {
    if(!b) {
        d = a, x = 1, y = 0;
    } else {
        ex_gcd(b, a%b, y, x, d);
        y -= x*(a/b);
    }
}

LL inv(LL t, LL p) {
    LL x, y, d;
    ex_gcd(t, p, x, y, d);
    return d == 1?(x%p+p)%p:-1;
}

LL gcd(LL a, LL b) {
    return b?gcd(b, a%b):a;
}

PLL linear(LL r[], LL a[], int n) { // x = r[i] (moda[i])
    LL x = 0, m = 1;
    for(int i = 0; i < n; ++i) {
        LL A = m, B = r[i] - x, d = gcd(a[i], A);
        if(B % d != 0) return PLL(0, -1);
        LL t = B/d * inv(A/d, a[i]/d) % (a[i]/d);
        x = x + m*t;
        m *= a[i]/d;
    }
    x = (x % m + m) % m;
    return PLL(x, m);
}

int main() {
    int n;
    while(scanf("%d", &n) != EOF) {
        for(int i = 0; i < n; ++i) {
            scanf("%lld%lld", &a[i], &r[i]);
        }
        PLL ans = linear(r, a, n);
        if(ans.second == -1) printf("-1\n");
        else printf("%lld\n", ans.first);
    }
    return 0;
}

View Code

https://www.cnblogs.com/linyujun/p/5199415.html

https://blog.csdn.net/acdreamers/article/details/8050018