洛谷P3868（中国剩余定理,快速乘）

题目链接

根据题意，有 n − a i ≡ 0 ( m o d 　 b i ) n-a_i\equiv 0(mod 　bi) n−ai​≡0(mod　bi)
根据同余变换法则，可以得到 n ≡ a i ( m o d 　 b i ) n\equiv a_i(mod 　bi) n≡ai​(mod　bi)，这就是一个中国剩余定理裸题了。

另外， 1 e 8 × 1 e 8 1e8\times 1e8 1e8×1e8爆 l o n g l o n g long long longlong的问题，可以使用快速乘解决，快速乘和快速幂原理相同，快速幂是 b b b个 a a a相乘，快速乘则是 b b b个 a a a相加，将乘法变换成加法处理，自然不会爆 l o n g l o n g long long longlong了。

AC代码：

/*
    * @Author: hesorchen
    * @Date: 2020-07-02 22:19:34
 * @LastEditTime: 2020-07-04 08:21:41
    * @Description: https://hesorchen.github.io/
    */
#include <map>
#include <set>
#include <list>
#include <queue>
#include <deque>
#include <cmath>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl '\n'
#define PI acos(-1)
#define PB push_back
#define ll long long
#define INF 0x3f3f3f3f
#define mod 10007
#define pll pair<ll, ll>
#define lowbit(abcd) (abcd & (-abcd))
#define max(a, b) ((a > b) ? (a) : (b))
#define min(a, b) ((a < b) ? (a) : (b))

#define IOS                      \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0);
#define FRE                              \
    {                                    \
        freopen("in.txt", "r", stdin);   \
        freopen("out.txt", "w", stdout); \
    }

inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
//==============================================================================

void exgcd(ll a, ll b, ll &x, ll &y)
{
    if (!b)
    {
        x = 1;
        y = 0;
        return;
    }
    exgcd(b, a % b, x, y);
    ll temp = x;
    x = y;
    y = temp - (a / b) * y;
}

ll ksc(ll a, ll b, ll p) //快速乘
{
    ll res = 0;
    while (b)
    {
        if (b & 1)
            res = (res + a) % p;
        a = (a + a) % p;
        b /= 2;
    }
    return res;
}

ll a[15], b[15];

int main()
{
    ll k, M = 1;
    cin >> k;
    for (int i = 1; i <= k; i++)
        cin >> a[i];
    for (int i = 1; i <= k; i++)
        cin >> b[i], M *= b[i];

    ll x, y, ans = 0, temp;

    for (int i = 1; i <= k; i++)
    {
        ll Mi = M / b[i];
        exgcd(Mi, b[i], x, y);
        ans += ksc((x % M + M) % M, ksc((a[i] % M + M) % M, Mi, M), M);
    }
    cout << (ans % M + M) % M << endl;

    return 0;
}