《洛谷P3868 [TJOI2009]猜数字》

将题意转化下，就是求满足$(n-a[i]) equiv 0 ~mod~(b[i])$的最小的非负整数x。

转化一下$(n-a[i]) equiv 0 ~mod~(b[i]) ightarrow  n equiv a[i]~ mod (b[i])$

那么，就是个CRT。

这里的话，中间会爆longlong，就算中间取模还是会爆，所以套个快速乘来防止爆longlong

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 2e3+5;
const int M = 2e6+5;
const LL Mod = 1e9+7;
#define rg register
#define pi acos(-1)
#define INF 1e9
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
    void print(int x){
        if(x < 0){x = -x;putchar('-');}
        if(x > 9) print(x/10);
        putchar(x%10+'0');
    }
}
using namespace FASTIO;
void FRE(){/*freopen("data1.in","r",stdin);
freopen("data1.out","w",stdout);*/}

LL a[15],b[15];
LL exgcd(LL a,LL b,LL &x,LL &y)
{
    if(b == 0)
    {
        x = 1,y = 0;
        return a;
    }
    LL t = exgcd(b,a%b,y,x);
    y -= a/b*x;
    return t;
}
LL Mul(LL a,LL b,LL p)
{
    LL re = 0;
    while(b)
    {
        if(b&1) re = (re+a)%p;
        a = (a+a)%p;
        b >>= 1;
    }
    return re;
}
int main()
{
    int n;n = read();
    LL M = 1;
    for(rg int i = 1;i <= n;++i) a[i] = read();
    for(rg int i = 1;i <= n;++i) b[i] = read(),M *= b[i];
    LL ans = 0;
    for(rg int i = 1;i <= n;++i)
    {
        LL Mi = M/b[i],x,y;
        LL gcd = exgcd(Mi,b[i],x,y);
        x = (x+b[i])%b[i];
        LL ma = Mul(x*a[i]%M,Mi,M);
        ans = (ans+ma)%M;
    }
    printf("%lld
",ans);
    system("pause");    
}