题意简述
给定(a[1],a[2],cdots,a[n]) 和 (b[1],b[2],cdots,b[n]),其中(b)中元素两两互素。
求最小的非负整数(n)，满足对于任意的(i)，(n - a[i])能被(b[i])整除。
题解思路
变形后用中国剩余定理即可，注意要快速乘。
代码
#include <cstdio>
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll n,x,y,lcm=1,ans;
ll a[15],p[15];
inline ll _mul(ll x,ll y,ll p) {
ll z=(ld)x/p*y;
ll res=(ull)x*y-(ull)z*p;
return (res+p)%p;
}
void exgcd(const ll& a,const ll& b,ll& x,ll& y) {
if (!b) { x=1; y=0; return; }
exgcd(b,a%b,x,y);
ll tmp=y; y=x-(a/b)*y; x=tmp;
}
int main() {
scanf("%lld",&n);
for (register int i=1;i<=n;++i) scanf("%lld",&a[i]);
for (register int i=1;i<=n;++i) scanf("%lld",&p[i]),lcm*=p[i];
for (register int i=1;i<=n;++i) a[i]=(a[i]%p[i]+p[i])%p[i];
for (register int i=1;i<=n;++i) {
ll xx=lcm/p[i]; exgcd(xx,p[i],x,y);
x=(x%p[i]+p[i])%p[i];
(ans+=_mul(_mul(x,xx,lcm),a[i],lcm))%=lcm;
}
printf("%lld
",ans);
}