中国剩余定理与与拓展

#include
using namespace std;
int n,lcm=1,A,M,a[105],m[105];
int exgcd(int a,int b,int&x,int&y)
{
	if(b==0) {
		x=1,y=0;return a;
	}
	int d=exgcd(b,a%b,x,y);
	int k=x;x=y,y=k-(a/b)*y;
	return d;
}
int work()
{
	int re=0,x,y;
	for(int i=1;i<=n;i++) lcm*=m[i];
	for(int i=1;i<=n;i++){
		M=lcm/m[i];
		int d=exgcd(M,m[i],x,y);
		x=(x%m[i]+m[i])%m[i];
		re=(re+x*M*a[i])%lcm;
	}
	return re;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d%d",&m[i],&a[i]);
	printf("%d\n",work());
	return 0;
}

#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long lt;

lt read()
{
    lt f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}

const int maxn=100010;
int n;
lt ai[maxn],bi[maxn];

lt mul(lt a,lt b,lt mod)
{
    lt res=0;
    while(b>0)
    {
        if(b&1) res=(res+a)%mod;
        a=(a+a)%mod;
        b>>=1;
    }
    return res;
}

lt exgcd(lt a,lt b,lt &x,lt &y)
{
    if(b==0){x=1;y=0;return a;}
    lt gcd=exgcd(b,a%b,x,y);
    lt tp=x;
    x=y; y=tp-a/b*y;
    return gcd;
}

lt excrt()
{
    lt x,y,k;
    lt M=bi[1],ans=ai[1];//第一个方程的解特判
    for(int i=2;i<=n;i++)
    {
        lt a=M,b=bi[i],c=(ai[i]-ans%b+b)%b;//ax≡c(mod b)
        lt gcd=exgcd(a,b,x,y),bg=b/gcd;
        if(c%gcd!=0) return -1; //判断是否无解，然而这题其实不用

        x=mul(x,c/gcd,bg);//把x转化为最小非负整数解
        ans+=x*M;//更新前k个方程组的答案
        M*=bg;
        ans=(ans%M+M)%M;
    }
    return (ans%M+M)%M;
}

int main()
{
    n=read();
    for(int i=1;i<=n;++i)
    bi[i]=read(),ai[i]=read();
    printf("%lld",excrt());
    return 0;
}