NOIP 考前 数论复习

POJ 2891

x=r1 (mod a1)

x=r2 (mod a2)

x=a1*x+r1,x=a2*y+r2;

a1*x-a2*y=r2-r1; 

用Extend_Gcd求出m1*x+m2*y=d; d=Gcd(x,y);

那么就可以解出原来的x=(x*(r2-r1)/d)

那么代入原式r1=a1*x+r1 新的a1=lcm(a1,a2);

#include <cstdio>
#define LL long long
LL a1,a2,r1,r2,x,y,n; 
LL Extend_Gcd(LL a,LL b,LL &x,LL &y)
{
    if (b==0) {x=1,y=0; return a;}
    LL Ret=Extend_Gcd(b,a%b,x,y);
    LL Tmp=x;
    x=y; y=(Tmp-a/b*y);
    return Ret;
}
int main()
{
    // freopen("c.in","r",stdin);
    while (scanf("%lld",&n)!=EOF)
    {
        scanf("%lld%lld",&a1,&r1); bool Ok=true;
        for (LL i=1;i<n;i++)
        {
            scanf("%lld%lld",&a2,&r2);
            LL a=a1,b=a2,m=r2-r1;
            LL d=Extend_Gcd(a,b,x,y);
            if (m%d) Ok=false;
            x=(((x*m/d))%(b/d)+(b/d))%(b/d);
            r1=x*a1+r1;
            a1=(a1*a2)/d;
        }
        if (Ok) printf("%lld
",r1); else puts("-1");
    }
    return 0;
}

 POJ 2407 欧拉函数

#include <cstdio>
int n;
int main()
{
    while (scanf("%d",&n)!=EOF)
    {
        if (n==0) break;
        int Ans=1; int Tmp=n;
        for (int i=2;i*i<=Tmp;i++)
            if (n%i==0)
            {
                Ans=Ans*(i-1);
                n=n/i;
                while (n%i==0) Ans=Ans*i,n=n/i;
            }
        if (n>1) Ans=Ans*(n-1);
        printf("%d
",Ans);
    }
    return 0;
}