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==) {x=,y=; 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=;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\n",r1); else puts("-1");
     }
     return ;
 }

POJ 2891

POJ 2407 欧拉函数

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

POJ 2407