Regionals 2015 >> Asia - EC Final>> Multiplication Table

/*由两个点的值及相对位置求解左上角点的坐标，解的判断很麻烦,wr*/
#include
#include
#include
using namespace std;
typedef long long ll;
const ll M=1e9,N=1e3+10;
struct point
{
    int x,y,v;
}pq[N*N];
int n,m,k;
int jud(ll x,ll y)
{
    int x0=pq[0].x,y0=pq[0].y;
    for(int i=2;i='0'&&ch<='9')) break;
            num=num*10+(ch-'0');
        }
        pq[k].x=i;pq[k].y=j;pq[k++].v=num;
      }
}
int main()
{

   int t,ca=1;
   scanf("%d",&t);
   while(t--)
   {
       read();
       printf("Case #%d: ",ca++);
        if(k==0) {printf("Yes\n");continue;}
        if(k==1)
        {
          int flag=0;
          for(int i=1;i*i<=pq[0].v;i++)
          {
              if(pq[0].v%i==0&&pq[0].v/i>=pq[0].y&&i>=pq[0].x) {flag=1;printf("Yes\n");break;}
              if(pq[0].v%i==0&&pq[0].v/i>=pq[0].x&&i>=pq[0].y){flag=1;printf("Yes\n");break;}
          }
          if(flag) continue;
          printf("No\n");continue;
        }

        int n1=pq[1].x-pq[0].x,m1=pq[1].y-pq[0].y;
        int c1=pq[0].v,c2=pq[1].v;
        ll c=(c2-c1-n1*m1)*(c2-c1-n1*m1)-4*n1*m1*c1;//b^2-4ac
        if(c<0) {printf("No\n");continue;}
        double temp=sqrt(c);
        if(temp-(ll)temp>0.00000001) {printf("No\n");continue;}
        ll x1,y1,x2,y2;
        if(pq[0].x==pq[1].x)
        {
            if((c2-c1)>=m&&(c2-c1)%m1==0)
            {
               x1=(c2-c1)/m1;
               if(c2>=x1&&c2%x1==0)
               {
                   y1=c2/x1;
                   if(jud(x1,y1)){printf("Yes\n");continue;}
               }
            }
            printf("No\n");
            continue;
        }
        if(pq[0].y==pq[1].y)
        {
            if((c2-c1)>=n1&&(c2-c1)%n1==0)
            {
               y1=(c2-c1)/n1;
               if(c2>=y1&&c2%y1==0)
               {
                   x1=c2/y1;
                   if(jud(x1,y1)){printf("Yes\n");continue;}
               }
            }
             printf("No\n");
             continue;
        }
        y1=c2-c1-n1*m1+temp;y2=c2-c1-n1*m1-temp;//y1,y2
        if(y1>=1&&y1<=M&&y1>=(2*n1)&&y1%(2*n1)==0)
        {
            y1=y1/(2*n1);
            if(c1>=y1&&c1%y1==0)
            {
                x1=c1/y1;
               if(jud(x1,y1)) {printf("Yes\n");continue;}
            }
        }
        if(y2>=1&&y2<=M&&y2>=(2*n1)&&y2%(2*n1)==0)
        {
            y2=y2/(2*n1);
            if(c1>=y2&&c1%y2==0)
            {
                x2=c2/y2;
                if(jud(x2,y2)){printf("Yes\n");continue;}
            }
        }
        printf("No\n");
   }
    return 0;
}

/*暴力枚举sqrt级ac*/
#include
#include
#include
using namespace std;
const int N=1000;
struct Node
{
  int x,y,v;
}a[N*N+100];
int jud(int p)
{
             int flag;
             for(int i=1;i*i<=a[0].v;i++)
             {
                 if(a[0].v%i==0)
                 {
                     if(a[0].v/i>=a[0].x&&i>=a[0].y)
                     {
                         flag=0;
                         for(int k=1;k=a[0].y&&i>=a[0].x)
                     {
                         flag=0;
                         for(int k=1;k=a[0].x&&i>=a[0].y){flag=1;break;}
                     if(a[0].v/i>=a[0].y&&i>=a[0].x){flag=1;break;}
                 }
             }
             if(flag) {printf("Yes\n");}
             else printf("No\n");
         }
         else
         {
             if(jud(p)) printf("Yes\n");
             else printf("No\n");
         }
    }
    return 0;
}

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