P5596 洛谷月赛 题 题解

因为a>=0,b>=0,所以y^2-x^2>=0,所以y>x,因为都是自然数设y=x+k,化简得x=b-k^2/2*k-a;可知x仅当b-k^2%2*k-a==0且b-k^2与2*k-a同号时为一个解.枚举k即可,易知上界为b^1/2或a/2

 #include <iostream>
 #include <cstdio>
 #include <memory.h>
 #include <cmath>
 #include <stdlib.h>
 #define int ll
 #define re register
 #define r(x) x=read()
 using namespace std;
 typedef long long ll;
 int MAX(int a,const int &b){if(a<b)a=b;return a;}
 int MIN(int a,const int &b){if(a>b)a=b;return b;}
 ll read()
 {
   char ch=;ll w=,ff=;
   while(ch<''||ch>''){if(ch=='-')ff=-;ch=getchar();}
   while(ch>=''&&ch<=''){w=w*+ch-'';ch=getchar();}
   return ff*w;
 }
 ll a,b,k,cnt,cnt2;
 int ans;
 signed main()
 {
   r(a),r(b);
   if(a==&&b==) printf("inf"),exit();
   if(*b==a*a) printf("inf"),exit();
   cnt2=a/,cnt=sqrt(b);
   if(cnt2>cnt) swap(cnt,cnt2);
   cnt2--,cnt++;
   for(re ll i=cnt2,x,y;i<=cnt;++i)
   {
     if(i<) continue;
     x=b-i*i,y=*i-a;
     if(y!=&&(x<=&&y<=||x>=&&y>=)&&x%y==) ans++;
   }
   printf("%lld",ans);
   return ;
 }