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b1/2或a/2
1 #include <iostream>
2 #include <cstdio>
3 #include <memory.h>
4 #include <cmath>
5 #include <stdlib.h>
6 #define int ll
7 #define re register
8 #define r(x) x=read()
9 using namespace std;
10 typedef long long ll;
11 int MAX(int a,const int &b){if(a<b)a=b;return a;}
12 int MIN(int a,const int &b){if(a>b)a=b;return b;}
13 ll read()
14 {
15 char ch=0;ll w=0,ff=1;
16 while(ch<'0'||ch>'9'){if(ch=='-')ff=-1;ch=getchar();}
17 while(ch>='0'&&ch<='9'){w=w*10+ch-'0';ch=getchar();}
18 return ff*w;
19 }
20 ll a,b,k,cnt,cnt2;
21 int ans;
22 signed main()
23 {
24 r(a),r(b);
25 if(a==0&&b==0) printf("inf"),exit(0);
26 if(4*b==a*a) printf("inf"),exit(0);
27 cnt2=a/2,cnt=sqrt(b);
28 if(cnt2>cnt) swap(cnt,cnt2);
29 cnt2--,cnt++;
30 for(re ll i=cnt2,x,y;i<=cnt;++i)
31 {
32 if(i<0) continue;
33 x=b-i*i,y=2*i-a;
34 if(y!=0&&(x<=0&&y<=0||x>=0&&y>=0)&&x%y==0) ans++;
35 }
36 printf("%lld",ans);
37 return 0;
38 }