预处理全部q2+p2有20000多个。
枚举任意两个的差,并以这个差扩展找到连续的有几个
为了避免重复浪费时间,只需判断无法向左扩展的一对数,具体见代码。
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("d:\in1.txt","r",stdin); freopen("d:\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!=' ')return ch; } return EOF; } const int M = 255; int Set[M*M*2]; int bisquares[M*M]; int n,m; struct result { int a,b; result(int a=0,int b=0):a(a),b(b){} bool operator < (const result &ant) const { if(b!=ant.b)return b<ant.b; else return a<ant.a; } }res[11000]; int init() { memset(Set,0,sizeof(Set)); for(int i=0;i<=m;i++) { for(int j=0;j<=m;j++) Set[i*i+j*j]=1; } int num=0; for(int i=0;i<=m*m*2;i++) { if(Set[i]) bisquares[num++]=i; } return num; } int check(int a,int b) { while(a+b<=m*m*2&&Set[a+b]) { a+=b; } return a; } int main() { freopen("ariprog.in","r",stdin); freopen("ariprog.out","w",stdout); while(scanf("%d%d",&n,&m)!=EOF) { int num = init(); int k=0; for(int i=0;i<num;i++) { for(int j=i+1;j<num;j++) { int b=bisquares[j]-bisquares[i]; int pre=bisquares[i]-b; if(pre>=0&&Set[pre])continue; int x=check(bisquares[j],b); for(int a=bisquares[i];a+(n-1)*b<=x;a+=b) res[k++]=result(a,b); } } sort(res,res+k); if(k==0) printf("NONE "); else for(int i=0;i<k;i++) printf("%d %d ",res[i].a,res[i].b); } return 0; }