问满足a^3 + b^3 + c^3 + n = (a+b+c)^3 的 (a,b,c)的个数
可化简为 n = 3*(a + b) (a + c) (b + c)
于是 n / 3 = (a + b)(a + c) (b + c)
令x = a + b,y = a + c,z = b + c,s = n / 3
s = xyz
并且令x <= y <= z，于是我们解s = xyz这个方程，可以枚举x,y得到z。
得到(x,y,z)后便可以得到a,b,c但可能有不符合条件的三元组，化简系数矩阵
1 1 0 由于枚举时已满足x <= y <= z 2 0 0 x + y - z >= 0，即 x + y >= z
1 0 1 1 0 1
0 1 1 0 1 1
另外如果x = y = z时此时只贡献了一个答案，如果x = y || y = z 答案只贡献了3个
其余贡献了6个
#include <cmath> #include <cstdio> #include <cstdlib> #include <cassert> #include <cstring> #include <set> #include <map> #include <list> #include <queue> #include <string> #include <iostream> #include <algorithm> #include <functional> #include <stack> using namespace std; typedef long long ll; #define T int t_;Read(t_);while(t_--) #define dight(chr) (chr>='0'&&chr<='9') #define alpha(chr) (chr>='a'&&chr<='z') #define INF (0x3f3f3f3f) #define maxn (2000005) #define maxm (10005) #define mod 1000000007 #define ull unsigned long long #define repne(x,y,i) for(int i=(x);i<(y);++i) #define repe(x,y,i) for(int i=(x);i<=(y);++i) #define repde(x,y,i) for(int i=(x);i>=(y);--i) #define repdne(x,y,i) for(int i=(x);i>(y);--i) #define ri register int inline void Read(int &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if(chr=='-')sign=-1; for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;} inline void Read(ll &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if (chr=='-')sign=-1; for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;} ll n; int powx(int c,ll t){ ll l = 1,r = sqrt(t); while(l <= r){ ll mid = (l + r) >> 1,s = 1; repe(1,c,i) s *= mid; if(s > t) r = mid - 1; else if(s < t) l = mid + 1; else return (int)mid; } return (int)r; } int main() { //freopen("a.in","r",stdin); //freopen("b.out","w",stdout); Read(n); if(n % 3){ puts("0"); return 0; } n /= 3; int lix = powx(3,n); int ans = 0; repe(2,lix,x){ if(n % x) continue; ll nx = n / x; int liy = powx(2,nx); repde(liy,x,y){ if(nx % y) continue; int z = nx / y; if(x + y <= z) break; if((x + y + z) & 1) continue; if(x == y && y == z) ++ans; else if(x == y || y == z) ans += 3; else ans += 6; } } cout << ans << endl; return 0; /* t = a + b (a-b)^2 + 4s/t = k^2 */ }