很容易得到n × m的方块数是

然后就是个求和的问题了,枚举两者中小的那个n ≤ m。

然后就是转化成a*m + c = x了。a,m≥0,x ≥ c。最坏是n^3 ≤ x,至于中间会不会爆,测下1e18就好。

#include<bits/stdc++.h>
using namespace std; typedef long long ull; vector<ull> ns;
vector<ull> ms; //#define LOCAL
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif
ull x, t, c, a, n, m; cin>>x;
int k = ;
//ns.push_back(1); ms.push_back(x);
int equ = ;
for(n = ; ; n++){
t = n*(n+)/;
c = (n)*(n+)*(*n+)/ - n*t;
if(c > x) break;
a = n*(n+) - t;
if((x - c) % a == ) {
m = (x-c)/a;
if(m < n) break;
ns.push_back(n);
ms.push_back(m);
k++;
if(m == n){
equ = ; break;
}
}
} if(equ){
k = *k-;
printf("%d\n", k);
int sz = ns.size();
for(int i = ; i < sz; i++){
printf("%I64d %I64d\n", ns[i], ms[i]);
}
for(int i = sz-; i >= ; i--){
printf("%I64d %I64d\n", ms[i], ns[i]);
}
}
else {
k = *k;
printf("%d\n", k);
int sz = ns.size();
for(int i = ; i < sz; i++){
printf("%I64d %I64d\n", ns[i], ms[i]);
}
for(int i = sz-; i >= ; i--){
printf("%I64d %I64d\n", ms[i], ns[i]);
}
} return ;
}

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