题解 \(by\;zj\varphi\)
相当于是问是哪一段连乘等于给定的数。
发现连乘的数的个数一定不会很多,最多不会超过 \(20\)
所以可以枚举是多少数连乘,有多少数连乘就开几次方,这样得到的数最多和答案连乘的中间数不会超过 \(5\),直接暴力即可。
Code
#include<bits/stdc++.h>
#define ri signed
#define pd(i) ++i
#define bq(i) --i
#define func(x) std::function<x>
namespace IO{
char buf[1<<21],*p1=buf,*p2=buf;
#define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
#define debug1(x) std::cerr << #x"=" << x << ' '
#define debug2(x) std::cerr << #x"=" << x << std::endl
#define Debug(x) assert(x)
struct nanfeng_stream{
template<typename T>inline nanfeng_stream &operator>>(T &x) {
bool f=false;x=0;char ch=gc();
while(!isdigit(ch)) f|=ch=='-',ch=gc();
while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
return x=f?-x:x,*this;
}
}cin;
}
using IO::cin;
namespace nanfeng{
#define FI FILE *IN
#define FO FILE *OUT
template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
using ll=long long;
using ull=unsigned long long;
int T;
ll st[23],as[23],n;
ull jud;
auto check=[](ll sa,int len) {
ull res=1,lim=sa+len-1;
for (ull i(sa);i<=lim;pd(i)) res*=i;
return res;
};
inline int main() {
FI=freopen("factorial.in","r",stdin);
FO=freopen("factorial.out","w",stdout);
cin >> T;
for (ri z(1);z<=T;pd(z)) {
cin >> n;
if (n==1) {printf("-1\n");continue;}
ll res=1;
int cnt=1;
st[as[cnt]=1]=n;
for (ri i(2);i<=20;pd(i)) {
if (n/i<res) break;
res*=i;
ll tmp=ceil(pow(1.0*n,1.0/(1.0*i)));
for (ll j(tmp);j<=tmp+5;pd(j)) {
if (j-(i>>1)<=1) continue;
if ((jud=check(j-(i>>1),i))==n) {
st[as[++cnt]=i]=j-(i>>1);
break;
} else if (jud>n) break;
}
}
printf("%d\n",cnt);
for (ri i(cnt);i;bq(i))
printf("%lld %lld\n",st[as[i]]+as[i]-1,st[as[i]]-1);
}
return 0;
}
}
int main() {return nanfeng::main();}