NOIP 模拟 $89\; \rm 谜之阶乘$

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();}