Codeforces gym101612 L.Little Difference（枚举+二分）

传送：http://codeforces.com/gym/101612

题意：给定一个数n（<=1e18），将n分解为若干个数的成绩。要求这些数两两之间的差值不能大于1。

分析：

若n==2^k，则答案一定是-1。

然后，考虑若n==a^k，枚举k，二分求a。若n==a^x*(a+1)^y，枚举x，y，二分求解a。

注意：两数相乘可能>1e18，特判。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef pair<ll,ll> pll;
 typedef pair<pair<ll,ll>,ll> plll;
 const int maxn=1e6+;
 const ll inf=1e18+1e6;
 ll n;
 vector<pll> g1;
 vector<plll> g2;
 ll mul(ll a,ll b){
     if (a>=1.0*inf/b) return inf;
     else return a*b;
 }
 ll _pow(ll a,ll b){
     ll res=,base=a;
     while (b){
         if (b&) res=mul(res,base);
         base=mul(base,base);
         b>>=;
     }
     return res;
 }
 ll solve(int k){
     ll l=,r=n,ans,mid;
     while (l<=r){
         mid=(l+r)>>;
         ll tmp=_pow(mid,k);
         if (tmp==n) return mid;
         if (tmp<n) l=mid+; else r=mid-;
     }
     return ;
 }
 ll solve2(int x,int y){
     ll l=,r=n,ans,mid;
     while (l<=r){
         mid=(l+r)>>;
         ll tmp=_pow(mid,x),tmp2=_pow(mid+,y);
         if (mul(tmp,tmp2)==n) return mid;
         if (mul(tmp,tmp2)<n) l=mid+; else r=mid-;
     }
     return ;
 }
 int main(){
     //freopen("little.in","r",stdin);freopen("little.out","w",stdout);
     ios::sync_with_stdio(false);
     cin >> n;
     if (n==(n&(-n))) return cout << - << endl,;
     g1.clear(); g2.clear();
     ll num=;
     // a^k
     for(int i=;i<=;i++){
         ll a=solve(i);
         if (_pow(a,i)==n){
             g1.push_back({i,a});
         }
     }
     // a^x * (a+1)^y
     for(int i=;i<=;i++){
         for (int j=;j<=;j++){
             ll a=solve2(i,j);
             ll tmp=_pow(a,i),tmp2=_pow(a+,j);
             if (mul(tmp,tmp2)==n){
                 g2.push_back({{i,j},a});
             }
         }
     }
     cout << g1.size()+g2.size() << endl;
     for (auto i:g1){
         ll tmp=i.first;
         cout << tmp;
         for (int j=;j<tmp;j++) cout << " " << i.second;
         cout << endl;
     }
     for (auto i:g2){
         ll tmp=i.first.first+i.first.second;
         cout << tmp;
         tmp=i.first.first;
         for (int j=;j<tmp;j++) cout << " " << i.second;
         tmp=i.first.second;
         for (int j=;j<tmp;j++) cout << " " << i.second+;
         cout << endl;
     }
     return ;
 }