博客介绍了三道算法题。“Game 23”需将数 n 经乘 2 或乘 3 变换为 m，求最少步数；“Polycarp Restores Permutation”要根据差值数组还原排列；“Vova and Train”计算火车行驶中可见灯笼数量。还给出了各题的正确解法。
A. Game 23
Description
Polycarp plays "Game 23". Initially he has a number ?n and his goal is to transform it to ?m. In one move, he can multiply ?n by 22 or multiply ?n by 33. He can perform any number of moves.
Print the number of moves needed to transform ?n to ?m. Print -1 if it is impossible to do so.
It is easy to prove that any way to transform ?n to ?m contains the same number of moves (i.e. number of moves doesn't depend on the way of transformation).
Input
The only line of the input contains two integers ?n and ?m (1≤?≤?≤5⋅1081≤n≤m≤5⋅108).
Output
Print the number of moves to transform ?n to ?m, or -1 if there is no solution.
Examples
Input
120 51840
Output
7
C. Polycarp Restores Permutation
Description
An array of integers ?1,?2,…,??p1,p2,…,pn is called a permutation if it contains each number from 11 to ?n exactly once. For example, the following arrays are permutations: [3,1,2][3,1,2], [1][1], [1,2,3,4,5][1,2,3,4,5] and [4,3,1,2][4,3,1,2]. The following arrays are not permutations: [2][2], [1,1][1,1], [2,3,4][2,3,4].
Polycarp invented a really cool permutation ?1,?2,…,??p1,p2,…,pn of length ?n. It is very disappointing, but he forgot this permutation. He only remembers the array ?1,?2,…,??−1q1,q2,…,qn−1 of length ?−1n−1, where ??=??+1−??qi=pi+1−pi.
Given ?n and ?=?1,?2,…,??−1q=q1,q2,…,qn−1, help Polycarp restore the invented permutation.
Input
The first line contains the integer ?n (2≤?≤2⋅1052≤n≤2⋅105) — the length of the permutation to restore. The second line contains ?−1n−1 integers ?1,?2,…,??−1q1,q2,…,qn−1 (−?<??<?−n<qi<n).
Output
Print the integer -1 if there is no such permutation of length ?n which corresponds to the given array ?q. Otherwise, if it exists, print ?1,?2,…,??p1,p2,…,pn. Print any such permutation if there are many of them.
Examples
Input
3
-2 1
Output
3 1 2
正确解法：
设ans[1]=x,则ans[2]=x+p[1],ans[3]=x+p[1]+p[2],ans[4]=x+p[1]+p[2]+p[3];
则其中最小的就是1.因为是1-n的序列。
我们手上有p的序列，然后求前缀和。得到 { 0, p[1], p[1]+p[2],.....}
最小的与1求差值。 x+minn=1. x=1-minn。
然后所有都加上x就是答案。
1 include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 using namespace std; 7 typedef long long ll; 8 const int inf=0x7fffffff; 9 const int N=200000+100; 10 int n,a[N],b[N],minn=0; 11 int main() 12 { 13 scanf("%d",&n); 14 for(int i=2;i<=n;i++) 15 { 16 scanf("%d",&a[i]); 17 a[i]=a[i-1]+a[i]; 18 minn=min(minn,a[i]); 19 } 20 minn=-minn+1; 21 for(int i=1;i<=n;i++) 22 { 23 a[i]+=minn; 24 b[i]=a[i]; 25 } 26 sort(b+1,b+n+1); 27 for(int i=1;i<=n;i++) 28 if(b[i]!=i) 29 { 30 cout<<-1<<endl; 31 return 0; 32 } 33 for(int i=1;i<=n;i++) 34 cout<<a[i]<<" "; 35 cout<<endl; 36 37 return 0; 38 }
A. Vova and Train
Description
Vova plans to go to the conference by train. Initially, the train is at the point 11 and the destination point of the path is the point LL. The speed of the train is 11 length unit per minute (i.e. at the first minute the train is at the point 11, at the second minute — at the point 22 and so on).
There are lanterns on the path. They are placed at the points with coordinates divisible by vv (i.e. the first lantern is at the point vv, the second is at the point 2v2v and so on).
There is also exactly one standing train which occupies all the points from ll to rr inclusive.
Vova can see the lantern at the point pp if pp is divisible by vv and there is no standing train at this position (p∉[l;r]p∉[l;r]). Thus, if the point with the lantern is one of the points covered by the standing train, Vova can't see this lantern.
Your problem is to say the number of lanterns Vova will see during the path. Vova plans to go to tt different conferences, so you should answer tt independent queries.
Input
The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of queries.
Then tt lines follow. The ii-th line contains four integers Li,vi,li,riLi,vi,li,ri (1≤L,v≤1091≤L,v≤109, 1≤l≤r≤L1≤l≤r≤L) — destination point of the ii-th path, the period of the lantern appearance and the segment occupied by the standing train.
Output
Print tt lines. The ii-th line should contain one integer — the answer for the ii-th query.
Examples
Input
4
10 2 3 7
100 51 51 51
1234 1 100 199
1000000000 1 1 1000000000
Output
3
0
1134
0
A. Vova and Train
Description
Vova plans to go to the conference by train. Initially, the train is at the point 11 and the destination point of the path is the point LL. The speed of the train is 11 length unit per minute (i.e. at the first minute the train is at the point 11, at the second minute — at the point 22 and so on).
There are lanterns on the path. They are placed at the points with coordinates divisible by vv (i.e. the first lantern is at the point vv, the second is at the point 2v2v and so on).
There is also exactly one standing train which occupies all the points from ll to rr inclusive.
Vova can see the lantern at the point pp if pp is divisible by vv and there is no standing train at this position (p∉[l;r]p∉[l;r]). Thus, if the point with the lantern is one of the points covered by the standing train, Vova can't see this lantern.
Your problem is to say the number of lanterns Vova will see during the path. Vova plans to go to tt different conferences, so you should answer tt independent queries.
Input
The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of queries.
Then tt lines follow. The ii-th line contains four integers Li,vi,li,riLi,vi,li,ri (1≤L,v≤1091≤L,v≤109, 1≤l≤r≤L1≤l≤r≤L) — destination point of the ii-th path, the period of the lantern appearance and the segment occupied by the standing train.
Output
Print tt lines. The ii-th line should contain one integer — the answer for the ii-th query.
Examples
Input
4
10 2 3 7
100 51 51 51
1234 1 100 199
1000000000 1 1 1000000000
Output
3
0
1134
0
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