Codeforces Round499-div2
Natasha is going to fly on a rocket to Mars and return to Earth. Also, on the way to Mars, she will land on n−2 intermediate planets. Formally: we number all the planets from 1 to n. 1 is Earth, n is Mars. Natasha will make exactly n flights: 1→2→…n→1.
Flight from x to y consists of two phases: take-off from planet x and landing to planet y. This way, the overall itinerary of the trip will be: the 1-st planet → take-off from the 1-st planet → landing to the 2-nd planet → 2-nd planet → take-off from the 2-nd planet → … → landing to the n-th planet → the n-th planet → take-off from the n-th planet → landing to the 1-st planet → the 1-st planet.
The mass of the rocket together with all the useful cargo (but without fuel) is m tons. However, Natasha does not know how much fuel to load into the rocket. Unfortunately, fuel can only be loaded on Earth, so if the rocket runs out of fuel on some other planet, Natasha will not be able to return home. Fuel is needed to take-off from each planet and to land to each planet. It is known that 1 ton of fuel can lift off aitons of rocket from the i-th planet or to land bi tons of rocket onto the i-th planet.
For example, if the weight of rocket is 9 tons, weight of fuel is 3 tons and take-off coefficient is 8 (ai=8), then 1.5 tons of fuel will be burnt (since 1.5⋅8=9+3). The new weight of fuel after take-off will be 1.5 tons.
Please note, that it is allowed to burn non-integral amount of fuel during take-off or landing, and the amount of initial fuel can be non-integral as well.
Help Natasha to calculate the minimum mass of fuel to load into the rocket. Note, that the rocket must spend fuel to carry both useful cargo and the fuel itself. However, it doesn't need to carry the fuel which has already been burnt. Assume, that the rocket takes off and lands instantly.
The first line contains a single integer n (2≤n≤1000) — number of planets.
The second line contains the only integer m (1≤m≤1000) — weight of the payload.
The third line contains n integers a1,a2,…,an (1≤ai≤1000), where ai is the number of tons, which can be lifted off by one ton of fuel.
The fourth line contains n integers b1,b2,…,bn (1≤bi≤1000), where bi is the number of tons, which can be landed by one ton of fuel.
It is guaranteed, that if Natasha can make a flight, then it takes no more than 109 tons of fuel.
If Natasha can fly to Mars through (n−2) planets and return to Earth, print the minimum mass of fuel (in tons) that Natasha should take. Otherwise, print a single number −1.
It is guaranteed, that if Natasha can make a flight, then it takes no more than 109 tons of fuel.
The answer will be considered correct if its absolute or relative error doesn't exceed 10−6. Formally, let your answer be p, and the jury's answer be q. Your answer is considered correct if |p−q|max(1,|q|)≤10−6.
2
12
11 8
7 5
10.0000000000
3
1
1 4 1
2 5 3
-1
6
2
4 6 3 3 5 6
2 6 3 6 5 3
85.4800000000
Let's consider the first example.
Initially, the mass of a rocket with fuel is 22 tons.
- At take-off from Earth one ton of fuel can lift off 11 tons of cargo, so to lift off 22 tons you need to burn 2 tons of fuel. Remaining weight of the rocket with fuel is 20 tons.
- During landing on Mars, one ton of fuel can land 5 tons of cargo, so for landing 20 tons you will need to burn 4 tons of fuel. There will be 16 tons of the rocket with fuel remaining.
- While taking off from Mars, one ton of fuel can raise 8 tons of cargo, so to lift off 16 tons you will need to burn 2 tons of fuel. There will be 14 tons of rocket with fuel after that.
- During landing on Earth, one ton of fuel can land 7 tons of cargo, so for landing 14 tons you will need to burn 2 tons of fuel. Remaining weight is 12 tons, that is, a rocket without any fuel.
In the second case, the rocket will not be able even to take off from Earth.
题意:
![](https://images2018.cnblogs.com/blog/1428967/201808/1428967-20180801170632149-542556130.png)
可得方程
m[i-1] = m[i] / ((1 - 1 / a[i-1])*(1 - 1 / b[i]));
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#define _CRT_SECURE_NO_WARNINGS #include<iostream> #include<iomanip> #include<algorithm> #include<cstring> #include<cstdio> #include<cstdlib> #include<queue> #include<cmath> using namespace std; typedef long long ll; typedef pair<int, int> pii; double a[1050]; double b[1050]; double c[1050]; int main() { int n, m; double res; cin >> n >> m; for (int i = 0; i < n; i++) { cin >> a[i]; if (a[i] <= 1) { cout << "-1"; return 0; } } for (int i = 0; i < n; i++) { cin >> b[i]; if (b[i] <= 1) { cout << "-1"; return 0; } } b[n] = b[0]; res = m; for (int i = n; i >= 1; i--) { res = res / ((1 - 1 / a[i-1])*(1 - 1 / b[i])); } cout << setiosflags(ios::fixed) << setprecision(14) << res - m; return 0; }
Description
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are nn banknote denominations on Mars: the value of ii-th banknote is aiai. Natasha has an infinite number of banknotes of each denomination.
Martians have kk fingers on their hands, so they use a number system with base kk. In addition, the Martians consider the digit dd (in the number system with base kk) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base kk is dd, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values dd Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers nn and kk (1≤n≤1000001≤n≤100000, 2≤k≤1000002≤k≤100000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values dd for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
2 8
12 20
2
0 4
3 10
10 20 30
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 1212, you will get 148148 in octal system. The last digit is 4848.
If you take one banknote with the value of 1212 and one banknote with the value of 2020, the total value will be 3232. In the octal system, it is 408408. The last digit is 0808.
If you take two banknotes with the value of 2020, the total value will be 4040, this is 508508 in the octal system. The last digit is 0808.
No other digits other than 0808 and 4848 can be obtained. Digits 0808 and 4848 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
题意
给定n个数字,每个数字可以选取无限次组成一个和,求这个和对k取余可能有多少种结果。
思路
由Bezout定理,只需要求出所有数字的最大公约数,然后求这个最大公约数有多少个不同的比k小的非负倍数即可。
【Bezout定理(Bézout's identity):若gcd(a,b)=d,则存在x,y使得ax+by=d;一般地,ax+by是d的整数倍。x,y叫做(a,b)的贝祖系数(Bézout coefficients),可由扩展欧几里德算法算出】
题解
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#define _CRT_SECURE_NO_WARNINGS #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<cstdlib> #include<queue> #include<cmath> using namespace std; typedef long long ll; typedef pair<int, int> pii; int gcd(int x, int y) { return y ? gcd(y, x%y) : x; } int a[100050]; bool tf[100050]; int main() { ios::sync_with_stdio(0); int n, k, res, tmp; res = 0; tmp = 0; cin >> n >> k; for (int i = 1; i <= n; i++) { cin >> a[i]; a[i] %= k; a[1] = gcd(a[1], a[i]); } while (true) { tmp = (tmp + a[1]) % k; if (tf[tmp] == true) break; tf[tmp] = true; } for (int i = 0; i < k; i++)if (tf[i])++res; cout << res << endl; for (int i = 0; i < k; i++)if (tf[i])cout << i << ' '; return 0; }