倒水问题

题目描述

There are N cups of water which are numbered from 1 to N. The i^th cup contains A_i liters of water, and the magical value of the i^th cup is B_i.
The 1^st operation will pour B₁ liters of water from the 1^st cup to the 2^nd cup.

The 2^nd operation will pour B₂ liters of water from the 2^nd cup to the 3^rd cup.

......

The N^th operation will pour B_N liters of water from the N^th cup to the 1^st cup.
The (N + 1)^th operation will pour B₁ liters of water from the 1^st cup to the 2^nd cup.
......
If the water in the cup is not enough to perform the operation, the game will stop immediately. The question is if this game will keep running forever? If not, how many times will it be operated?

输入描述:

The first line contains an integer T, where T is the number of test cases. T test cases follow. For each test case, the first line contains one integer N, where N is the number of cups.

The second line contains N integers A₁, A₂, ... , A_N, where A_i is the volume of water in the i^th cup.

The third line contains N integers B₁, B₂, ... , B_N, where B_i is the magical value of the i^th cup.

输出描述:

For each test case, output one line containing “Case #x: y”, where x is the test case number (startingfrom 1). If the game will keep running forever, y is “INF”. Otherwise, y is the operation times.

• 1 ≤ T ≤ 10². 

• 2 ≤ N ≤ 10². 

• 0 ≤ A_i ≤ 10⁹. 

• 1 ≤ B_i ≤ 10⁹.

示例1

输入

2
3
4 5 6
1 3 6
3
2 0 1
1 1 1

输出

Case #1: 7
Case #2: INF

备注:

For the first case, this game’s operation times is 7.
[4,5,6] ⇒ [3,6,6] ⇒ [3,3,9] ⇒ [9,3,3] ⇒ [8,4,3] ⇒ [8,1,6] ⇒ [14,1,0] ⇒ [13,2,0]
For the second case, this game will keep running forever.

    题目很简单，大意就是轮流到水，第一行是中的每个数是每个水杯中的初始水量，下面的那行数字是每次倒出去的水量，问最多能到多少次，最后一个水杯里的水倒到第一个被子中，如果可以无限循环的话输出inf 详情看备注。

思路：如果最小排水量和最大排水量一样那么就是无限循环。然后找出每次进水量和出水量的差值。第一个需要特判 之后的规律一样 另外需要注意maxx的数据范围。

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#define  N 0x3f3f3f3f
using namespace std;
int a[110],b[110],c[110],d[110];
int main()
{
	int t;
	cin>>t;
	for(int x=1;x<=t;x++)
	{
		memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
		int n;
		cin>>n;
		for(int i=1;i<=n;i++)
		{
			cin>>a[i];
		}
		for(int i=1;i<=n;i++)
		{
			cin>>b[i];
			c[i]=b[i];
		}
		int f=0;
		sort(c+1,c+n+1);
		if(c[1]==c[n])
		{
			f=1;
		}
		for(int i=1;i<=n;i++)
		{
			if(i==n)
				d[1]=b[n]-b[1];
			else
			d[i+1]=b[i]-b[i+1];
		}
		long long maxx=N;
		if(d[1]<0)
		{
			maxx=(a[1]-b[1])/(b[1]-b[n])+1;
		}
		int q,w=0;
		for(int i=2;i<=n;i++)
		{
			if(d[i]<0)
			{
				d[i]=-d[i];
				q=(a[i]/d[i]);
				if(maxx>q)
				maxx=q,w=i-1;
				
			}
		}
		if(a[1]<b[1])
		cout<<"Case #"<<x<<": 0"<<endl;
		else if(f==0) cout<<"Case #"<<x<<": "<<maxx*n+w<<endl;
		else cout<<"Case #"<<x<<": INF"<<endl;
	}
	return 0;
}