D. Mixing Milk

GDUT 2020寒假训练 排位赛四 D

原题链接

• D. Mixing Milk

题目

Farming is competitive business – particularly milk production. Farmer John figures that if he doesn’t innovate in his milk production methods, his dairy business could get creamed! Fortunately, Farmer John has a good idea. His three prize dairy cows Bessie, Elsie, and Mildred each produce milk with a slightly different taste, and he plans to mix these together to get the perfect blend of flavors.

To mix the three different milks, he takes three buckets containing milk from the three cows. The buckets may have different sizes, and may not be completely full. He then pours bucket 1 into bucket 2, then bucket 2 into bucket 3, then bucket 3 into bucket 1, then bucket 1 into bucket 2, and so on in a cyclic fashion, for a total of 100 pour operations (so the 100th pour would be from bucket 1 into bucket 2). When Farmer John pours from bucket a into bucket b, he pours as much milk as possible until either bucket a becomes empty or bucket b becomes full.

Please tell Farmer John how much milk will be in each bucket after he finishes all 100 pours.

Input
The first line of the input file contains two space-separated integers: the capacity c1 of the first bucket, and the amount of milk m1 in the first bucket. Both c1 and m1 are positive and at most 1 billion, with c1≥m1. The second and third lines are similar, containing capacities and milk amounts for the second and third buckets.

Output
Please print three lines of output, giving the final amount of milk in each bucket, after 100 pour operations.

样例

input

10 3
11 4
12 5

output
0
10
2

题目大意

有三个容量不一的桶，装了不一样的多的奶。将第一桶的奶倒入第二桶中，再将第二桶的奶倒入第三桶中，再将第三桶的奶倒入第一桶中。每次倒奶如果目标桶装得下就全倒进去，目标桶装不下的话就先装满目标桶，剩下的部分留在原桶中。
倒100次，求100次后每个桶中有多少奶。

思路

模拟
真就是模拟，判断目标的桶能不能装得下，装得下就都装进去，装不下就剩下。

代码

#include<iostream>
#include<cstdio>
#include<memory.h>
using namespace std;
long long b[3];
long long w[3];
int main()
{
	for(int i=0;i<3;i++)
	{
		cin>>b[i]>>w[i];
	}
	for(int i=0;i<100;i++)
	{
		if(i%3==0)//0->1
		{
			int t=b[1]-w[1];//还能装下 
			if(w[0]<t)//装满 
			{
				w[1]+=w[0];
				w[0]=0;
			}
			else//剩余 
			{
				w[1]=b[1];
				w[0]-=t;
			}
		}
		else if(i%3==1)//1-->2
		{
			int t=b[2]-w[2];//还能装下 
			if(w[1]<t)//装满 
			{
				w[2]+=w[1];
				w[1]=0;
			}
			else//剩余 
			{
				w[2]=b[2];
				w[1]-=t;
			}
		}
		else//2-->0
		{
			int t=b[0]-w[0];//还能装下 
			if(w[2]<t)//装满 
			{
				w[0]+=w[2];
				w[2]=0;
			}
			else//剩余 
			{
				w[0]=b[0];
				w[2]-=t;
			}
		}
	}
	cout<<w[0]<<endl<<w[1]<<endl<<w[2]<<endl;
	return 0;
}