2019 GDUT Rating Contest I : Problem H. Mixing Milk
题面:
H. Mixing Milk
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.
- Pour 1->2: 0 7 5
- Pour 2->3: 0 0 12
- Pour 3->1: 10 0 2
- Pour 1->2: 0 10 2
- Pour 2->3: 0 0 12
题目描述:
题目分析:







1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 using namespace std;
5 int capa[5], leave[5];
6
7 void pour(int a, int b){
8 if(leave[a]+leave[b] <= capa[b]){ //第一种情况
9 leave[b] += leave[a];
10 leave[a] = 0;
11 }
12 else{ //第二种情况
13 leave[a] -= capa[b]-leave[b];
14 leave[b] = capa[b];
15 }
16 }
17
18 int main(){
19 for(int i = 1; i <= 3; i++){
20 cin >> capa[i] >> leave[i];
21 }
22
23 for(int i = 0; i < 33; i++){
24 pour(1, 2); //桶1倒进桶2
25 pour(2, 3); //桶2倒进桶3
26 pour(3, 1); //桶3倒进桶1
27 }
28
29 pour(1, 2); //最后别忘这个
30
31 for(int i = 1; i <= 3; i++){
32 cout << leave[i] << endl;
33 }
34 return 0;
35 }
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