hdoj--5625--Clarke and chemistry（枚举）

Clarke and chemistry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 173    Accepted Submission(s): 95

Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke turned into a junior student and took a chemistry exam.


But he did not get full score in this exam. He checked his test paper and found a naive mistake, he was wrong with a simple chemical equation balancer.


He was unhappy and wanted to make a program to solve problems like this.

This chemical equation balancer follow the rules:

Two valences A
combined by |A|
elements and B
combined by |B|
elements.

We get a new valence C
by a combination reaction and the stoichiometric coefficient of
C
is 1.
Please calculate the stoichiometric coefficient a
of A
and b
of B
that aA+bB=C,  a,b∈N∗.

 

Input

The first line contains an integer
T(1≤T≤10),
the number of test cases.

For each test case, the first line contains three integers
A,B,C(1≤A,B,C≤26),
denotes |A|,|B|,|C|
respectively.

Then A+B+C
lines follow, each line looks like X c,
denotes the number of element X
of A,B,C
respectively is c.
(X
is one of 26
capital letters, guarantee X
of one valence only appear one time, 1≤c≤100)

 

Output

For each test case, if we can balance the equation, print
a
and b.
If there are multiple answers, print the smallest one,
a
is smallest then b
is smallest. Otherwise print NO.

 

Sample Input

2
2 3 5
A 2
B 2
C 3
D 3
E 3
A 4
B 4
C 9
D 9
E 9
2 2 2
A 4
B 4
A 3
B 3
A 9
B 9

 

Sample Output

2 3
NO

Hint:
The first test case, $a=2, b=3$ can make equation right.
The second test case, no any answer.

 

Source

BestCoder Round #72 (div.2)

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int A[27],B[27],C[27];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		char op[2];
		bool flag=true;
		int d,maxx=0;
		memset(B,0,sizeof(B));
		memset(C,0,sizeof(C));
		memset(A,0,sizeof(A));
		for(int i=0;i<a;i++)
		{
			scanf("%s%d",op,&d);
			A[op[0]-'A']=d;
		}
		for(int i=0;i<b;i++)
		{
			scanf("%s%d",op,&d);
			B[op[0]-'A']=d;
		}
		for(int i=0;i<c;i++)
		{
			scanf("%s%d",op,&d);
			C[op[0]-'A']=d;
//			maxx=max(maxx,d/A[op[0]-'A']);
//			maxx=max(maxx,d/B[op[0]-'A']);
		}
		int i,j,k;
		int min,x,y;
		for(i=1;i<=1000&&flag;i++)
		for(j=1;j<=1000;j++)
		{
			int s;
			for(s=0;s<26;s++)
			{
				x=i*A[s]+j*B[s];
				y=C[s];
				if(x==y)
				continue;
				else break;
			}
			if(s==26) flag=false;
			if(!flag) break;
		}
		if(!flag) printf("%d %d\n",i-1,j);
		else printf("NO\n");
	}
	return 0;
}