HDU 5625 Clarke and chemistry

Clarke and chemistry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 58    Accepted Submission(s): 30

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.
  
  

 

#include<cstring>
#include<iostream>
using namespace std;

int num1[30];
int num2[30];
int sum[30];
bool check(int a,int b){
    int i;
    for(i=0;i<26;i++){
        if(a*num1[i]+b*num2[i]!=sum[i]){
            return false;
        }
    }
    return true;
}

int main(){
    int T,t,flag,a,b,c,i,j;
    char x;
    cin>>T;
    while(T--){
        flag=0;
        memset(num1,0,sizeof(num1));
        memset(num2,0,sizeof(num2));
        memset(sum,0,sizeof(sum));
        cin>>a>>b>>c;
        while(a--){
            cin>>x>>t;
            num1[x-'A']=t;
        }
        while(b--){
            cin>>x>>t;
            num2[x-'A']=t;
        }
        while(c--){
            cin>>x>>t;
            sum[x-'A']=t;
        }
        for(i=1;i<=100;i++){
            for(j=1;j<=100;j++){
                if(check(i,j)){
                    flag=1;
                    a=i;
                    b=j;
                    break;
                }
            }
            if(flag){
                break;
            }
        }
        if(flag){
            cout<<a<<" "<<b<<endl;
        }
        else{
            cout<<"NO"<<endl;
        }
    }
    return 0;
}