bestcoder 72 Clarke and chemistry

Clarke and chemistry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 195    Accepted Submission(s): 110

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.
   
   
   
   

 

枚举加上判断就可以了

#include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<stack> #include<map> using namespace std; typedef long long ll; const ll inf = 0x3f3f3f3f; int a[30], b[30], c[30]; char ch[2]; int an, bn, cn; void solve(){ scanf("%d%d%d", &an, &bn, &cn); for (int i = 0; i <= 27; i++){ a[i] = b[i] = c[i] = 0; } int ind; for (int i = 0; i < an; i++){ scanf("%s", ch); scanf("%d", &ind); a[ch[0] - 'A'] += ind; } for (int i = 0; i < bn; i++){ scanf("%s", ch); scanf("%d", &ind); b[ch[0] - 'A'] += ind; } for (int i = 0; i < cn; i++){ scanf("%s", ch); scanf("%d", &ind); c[ch[0] - 'A'] += ind; } int resa = -1, resb = -1; for (int i = 1; i < 101; i++){ for (int j = 1; j < 101; j++){ int flag = 1; for (int k = 0; k < 26; k++){ if (a[k] * i + b[k] * j != c[k]){ flag = 0; break; } } if (flag == 1){ resa = i, resb = j; } } if (resa != -1)break; } if (resa == -1 || resb == -1){printf("NO\n");return ;} printf("%d %d\n", resa, resb); } int main(){ int t; scanf("%d", &t); while (t--){ solve(); } return 0; } </map></stack></queue></cmath></cstring></algorithm></cstdio>