hdu5625 Clarke and chemistry

Clarke and chemistry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 321    Accepted Submission(s): 173

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
  
  
  
  


  
  
  
  


  
  
  
  

   
   暴力三层for循环，cou计数
  
  
  
  


  
  
  
  

   
   
#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

int T, A, B, C;
int a[26], b[26], c[26];
int main()
{
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d%d", &A, &B, &C);
		getchar();
		char ch;
		int n;
		memset(a, 0, sizeof(a));
		memset(b, 0, sizeof(b));
		memset(c, 0, sizeof(c));
		for (int i = 0; i < A; i++) {
			scanf("%c %d", &ch, &n);
			a[ch - 'A'] = n;
			getchar();
		}
		for (int i = 0; i < B; i++) {
			scanf("%c %d", &ch, &n);
			b[ch - 'A'] = n;
			getchar();
		}
		for (int i = 0; i < C; i++) {
			scanf("%c %d", &ch, &n);
			c[ch - 'A'] = n;
			getchar();
		}

		int cou, j, k;
		for (j = 1; j <= 100; j++) {
			for (k = 1; k <= 100; k++) {
				cou = 0;
				for (int m = 0; m < 26; m++) {
					if (j * a[m] + k * b[m] == c[m]) cou++;
				}
				if (cou == 26) break;
			}
			if (cou == 26) break;
		}
		if (cou == 26) {
			printf("%d %d\n", j, k);
		}
		else {
			puts("NO");
		}
	}
	return 0;
}