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;
}