Clarke and chemistry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 63 Accepted Submission(s): 33
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 AA
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
. Please calculate the stoichiometric coefficient a
of A
and b
of B
that aA + bB = C,\ \ a, b \in \text{N}^*
.
, the number of test cases.
For each test case, the first line contains three integers A, B, C(1 \le A, B, C \le 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
capital letters, guarantee X
of one valence only appear one time, 1 \le c \le 100
)
and b
. If there are multiple answers, print the smallest one, a
is smallest then b
is smallest. Otherwise print NO.
NO
Hint:
The first test case, $a=2, b=3$ can make equation right.
The second test case, no any answer.
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<stack>
#include<queue>
#define LL __int64
using namespace std;
int t;
int a,b,c;
char ceshi[30]="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char aa[30],bb[30],cc[30];
int aaa[30],bbb[30],ccc[30];
map<char,int> mp1;
map<char,int> mp2;
map<char,int> mp3;
int main()
{
while(scanf("%d",&t)!=EOF)
{
for(int i=1;i<=t;i++)
{
mp1.clear();
mp2.clear();
mp3.clear();
scanf("%d%d%d",&a,&b,&c);
getchar();
for(int j=1;j<=a;j++)
{
scanf("%c %d",&aa[j],&aaa[j]);
mp1[aa[j]]=aaa[j];getchar();
}
for(int j=1;j<=b;j++)
{
scanf("%c %d",&bb[j],&bbb[j]);
mp2[bb[j]]=bbb[j];getchar();
}
for(int j=1;j<=c;j++)
{
scanf("%c %d",&cc[j],&ccc[j]);
mp3[cc[j]]=ccc[j];getchar();
}
int k=0,g=0,ans;
int flag=0;
int ggg1,ggg2;
for(k=1;k<=99;k++)
{
for(g=1;g<=99;g++)
{
ans=0;
for(int kk=0;kk<=25;kk++)
{
if(mp1[ceshi[kk]]*k+mp2[ceshi[kk]]*g==mp3[ceshi[kk]]&&mp3[ceshi[kk]]!=0)
ans++;
}
if(ans==c)
{
ggg1=k;
ggg2=g;
flag=1;
break;
}
}
if(flag)
break;
}
if(flag)
printf("%d %d\n",ggg1,ggg2);
else
printf("NO\n"); }
}
return 0;
}