hdu 5625

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

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 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}^*

.

 

Input

The first line contains an integer T(1 \le T \le 10)

, 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

)

 

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.

 

Source

BestCoder Round #72 (div.2)

 

 

 http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=671&pid=1001 中文题意

枚举 a，b      代码长时间不写 手糙了

今天02网上找模板水过 明天补看

#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;
}