hdu 5625 Clarke and chemistry

Posted 2020-06-13

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.

 

 

Source

BestCoder Round #72 (div.2)

 

 

  1 #pragma comment(linker, "/STACK:1024000000,1024000000")
  2 #include<iostream>
  3 #include<cstdio>
  4 #include<cstring>
  5 #include<cmath>
  6 #include<math.h>
  7 #include<algorithm>
  8 #include<queue>
  9 #include<set>
 10 #include<bitset>
 11 #include<map>
 12 #include<vector>
 13 #include<stdlib.h>
 14 using namespace std;
 15 #define ll long long
 16 #define eps 1e-10
 17 #define MOD 1000000007
 18 #define N 1000000
 19 #define inf 1e12
 20 int a,b,c;
 21 int mp[3][27];
 22 int mpp[3][27];
 23 int vis[27];
 24 int main()
 25 {
 26     int t;
 27     scanf("%d",&t);
 28     while(t--){
 29         memset(mpp,0,sizeof(mpp));
 30         
 31         scanf("%d%d%d",&a,&b,&c);
 32         char s[2];
 33         int cnt;
 34         for(int i=0;i<a;i++){
 35             scanf("%s%d",s,&cnt);
 36             int index = s[0]-‘A‘;
 37             mpp[0][index]+=cnt;
 38             //printf("1---%d\n",mp[0][index]);
 39         }
 40         for(int i=a;i<a+b;i++){
 41             scanf("%s%d",s,&cnt);
 42             int index = s[0]-‘A‘;
 43             mpp[1][index]+=cnt;
 44             //printf("2---%d\n",mp[1][index]);
 45         }
 46         for(int i=a+b;i<a+b+c;i++){
 47             scanf("%s%d",s,&cnt);
 48             int index = s[0]-‘A‘;
 49             mpp[2][index]+=cnt;
 50             //printf("3---%d\n",mp[2][index]);
 51         }
 52         
 53         int i,j;
 54         int flag=0;
 55         for(i=1;i<=1006;i++){
 56             for(j=1;j<=1006;j++){
 57                 for(int k=0;k<26;k++){
 58                     mp[0][k]=mpp[0][k];
 59                     mp[1][k]=mpp[1][k];
 60                     mp[2][k]=mpp[2][k];
 61                 }
 62                 memset(vis,0,sizeof(vis));
 63                 for(int k=0;k<26;k++){
 64                     mp[0][k]*=i;
 65                     mp[1][k]*=j;
 66                 }
 67                 int w=1;
 68                 for(int k=0;k<26;k++){
 69                     if(mp[2][k]){
 70                         if(mp[0][k]+mp[1][k] == mp[2][k]){
 71                             vis[k]=1;
 72                         }else{
 73                             w=0;
 74                             break;
 75                         }
 76                     }
 77                 }
 78     
 79                 if(w==1){
 80                     for(int k=0;k<26;k++){
 81                         if(vis[k]==0 && (mp[0][k] || mp[1][k] || mp[2][k])){
 82                             w=0;
 83                             break;
 84                         }
 85                     }
 86                 }
 87                 if(w==1){
 88                     flag=1;
 89                     break;
 90                 }
 91             }
 92             if(flag==1) break;
 93         }
 94         if(flag==1){
 95             printf("%d %d\n",i,j);
 96         }else{
 97             printf("NO\n");
 98         }
 99     }
100     return 0;
101 }
102 
103 //A 6 2 12
104 //B 6  3 18 
105 //C 12  5 60 

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