Clarke and chemistry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 145 Accepted Submission(s): 75
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.
题意:物质A和物质B生成物质C,暴力求解进行模拟即可
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define N 105
int A[N], B[N], C[N];
int main()
{
// freopen("E:\input.txt", "r", stdin);
int i, j, n;
int t;
int a, b, c;
char s[5];
int num; //开始将num定义成了char类型,导致s一直读不进去
scanf("%d", &t);
while (t--)
{
memset(A, 0, sizeof(A));
memset(B, 0, sizeof(B));
memset(C, 0, sizeof(C));
scanf("%d %d %d", &a, &b, &c);
for (i = 0; i < a; i++)
{
//字符数组s读不进去
scanf("%s %d", s, &num);
//cin >> s >> num;
A[s[0] - 'A'] = num;
}
for (i = 0; i < b; i++)
{
scanf("%s %d", s, &num);
B[s[0] - 'A'] = num;
}
for (i = 0; i < c; i++)
{
scanf("%s %d", s, &num);
C[s[0] - 'A'] = num;
}
int a1, b1;
bool flag = true;
for (a1 = 1; a1 < N; a1++)
{
for (b1 = 1; b1 < N; b1++)
{
flag = true;
for (i = 0; i < 26 && flag; i++)
{
if (A[i] * a1 + B[i] * b1 != C[i])
{
flag = false;
}
}
if (flag)
{
break;
}
}
if (flag)
{
break;
}
}
if (flag)
{
printf("%d %d\n", a1, b1);
}
else
{
printf("NO\n");
}
}
return 0;
}