HDU 5625 Clarke and chemistry

题意:给一个a,b,c 代表A,B,C集合内元素个数，然后a+b+c行，每行有X C 代表该集合有C个X元素
问你是否存在整数i,j满足iA+jB=C
思路:直接遍历所有a,b就OK

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<math.h>
#include<string>
#include<vector>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm> 
using namespace std;
typedef long long LL;
#define maxn 1005
#define f(x) (x*1.0)
#define inf 0x3f3f3f3f
#define maxm maxn*maxn
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
#define lowbit(x) (x&(-x))
#define cheak(i) printf("%d ",i)
#define lson(x) (splay[x].son[0])
#define rson(x) (splay[x].son[1])
#define rfor(i,a,b) for(i=a;i<=b;++i)
#define lfor(i,a,b) for(i=a;i>=b;--i)
#define mem(a,b) memset(a,b,sizeof(a))
#define mec(a,b) memcpy(a,b,sizeof(b))
int mark1[maxn],mark2[maxn],mark3[maxn];
int main()
{
    char str[2];
    int T,i,j,k,a,b,c,x;
    scanf("%d",&T);
    while(T--)
    {
        mem(mark1,0);
        mem(mark2,0);
        mem(mark3,0);
        scanf("%d%d%d",&a,&b,&c);
        rfor(i,1,a)
        {
            scanf("%s%d",str,&x);
            mark1[str[0]-'A']+=x;
        }
        rfor(i,1,b)
        {
            scanf("%s%d",str,&x);
            mark2[str[0]-'A']+=x;
        }
        rfor(i,1,c)
        {
            scanf("%s%d",str,&x);
            mark3[str[0]-'A']+=x;
        }
        int t=0;
        rfor(i,1,100)
        {
            rfor(j,1,100)
            {
                int flag=0;
                rfor(k,0,25)
                if(i*mark1[k]+j*mark2[k]!=mark3[k])
                flag=1;
                if(!flag)
                {
                    printf("%d %d\n",i,j);
                    t=1;
                    break;
                }
            }
            if(t) break;
        }
        if(!t) printf("NO\n");
    }
    return 0;
}