View Code of Problem 7

#include <stdio.h>
#define MAX 2000
int main()
{
    int test[MAX];//测试数据
    int eq[MAX];//equalties 等式
    int eq1[MAX];//equalties 等式
    int t, n, m;
    int i, j, k;
    int a, b;
    int flag;//非0时有可转换的数据

    scanf("%d",&t);//T组输入
    for (i=0; i<t; i++)
    {
        //读入等式
        scanf("%d", &n);//n行等式
        for (j=0; j<n; j++)
        {
            scanf("%d=%d", &a, &b);
            eq[j] = a;
            eq1[j] = b;
        }

        //读入测试数据
        scanf("%d",&m);//m组数据
        for (j=0; j<m; j++)
            scanf("%d",&test[j]);

        //计算
        for (j=0; j<m; j++)
        {
            flag = 0;

            for (k=0; k<n*2; k++)
            {
                if (test[j] == eq[k])
                {
                    flag = 1;
                    printf("%d\n", eq1[k]);
                    break;
                }
                
                if (test[j] == eq1[k])
                {
                    flag = 1;
                    printf("%d\n", eq[k]);
                    break;
                }

            }

            if (flag==0)
                    printf("UNKNOW\n");
        }
        printf("\n");
    }
    return (0);
}

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