9 degrees OJ 1069 student information (simulation), oj1069

9 degrees OJ 1069 student information (simulation), oj1069

Question 1069: Find Student Information

Time Limit: 1 second

Memory limit: 32 MB

Special question: No

Submit: 7836

Solution: 2116

Description:

Enter information about N students and then query them.

Input:

The first Act N entered, that is, the number of students (N <= 1000)

The next N rows contain information about N students. The information format is as follows: 01 Li Jiangnan 2102 Liu Tang male 2303 Zhang Jun male 1904 Wang na female 19 then input a M (M <= 10000 ), next, there will be M rows, representing M queries. Each row will enter a student ID in the following format: 02030104

Output:

Output M rows. Each row contains the information of a student corresponding to the query.

If No student information exists, "No Answer!" Is output !"

Sample input:

401 Li Jiang male 2102 Liu Tang male 2303 Zhang Jun male 1904 Wang na female 1950203010403

Sample output:

02 Liu Tang male 2303 Zhang Jun male 1901 Li Jiang male 2104 Wang na female 1903 Zhang Jun male 19

#include<stdio.h>#include<string.h>typedef struct node{    char id[100];    char name[200];    char sex[10];    int age;}node;node data[1001];int main(int argc, char *argv[]){    int N;    int M;    while(~scanf("%d",&N))    {        for(int i=0;i<N;++i)        {            scanf("%s%s%s%d",data[i].id,data[i].name,data[i].sex,&data[i].age);        }        scanf("%d",&M);        char queryid[100];        int j;        for(int i=0;i<M;++i)        {            scanf("%s",queryid);            for(j=0;j<N;++j)            {                if(strcmp(data[j].id,queryid)==0)                {                    printf("%s %s %s %d\n",data[j].id,data[j].name,data[j].sex,data[j].age);                    break;                }            }            if(j==N)                printf("No Answer!\n");        }    }    return 0;}