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https://vpn.bupt.edu.cn/http/10.105.242.80/problem/p/98/
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IP头部
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TCP头部
数据解析
45 00 00 34 //0034表示总长度, 即52B, 5表示首部长度, 5*4B
7a 67 40 00
40 06 63 5a
0a cd 0a f4 //源地址 10.205.10.244
7d 38 ca 09 //目的地址 125.56.202.9
cd f6 00 50 //源端口, 目的端口
b4 d7 ae 1c
9b cf f2 40
80 10 ff 3d
fd d0 00 00
Case #1
Total length = 52 bytes
Source = 10.205.10.244
Destination = 125.56.202.9
Source Port = 52726
Destination Port = 80
AC代码
有点意思, 码量较大
需要训练做模拟题的速度
#include <bits/stdc++.h>
#define FF(a,b) for(int a=0;a<b;a++)
#define F(a,b) for(int a=1;a<=b;a++)
#define LEN 100
#define INF 1000000
#define bug(x) cout<<#x<<"="<<x<<endl;
using namespace std;
typedef long long ll;
char buf[1000];
char str[100][10];
int hex2int(char ch){
if(ch>='a'){
return ch-'a'+10;
}else{
return ch-'0';
}
}
int hex2int(char *ch){
int n=strlen(ch);
int ans=0;
int base=1;
for(int i=n-1,j=0
;i>=0;
i--,j++) //脑子秀逗了, 其实写个单循环就好了
{
ans+=hex2int(ch[i])*base;
base*=16;
}
return ans;
}
int main()
{
// freopen("./in","r",stdin);
int N;
scanf("%d",&N);
FF(i,N){
//gets(buf); 可以用gets读入带空格的一行
int n=0;
while(1){
scanf("%s",str[n++]);
char pd=getchar();
if(pd=='\n' || pd==EOF)
break;
}
int IPlen=hex2int(str[0][1]);
strcat(str[2],str[3]);
int TOTlen=hex2int(str[2]);
printf("Case #%d\n",i+1);
printf("Total length = %d bytes\n",TOTlen);
printf("Source = %d.%d.%d.%d\n",hex2int(str[12]),hex2int(str[13]),hex2int(str[14]),hex2int(str[15]));
printf("Destination = %d.%d.%d.%d\n",hex2int(str[16]),hex2int(str[17]),hex2int(str[18]),hex2int(str[19]));
strcat(str[IPlen*4],str[IPlen*4+1]);
strcat(str[IPlen*4+2],str[IPlen*4+3]);
printf("Source Port = %d\n",hex2int(str[IPlen*4]));
printf("Destination Port = %d\n",hex2int(str[IPlen*4+2]));
puts("");
}
return 0;
}
//Case #1
//Total length = 52 bytes
//Source = 10.205.10.244
//Destination = 125.56.202.9
//Source Port = 52726
//Destination Port = 80