POJ NO.2105 IP Address(POJ 2105 IP地址转换)
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POJ NO.2105 IP Address
Description
Suppose you are reading byte streams from any device, representing IP addresses. Your task is to convert a 32 characters long sequence of '1s' and '0s' (bits) to a dotted decimal format. A dotted decimal format for an IP address is form by grouping 8 bits at a time and converting the binary representation to decimal representation. Any 8 bits is a valid part of an IP address. To convert binary numbers to decimal numbers remember that both are positional numerical systems, where the first 8 positions of the binary systems are:
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
128 64 32 16 8 4 2 1
Input
The input will have a number N (1<=N<=9) in its first line representing the number of streams to convert. N lines will follow.
Output
The output must have N lines with a doted decimal IP address. A dotted decimal IP address is formed by grouping 8 bit at the time and converting the binary representation to decimal representation.
Sample Input
4
00000000000000000000000000000000
00000011100000001111111111111111
11001011100001001110010110000000
01010000000100000000000000000001
Sample Output
0.0.0.0
3.128.255.255
203.132.229.128
80.16.0.1
-------------------------------------------------------------------------
大概的意思是把32位的二进制数转换成用小数点(.)隔开的十进制表示的IP地址。
解答一:每次读入8个字符后转换成十进制数。一行是32个字符,需要处理4次,中间还要添加3次小数点。
#include<stdio.h>int main(){//freopen("input.txt","r",stdin);int i,j,k;int n;scanf("%d\n",&n);for(i=0;i<n;i++){for(j=0;j<4;j++){int d=128,t=0;for(k=0;k<8;k++){char c=getchar()%48;//c=1 or 0,查ASCII码可知t+=d*c;d/=2;}if(j<3)printf("%d.",t);elseprintf("%d\n",t);}}return 0;}
解答二:使用strtol函数,将字符数组直接转换成十进制。
#include<stdio.h>#include<stdlib.h>int main(){//freopen("input.txt","r",stdin);int i,j;int n;scanf("%d\n",&n);for(i=0;i<n;i++){for(j=0;j<4;j++){char str[9];int t;scanf("%8s",str);t=strtol(str,0,2);if(j<3)printf("%d.",t);elseprintf("%d\n",t);}}return 0;}
扩展:将一个32位的无符号整型数变换成IP地址,比如
1988,表示成无符号整型为00000000 00000000 00000111 11000100,转换成IP:0.0.7.196
#include<stdio.h>int main(){//freopen("input.txt","r",stdin);int i,j;unsigned int n,a,t;scanf("%u\n",&n);for(i=0;i<n;i++){scanf("%u",&a);for(j=0;j<4;j++){t=(a&0xff000000)>>24;//将32位二进制从左到右每次取8位a<<=8;if(j<3)printf("%u.",t);elseprintf("%u\n",t);}}return 0;}
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