HUNAN UNIVERSITY ACM/ICPC Judge Online —— Problem 10049 IP Address

这次注意到了两个问题：首先是连续输入二进制数的时候使用整型的数组是不方便的，所以后来改用了char。第二就是char型数组需要预留一个元素用来存储'/0'的结束字符，我之前的几次提交失败的原因就是在这。

问题描述：

IP Address Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB Total submit users: 405, Accepted users: 388 Problem 10049 : No special judgement Problem 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: 27 26 25 24 23 22 21 20 128 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 Problem Source MCA 2004 Submit Discuss Judge Status Problems Ranklist

三次提交AC！

// 10049.cpp:Definestheentrypointfortheconsoleapplication.
//

// #include"stdafx.h"
#include < iostream.h >
#include < math.h >
// usingnamespacestd;

int OutIP[ 9 ][ 4 ];

void ChangeToIp( const char Temp[], int j)
... {
for(intx=0;x<4;x++)
...{
inttemp=0;
for(inty=0;y<8;y++)
...{
if(Temp[8*x+y]=='1')
temp+=1*pow(2,7-y);
else//if(Temp[8*x+y]=='1')
temp+=0*pow(2,7-y);
}
OutIP[j][x]=temp;
}
}

int main()
... {
intN=0;
cin>>N;

if((N>0&&N<10)!=1)
return0;

charTempIn[33];
for(inti=0;i<N;i++)
...{
cin>>TempIn;
ChangeToIp(TempIn,i);
}

for(inta=0;a<N;a++)
...{
cout<<OutIP[a][0]<<"."<<OutIP[a][1]<<"."<<OutIP[a][2]<<"."<<OutIP[a][3]<<endl;
}

//cout<<TempIn<<endl;
return0;
}

附上“马牛不是人”的解法：

/* start17:43 */ /* end18:0825min */ #include < stdio.h > #include < string .h > main() { int n,i,j,p,dec = 0 ; char s[ 32 ]; scanf( " %d " , & n); for (i = 0 ;i < n;i ++ ){ scanf( " %s " ,s); for (j = 0 ,p = 7 ;j < 32 ;j ++ ,p -- ){ if (s[j] == ' 1 ' )dec += pow( 2 ,p); if (p == 0 ){ p = 8 ; printf( " %d " ,dec); dec = 0 ; if (j == 31 )printf( " " ); else printf( " . " ); } } } system( " PAUSE " ); }