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#include<iostream>
#include<math.h>
using namespace std;
#define N 2000
void Alter(char num[]) {
int i,len;
int sum=0;
int a[N],b[N];
int k=0;
len=strlen(num);
for(i=0;i<len;i++) { //先转换为十进制数
if(num[i]>'0'&&num[i]<'9')
sum+=(num[i]-'0')*pow(16,len-1-i);
if(num[i]>'A'&&num[i]<'F')
sum+=(num[i]-55)*pow(16,len-1-i);
}
//在转换为二进制
while(sum>0) {
a[k++]=sum%2;
sum/=2;
}
for(i=0;i<k;i++) {
b[i]=a[k-1-i];
cout<<b[i];
}
cout<<endl;
}
int main() {
char num[N];
int n;
int i;
cin>>n;
for(i=0;i<n;i++) {
cin>>num;
Alter(num);
}
return 0;
}
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <locale.h>
int main() {
int i,v;
char bs[33];
char b[33];
char hs[9];
char h[9];
char s[4];
char *e;
// 十进制整数转二进制串;
i=1024;
ltoa(i,b,2);
sprintf(bs,"%032s",b);
printf("i=%d,bs=%s\n",i,bs);
// 十进制整数转十六进制串;
i=1024;
ltoa(i,h,16);
sprintf(hs,"%08s",h);
printf("i=%d,hs=%s\n",i,hs);
// 十六进制字符串转成十进制数
strcpy(hs,"00000400");
sscanf(hs,"%x",&i);
printf("hs=%s,i=%d\n",hs,i);
// 二进制字符串转化为十六进制字符串;
strcpy(bs,"00000000000000000000010000000000");
i=strtol(bs,&e,2);
ltoa(i,h,16);
sprintf(hs,"%08s",h);
printf("bs=%s,hs=%s\n",bs,hs);
// 二进制字符串转化为十进制数;
strcpy(bs,"00000000000000000000010000000000");
i=strtol(bs,&e,2);
printf("bs=%s,i=%d\n",bs,i);
// 十六进制字符串转成二进制串
strcpy(hs,"00000400");
sscanf(hs,"%x",&i);
ltoa(i,b,2);
sprintf(bs,"%032s",b);
printf("hs=%s,bs=%s\n",hs,bs);
// ASC\GBK字符串转十六进制串
strcpy(s,"a汉");
i=0;
while (1) {
if (0==s[i]) break;
sprintf(hs+i*2,"%02X",(unsigned char)s[i]);
i++;
}
setlocale(LC_ALL,"chs");
printf("s=%s,hs=%s\n",s,hs);
// 十六进制字符串转成汉字(GBK)及字符(ASC)
strcpy(hs,"61BABA");
i=0;
while (1) {
if (1!=sscanf(hs+i*2,"%2x",&v)) break;
s[i]=(char)v;
i++;
}
s[i]=0;
printf("hs=%s,s=%s\n",hs,s);
return 0;
}
//i=1024,bs=00000000000000000000010000000000
//i=1024,hs=00000400
//hs=00000400,i=1024
//bs=00000000000000000000010000000000,hs=00000400
//bs=00000000000000000000010000000000,i=1024
//hs=00000400,bs=00000000000000000000010000000000
//s=a汉,hs=61BABA
//hs=61BABA,s=a汉
//不做输入检查,假定输入一定正确
#include <iostream>
void change(const char* str);
int main()
{
const int MaxStrLen = 2000;
int num;
std::cin >> num;
char (*strarr)[MaxStrLen+1] = new char[num][MaxStrLen+1];
for (int i = 0; i < num; i++)
{
std::cin >> strarr[i];
std::cin.get();
}
for (int i = 0; i < num; i++)
{
change(strarr[i]);
}
delete[] strarr;
return 0;
}
void change(const char* str)
{
const char Map[16][5] =
{
"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111",
"1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"
};
bool firstflag = true;
while (*str)
{
char c = *str;
int i = 0;
if (c >= '0' && c <= '9')
i = c - '0';
else
i = c - 'A' + 10;
//判断并处理首字符
if (firstflag)
{
int pos = 0;
for (; pos < 4; pos++)
{
if ( Map[i][pos] != '0')
break;
}
std::cout << Map[i] + pos;
firstflag = false;
}
else
std::cout << Map[i];
++str;
}
std::cout << std::endl;
}
#include<iostream>
#include<math.h>
#include <cstring>
using namespace std;
#define N 2000
void Alter(char num[])
{
int i,len;
int sum=0;
int a[32],b[32]; /* 对于一个整型数来讲(int),最多32位,对于long long是64位*/
int k=0;
len = strlen(num);
for(i=0;i<len;i++) { //先转换为十进制数
if(num[i]>'0'&&num[i]<'9')
sum+=(num[i]-'0')*pow(16,len-1-i);
if(num[i]>'A'&&num[i]<'F')
sum+=(num[i]-'A')*pow(16,len-1-i);
}
//在转换为二进制
while(sum > 0) {
a[k++]=sum%2;
sum/=2;
}
for(i=0;i<k;i++) {
b[i]=a[k-1-i];
cout<<b[i];
}
cout<<endl;
}
int main()
{
static char num[N]; /*数组较大,建议加上static*/
int n;
int i;
cin>>n;
for(i=0;i<n;i++) {
cin>>num;
Alter(num);
}
return 0;
}
大的数组建议在数据段或堆上申请;
存放二进制数的数组严重资源浪费,对于一个int类型的数,其二进制最多是32位,定义2000个长度,至少2000-32个元素用不到;