用c语言编程！要求输入1,输出A.输入26，输出Z。输入27输出AA.输入28输出AB。也就是一个 c语言编程,如果a用数字1表示,b用数字2表示……z用数字2...
c\u8bed\u8a00 \u5982\u4f55\u5c06\u5b57\u5e55\u4e0e\u6570\u5b57\u5bf9\u5e94\uff0c\u6bd4\u5982\u8f93\u5165A\uff0c\u8f93\u51fa1\uff1b\u8f93\u5165Z\uff0c\u8f93\u51fa26\uff1b\u8f93\u5165AA\uff0c\u8f93\u51fa27.\u5982\u6b64\u7c7b\u63a8\uff0c\u8c22\u8c22\uff01\u5176\u5b9e\u5c31\u662f\u4e0d\u540c\u8fdb\u5236\u4e4b\u95f4\u7684\u8f6c\u6362
icol *= 26;
icol += (*(str+i) - 'A'+1);
\u628a\u5b83\u7406\u89e3\u6210\u4e00\u4e2a26\u8fdb\u5236\u7684\u5c31\u597d\u4e86
\u4ee3\u7801\u5982\u4e0b
#include
#include
#include
char cord[0x1000];
void alph(char *str)
{
int i=0;
int icol=0;
for( i=0; i<strlen(str); ++i)
{
icol *= 26;
icol += (*(str+i) - 'A'+1);
}
printf("%d\n", icol);
}
int main()
{
while(scanf("%s", &cord))
{
alph(cord);
}
return 0;
}
\u5b57\u6bcd-'a'+1 \u8868\u793a 1-26
\u5141\u8bb82\u4f4d
#include
#include
int f26(char *s){
int i,L;
int sum=0;
L = strlen(s);
for (i=0;i<L;i++){
sum = sum + pow ( 26,i) * (s[L-i-1]-'a'+1);
}
return sum;
}
main(){
char x[3]="ab";
printf("%s == %d\n",x,f26(x));
}
此题可以做，代码如下：
//#include "stdafx.h"//If the vc++6.0, with this line.
#include "stdio.h"
int main(void){
char R[27]="ZABCDEFGHIJKLMNOPQRSTUVWXY",tmp[10];
int n,i,j;
while(1){
printf("Input n(int n>0)...
n=");
if(scanf("%d",&n) && n>0)
break;
printf("Error, n must >0: ");
}
for(i=0;n;i++){
if((tmp[i]=n%26)==0)
n--;
n/=26;
}
for(j=i-1;j>=0;j--)
printf("%c",R[tmp[j]]);
printf("
");
return 0;
}
但这并不是二十六进制问题。按照数系基数的定义，基数是多少，就得有多少个符号来表示该进制的数据，但必须包含符号0代表“没有”。A~Z共26个符号这没错，但没有包含0，且当输入26时二十六进制应该输出A0，这题中要求是输出Z，紧接着27却要输出AA，这就不符合二十六进制的定义了……供参考。
用短除法像求2进制那样取余，最后将余数倒置即可
enum
{
...
};
int main(void)
{
switch()
{
case:
{
}
break;
.
.
.
return 0;
}
框架都给你了，还不晓得写我也没办法。
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