poj 1220 NUMBER BASE CONVERSION

purplest — Fri, 12 Aug 2011 01:48:00 GMT

高精度的进制转换

#include<stdio.h>
#include<string.h>

int n, ori, to, t[1000], tonum[1000];
char str[1000], num[1000];

int main ()
{
    scanf("%d", &n);
    while (n--)
    {
        scanf("%d%d%s", &ori, &to, str);
        int i, j= 0, len= strlen(str), k= len;
        for (i=len-1; 0 <= i; i--)
            t[len-1-i]= str[i]-( 58 > str[i] ? '0' : 91 > str[i] ? 'A'-10 : 'a'-36 );
        while( k )
        {
            for ( i = k-1 ; i > 0; i-- )
            {
                t[i-1]+=t[i]%to*ori;
                t[i]/=to;
            }
            tonum[j++]=t[0]%to;
            t[0]/=to;
            while( 0 < k && !t[k-1]) k--;
        }
        for ( num[j] = i = 0 ; i < j ; i++ )
            num[j-1-i]=tonum[i]+( 10 > tonum[i] ? '0' : 36 > tonum[i] ? 'A'-10 :'a'-36);
        printf("%d %s\n%d %s\n\n", ori, str, to, num);
    }
    return 0;
}

purplest 2011-08-12 09:48 发表评论

poj 2273 An Excel-lent Problem

purplest — Fri, 05 Aug 2011 15:06:00 GMT

进制转换，注意要加getchar()

#include <stdio.h>
#include <string.h>

char s[]={"ZABCDEFGHIJKLMNOPQRSTUVWXY"};
int str[]={26,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25};
int r, c, sum[20];
char t1,t2;

int main()
{
    while ( scanf("%c%d%c%d", &t1, &r, &t2, &c) != EOF && r&&c )
    {
        memset(sum, 0, sizeof(sum));
        int k=0;
        getchar();
        while ( c )
        {
            sum[k++]= c%26;
            c=(c-str[sum[k-1]])/26;
        }
        for ( int i = k-1; i >= 0; i-- )
            printf("%c", s[sum[i]]);
        printf("%d\n", r);
    }
    return 0;
}

purplest 2011-08-05 23:06 发表评论

C++博客-purplest-随笔分类-进制转换

poj 1220 NUMBER BASE CONVERSION

poj 2273 An Excel-lent Problem