poj 1220 NUMBER BASE CONVERSION(字符串处理经典)

NUMBER BASE CONVERSION

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 4268 Accepted: 1919

Description

Write a program to convert numbers in one base to numbers in a second base. There are 62 different digits: 
{ 0-9,A-Z,a-z } 
HINT: If you make a sequence of base conversions using the output of one conversion as the input to the next, when you get back to the original base, you should get the original number. 

Input

The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines will have a (decimal) input base followed by a (decimal) output base followed by a number expressed in the input base. Both the input base and the output base will be in the range from 2 to 62. That is (in decimal) A = 10, B = 11, ..., Z = 35, a = 36, b = 37, ..., z = 61 (0-9 have their usual meanings). 

Output

The output of the program should consist of three lines of output for each base conversion performed. The first line should be the input base in decimal followed by a space then the input number (as given expressed in the input base). The second output line should be the output base followed by a space then the input number (as expressed in the output base). The third output line is blank. 

Sample Input

862 2 abcdefghiz10 16 123456789012345678901234567890123456789016 35 3A0C92075C0DBF3B8ACBC5F96CE3F0AD235 23 333YMHOUE8JPLT7OX6K9FYCQ8A23 49 946B9AA02MI37E3D3MMJ4G7BL2F0549 61 1VbDkSIMJL3JjRgAdlUfcaWj61 5 dl9MDSWqwHjDnToKcsWE1S5 10 42104444441001414401221302402201233340311104212022133030

Sample Output

62 abcdefghiz2 1101110000010001011111001001011001111100100110001101001000110 123456789012345678901234567890123456789016 3A0C92075C0DBF3B8ACBC5F96CE3F0AD216 3A0C92075C0DBF3B8ACBC5F96CE3F0AD235 333YMHOUE8JPLT7OX6K9FYCQ8A35 333YMHOUE8JPLT7OX6K9FYCQ8A23 946B9AA02MI37E3D3MMJ4G7BL2F0523 946B9AA02MI37E3D3MMJ4G7BL2F0549 1VbDkSIMJL3JjRgAdlUfcaWj49 1VbDkSIMJL3JjRgAdlUfcaWj61 dl9MDSWqwHjDnToKcsWE1S61 dl9MDSWqwHjDnToKcsWE1S5 421044444410014144012213024022012333403111042120221330305 4210444444100141440122130240220123334031110421202213303010 1234567890123456789012345678901234567890

Source

Greater New York 2002

分析：字符串处理，本题是一个不断求余的问题，仔细想一下可知，此大数最后一位即自己余所求进制，举例，121十进制转换为二进制，121%2=1，1即为最后一位，60%2=0,0即为倒数第二位，模拟此过程。每次求的余数d就将其push进stack中，最后输出即可。

新的问题又出来了，本题数据不可能直接拿来求余，解决这个问题就要用到数论的简单知识了，((a*n)+k)%n=(a+k%n)%n;

不知道此公式的同学找数字演算下，其实仔细推下也可以推出来，（（3*97）+15）%97 =  3+15%97(这里为了容易看出规律，省略了点东西)。

接下来，n进制数可表示为n*（第一位）+（剩余数位），再结合上面公式，再加一点耐心，就可以做出代码了。

给出AC代码：

#include<iostream>#include<cstdio>#include<cmath>#include<map>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;#define rep(i,a,b) for(int i=(a);i<(b);i++)#define rev(i,a,b) for(int i=(a);i>=(b);i--)#define clr(a,x) memset(a,x,sizeof a)const int maxn=1005;const int maxm=4005;const int mod=1e9 +7;char dit[maxn],r[maxn];void chang(char *ss,int n,char *r,int m){    char s[maxn];memcpy(s,ss,sizeof s);    map<char,int>mp;    char smp[]="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";    for(int i='0';i<='9';i++)        mp[i]=i-'0';    for(int i='A';i<='Z';i++)        mp[i]=i-'A'+10;    for(int i='a';i<='z';i++)        mp[i]=i-'a'+36;    int len=strlen(s),cnt=0;    for(int k=0;k!=len;)    {        int d=0;        for(int i=k;i<len;i++)        {            int cur=(mp[s[i]]+d*n);            d=cur%m;            s[i]=smp[cur/m];        }        r[cnt++]=smp[d];        while(s[k]=='0')            k++;    }    r[cnt]='\0';    //strrev(r);    for(int i=0,j=cnt-1;i<j;i++,j--)    {        char c=r[i];        r[i]=r[j];        r[j]=c;    }}int main(){    int t,n,m;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%s",&n,&m,dit);        chang(dit,n,r,m);        printf("%d %s\n%d %s\n\n",n,dit,m,r);    }    return 0;}