poj1220------高精度进制转换模板

?

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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int maxs = 1000; int oldBase,newBase;//原进制，新进制 char origin[maxs];//原串 int str[maxs],ans[maxs],goal[maxs];//被除数，除数,结果数组   //将单个字符转变为数字 int getNum(char ch) {     if(ch>='0'&&ch<='9')//数字         return ch-'0';     if(ch>='A'&&ch<='Z') //大写字母         return ch-'A'+10;     return ch-'a'+36;//小写字母 } //将数字转变为字符 char getchar(int num) {     if(num>=0&&num<=9)         return num+'0';     if(num>=10&&num<=35)         return num-10+'A';     return num-36+'a'; } void trans() {     memset(goal,0,sizeof(goal));//初始化结果数组     //初始化,把输入的字符串转变为对应的数字     int len = strlen(origin);     str[0] = len; //str[0]存储字符串的长度     for(int i=0;i<len;i++)         str[i+1]=getNum(origin[i]);     while(str[0]>=1)//被除数还没有为0,继续短除法     {         memset(ans,0,sizeof(ans));//ans[0]代表商的位数         int y=0,i;//余数         for(i=1;i<=str[0];i++)//手动模拟除法         {             y=y*oldBase+str[i];             ans[++ans[0]]=y/newBase;             y=y%newBase;         }         goal[++goal[0]]=y;//第一轮短除法结束，得到的余数           //得到第二轮短除法的被除数,也就是上一轮的商,但是要去除前导0         i=1;         while(i<=ans[0]&&ans[i]==0)             i++;         memset(str,0,sizeof(str));         for(;i<=ans[0];i++)             str[++str[0]]=ans[i];     }     //输出结果     for(int i=goal[0];i>=1;i--)         printf("%c",getchar(goal[i]));     printf("\n"); } int main() {     freopen("in.txt","r",stdin);     int T;     cin>>T;     while(T--)     {         scanf("%d%d",&oldBase,&newBase);         scanf("%s",origin);         printf("%d %s\n%d ",oldBase,origin,newBase);         trans();         printf("\n");     }   }