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  • UVALive 3958 Weird Numbers （负进制数）

    Weird Numbers

    题目链接：

    http://acm.hust.edu.cn/vjudge/contest/129733#problem/F

    Description

    Binary numbers form the principal basis of computer science. Most of you have heard of other systems, such as ternary, octal, or hexadecimal. You probably know how to use these systems and how to convert numbers between them. But did you know that the system base (radix) could also be negative? One assistant professor at the Czech Technical University has recently met negabinary numbers and other systems with a negative base. Will you help him to convert numbers to and from these systems? A number N written in the system with a positive base R will always appear as a string of digits between 0 and R - 1 , inclusive. A digit at the position P (positions are counted from right to left and starting with zero) represents a value of RP . This means the value of the digit is multiplied by RP and values of all positions are summed together. For example, if we use the octal system (radix R = 8 ), a number written as 17024 has the following value: 1*8 4 +7*8 3 +0*8 2 +2*8 1 +4*8 0 = 1*4096 + 7*512 + 2*8 + 4*1 = 7700 With a negative radix - R , the principle remains the same: each digit will have a value of (- R)P . For example, a negaoctal (radix R = - 8 ) number 17024 counts as: 1*(- 8) 4 +7*(- 8) 3 +0*(- 8) 2 +2*(- 8) 1 +4*(- 8) 0 = 1*4096 - 7*512 - 2*8 + 4*1 = 500 One big advantage of systems with a negative base is that we do not need a minus sign to express negative numbers. A couple of examples for the negabinary system (R = - 2) : You may notice that the negabinary representation of any integer number is unique, if no "leading zeros" are allowed. The only number that can start with the digit "0", is the zero itself.

    Input

    The input will contain several conversions, each of them specified on one line. A conversion from the decimal system to some negative-base system will start with a lowercase word "to" followed by a minus sign (with no space before it), the requested base (radix) R , one space, and a decimal number N .

    Output

    For each conversion, print one number on a separate line. If the input used a decimal format, output the same number written in the system with a base - R . If the input contained such a number, output its decimal value. Both input and output numbers must not contain any leading zeros. The minus sign "-" may only be present with negative numbers written in the decimal system. Any non-negative number or a number written in a negative-base system must not start with it.

    Sample Input

    ``` to-2 10 from-2 1010 to-10 10 to-10 -10 from-10 10 end ```

    Sample Output

    ``` 11110 -10 190 10 -10 ```

    Source

    2016-HUST-线下组队赛-2
    ##题意： 对给定的数，从指定负进制转成十进制，或从十进制转成负进制数.
    ##题解： from操作跟正进制一样，比较简单. 对于to操作：[负进制数](http://baike.baidu.com/link?url=hk4TMW9mCVKEAWfDVCECQAFJ_GXcQJTMgh1juL2R4lCjxL6vzVi4ufEHF9cn-aWSReWaorH0kFLheWTWhx1JzK) 数n转成正进制r的过程：不停地模r、除r，直到n为0. 得到的余数序列就是结果. 当r是负数时，进行上述过程可能会得到负余数，这不符合结果. 所以在得到负余数时，商加一，余数减radix，使得余数为正再进行下一步. 比赛时用的暴力枚举，根据当前位表示数的范围和唯一性来确定每一位数. 远不如这个简洁.
    ##代码： ``` cpp #include #include #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 25 #define mod 100000007 #define inf 0x3f3f3f3f #define mid(a,b) ((a+b)>>1) #define IN freopen("in.txt","r",stdin); using namespace std;

    int radix;
    char str[50];

    int main(int argc, char const *argv[])
    {
    //IN;

    while(scanf("%s", str) != EOF && str[0] != 'e')
{
    if(str[0] == 'f') {
        sscanf(str+4, "%d", &radix);
        char num[50];
        scanf("%s", num);

        int ans = 0;
        for(int i=0; i<strlen(num); i++) {
            ans = ans*radix + num[i] - '0';
        }
        printf("%d
", ans);
    }

    else {
        sscanf(str+2, "%d", &radix);
        int n; scanf("%d", &n);
        int ans[50] = {0}, cnt = 0;

        do {  // 处理 n=0
            int last = n % radix;
            n = n / radix;
            if(last < 0) last -= radix, n += 1;
            ans[cnt++] = last;
        } while(n);

        for(int i=cnt-1; i>=0; i--)
            printf("%d", ans[i]);
        printf("
");
    }
}

return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5799724.html
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