循环小数的循环节问题

来源：互联网发布：a神淘宝店编辑：程序博客网时间：2024/05/17 05:02
题目链接：https://www.bnuoj.com/v3/external/2/202.pdf
/*通过模拟手算竖式的过程，可以发现如果同一个余数在竖式中出现两次，那么小数就开始循环了，循环节的始末位置与这个余数出现在竖式中的位置有关这里就不详细给出分析过程了，网上可以找到详细的解题过程*/#include <cstdio>#include <cstring>const int maxn = 3010;//数组d用来存放商，数组m用来存放余数，book[k]统计余数k出现的次数int d[maxn], m[maxn], book[maxn];int main(){    int a, b;    while (scanf("%d %d", &a, &b) == 2) {        //head，tail用来锁定循环节的始末位置        int len, head, tail;        memset(book, 0, sizeof(book));        d[0] = a / b;        m[0] = a % b;        book[m[0]]++;        int i;        //余数重复出现，说明出现循环节，跳出循环        for (i = 0; book[m[i]] < 2; i++) {            int tem = m[i] * 10;            d[i+1] = tem / b;            m[i+1] = tem % b;            book[m[i+1]]++;   //统计余数出现的次数        }        tail = i;        //找到循环节长度        for (head = 0; m[head] != m[tail]; head++)            ;        len = tail - head;        //输出部分        printf("%d/%d = %d.", a, b, d[0]);        if (len <= 50) {            for (int k = 1; k <= head; k++) printf("%d", d[k]);            printf("(");            for (int k = head + 1; k <= tail; k++) printf("%d", d[k]);            printf(")\n");        } else {            for (int k = 1; k <= head; k++) printf("%d", d[k]);            printf("(");            for (int k = head + 1; k <= head + 50; k++) printf("%d", d[k]);            printf("...)\n");        }        printf("   %d = number of digits in repeating cycle\n\n", len);    }    return 0;}
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