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有一个n位数，它的各位数字翻转以后组成新的n位数，原来的n位数是新n位数的n倍，求所有可能的n位数。
如：
8712/4 = 2,178


鸟哥的代码：
#include <stdio.h> #include <stdint.h> static uint64_t revert(uint64_t n, int *width) { int i; char buf[64]; uint64_t ret = 0, weight = 1; *width = sprintf(buf, "%llu", (unsigned long long)n); for (i = 0; buf[i]; ++i) {   ret += (buf[i] - '0') * weight;   weight *= 10; } return ret; } int main() { int w; uint64_t n; for (n = 0; n <= 200000000; ++n) {   uint64_t r = revert(n, &w);   if (r == n * w) {    printf("%lld = %lld / %d\n", (unsigned long long)n,           (unsigned long long)r, w);   } } }
#make calcnumber
cc     calcnumber.c   -o calcnumber

#./calcnumber
0 = 0 / 1
1 = 1 / 1
2 = 2 / 1
3 = 3 / 1
4 = 4 / 1
5 = 5 / 1
6 = 6 / 1
7 = 7 / 1
8 = 8 / 1
9 = 9 / 1
2178 = 8712 / 4
108901089 = 980109801 / 9
109999989 = 989999901 / 9


http://live.aulddays.com/tech/11/linux-shell-too-great-base-error.htm


我用shell写了一个：
#cat *.sh
#!/bin/bash
for i in {1000..9999} ;do
    rev=$(echo $i | rev)
    div=$(echo "$rev/4" | bc)
    if [[ "10#$i" -eq "10#$div" ]];then
      echo "result="$i;
    fi
done

sh calcnumber.sh
result=1997  #这个是小数四舍五入后相等的，shell在处理小数这方面有些弱，得靠BC。
result=2178

作者：jackxiang@向东博客 专注WEB应用 构架之美 --- 构架之美，在于尽态极妍 | 应用之美，在于药到病除
地址：https://jackxiang.com/post/10269/
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