10楼: Originally posted by lurencyj at 2012-04-20 22:32:07:
不明白了。。。。
求讲解。。。。。

假设一个余数叫r，那么因为a ≡ b (mod n)，我们可以将a和b写成
a = k1 * n + r
b = k2 * n + r  【注意这里的r 然后就有
ac = c * k1 * n + r * c
bc = c * k2 * n + r * c
于是得到：
ac 模 n = r * c 模 n
bc 模 n = r * c 模 n
即为
ac ≡ bc (mod n)

#include
#include

using namespace std;

int main(int argc, char *argv[])
{
                long a = 1, n = 1;

                while(a%2011)
                {
                                cout << "n = " << n
                                                << ", a = " << a
                                                << endl;

                                if(n == 20)
                                                break;

                                n++;
                                a = 10*a+1;
                }
                cout << "a = " << a << endl;
                cout << "size of long = " << sizeof(long) << endl;
                cout << "maxmum of long = " << numeric_limits::max() << endl;
                return 0;
}

n = 1, a = 1
n = 2, a = 11
n = 3, a = 111
n = 4, a = 1111
n = 5, a = 11111
n = 6, a = 111111
n = 7, a = 1111111
n = 8, a = 11111111
n = 9, a = 111111111
n = 10, a = 1111111111
n = 11, a = 11111111111
n = 12, a = 111111111111
n = 13, a = 1111111111111
n = 14, a = 11111111111111
n = 15, a = 111111111111111
n = 16, a = 1111111111111111
n = 17, a = 11111111111111111
n = 18, a = 111111111111111111
n = 19, a = 1111111111111111111
n = 20, a = -7335632962598440505
a = -7335632962598440505
size of long = 8
maxmum of long = 9223372036854775807

2楼: Originally posted by lurencyj at 2012-04-20 16:06:14:
确实是溢出问题。
我的C++代码：

#include <iostream>
#include <limits>

using namespace std;

int main(int argc, char *argv[])
{
                long a = 1, n = 1;

                while(a%2011)
                {
         ...

5楼: Originally posted by lurencyj at 2012-04-20 16:33:22:
补充一下：
8位的long，最大数值为1.8447E19
4位的long，最大数值为4.295E9

  a ≡ a (mod n)
=> a ≡ a mod n (mod n)
=> 10a ≡ 10(a mod n) (mod n)
=> 10a+1 ≡ 10(a mod n)+1 (mod n)

#include

int main() {
        int a = 1, n = 1;
       
        while(a %= 2011){
                a = 10*a + 1;
                n++;
        }
       
        printf("n=%d\n", n);
        return 0;
}

8楼: Originally posted by sudo at 2012-04-20 22:22:36:
这个问题可以数学味一点

同余方程式

  a ≡ a (mod n)
=> a ≡ a mod n (mod n)
=> 10a ≡ 10(a mod n) (mod n)
=> 10a+1 ≡ 10(a mod n)+1 (mod n)


那么程序就是这样了

#include < ...