A water problem
Problem Description
Two planets named Haha and Xixi in the universe and they were created with the universe beginning.
There is 73 days in Xixi a year and 137 days in Haha a year.
Now you know the days N after Big Bang, you need to answer whether it is the first day in a year about the two planets.
Input
There are several test cases(about 5 huge test cases).
For each test, we have a line with an only integer N(0≤N), the length of N is up to 10000000.
Output
For the i-th test case, output Case #i: , then output “YES” or “NO” for the answer.
主要就是大数取模的问题，涉及到秦九韶算法,数论
数论的的基本定理：
(a + b) mod c =
((a mod c) + (b mod c)) mod c
把一个n次多项式
\[f(x) = {a_n}x^n + {a_{n-1}x^{n-1}} + \cdot\cdot\cdot+{a_1}x+a_0\]改写成如下形式：
\[\begin{eqnarray} f(x) &=& {a_n}x^n + {a_{n-1}x^{n-1}} + \cdot\cdot\cdot+{a_2}x^2+{a_1}x+a_0\nonumber \\ &=& ({a_n}x^{n-1} + {a_{n-1}x^{n-2}} + \cdot\cdot\cdot+{a_2}x^2+{a_1})x+a_0 \nonumber \\ &=& (({a_n}x^{n-2} + {a_{n-1}x^{n-3}} + \cdot\cdot\cdot+{a_2})x+{a_1})x+a_0\nonumber \\ &\vdots \nonumber \\ &=& ((\cdot\cdot\cdot({a_n}x + a_{n-1})x+a_{n-2})x\cdot\cdot\cdot+a_1)x+a_0\nonumber \end{eqnarray}\]这里把要测试的数看成是 x = 10，各位数字为a的多项式的和，结合数论的基本定理即可解之：
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define N 10000010
//#define N 1000
int cnt;
int main()
{
char m[N];
while (scanf("%s", m) != EOF)
{
int d = 0;
for (int i = 0; i<strlen(m); i++)
d = ((d * 10) % 10001 + m[i] - '0') % 10001;
if (d == 0)
printf("Case #%d: YES\n", ++cnt);
else
printf("Case #%d: NO\n", ++cnt);
memset(m, 0, sizeof(m));
}
return 0;
}
PS：
吐槽一下，实验了好多次终于知道怎么在这里面用latex写漂亮的数学公式了，但是真的好麻烦。。。。我就是按照这个在default.html里加句代码，然后如果对自己写的Latex代码信心不足，可以先在这个地方实验，这个网站是实时生成公式，挺方便的，另外可参考 How to supported latex in github-pages
