Sawtooth

Think about a plane:

● One straight line can divide a plane into two regions.
● Two lines can divide a plane into at most four regions.
● Three lines can divide a plane into at most seven regions.
● And so on...

Now we have some figure constructed with two parallel rays in the same direction, joined by two straight segments. It looks like a character “M”. You are given N such “M”s. What is the maximum number of regions that these “M”s can divide a plane ?

2
1
2

Case #1: 2
Case #2: 19

题解及代码：

         推导公式很简单：首先通过看图，我们可以知道，任意两个M都能相交出现16个交点，然后对M进行标号1--n，总的交点数就是∑16*i  (1=<i<n)，

然后根据：边数+点数+1=分成的区域数，可以算出公式是：8*n^2-7*n+1。因为数比较大，所以会想到用java，结果kuangbin大神卡的很紧Orz....

         那就用c++的大数模版来吧，通过观察我们发现：上述公式可以转换一下成n*(8*n-7)+1，这里8*n可以用位运算来算，我们让m=8*n-7，m最大也到

不了10^13，而n最大是10^12，这样我们可以利用大数的思想，将m分成两部分，分别与n进行相乘，最后简单处理一下输出就行了。

Time：187ms

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll mod=1000000;

int main()
{
    ll n,m,l_m,r_m;
    int cas;
    scanf("%d",&cas);
    for(int ca=1; ca<=cas; ca++)
    {
        scanf("%I64d",&n);
        //printf("%I64d\n",8*n*n-7*n+1);
        m=(n<<3)-7;
        l_m=m/mod;
        r_m=m%mod;
        l_m*=n;
        r_m=r_m*n+1;
        l_m=l_m+r_m/mod;
        r_m%=mod;
        printf("Case #%d: ",ca);
        if(l_m) printf("%I64d%06I64d\n",l_m,r_m);
        else printf("%I64d\n",r_m);
    }
    return 0;
}

hdu 5047 Sawtooth

原文：http://blog.csdn.net/knight_kaka/article/details/39614167