题意：给n个‘M'形，问最多能把平面分成多少区域
解法：推公式 ： f(n) = 4n(4n+1)/2 - 9n + 1 = (8n+1)(n-1)+2
前面部分有可能超long long，所以要转化一下，令a = 8n+1, b = n-1，将两个数都化为a1*10^8+b1的形式，则
(a1*10^8+b1)(a2*10^8+b2) =(a1a2*10^8 + a1b2 + a2b1)*10^8 + b1b2 + 2，由于a1,a2最多2为10^4左右，中间的数就都不会超过long long 了，先打印出前面，再打8位的后面即可。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#define lll __int64
#define ll long long
using namespace std; int main()
{
ll n;
int t,cs = ;
ll e = 100000000LL;
scanf("%d",&t);
getchar();
while(t--)
{
scanf("%I64d",&n);
ll a = 8LL*n+1LL;
ll b = n-1LL;
ll a1 = a/e;
ll b1 = a%e;
ll a2 = b/e;
ll b2 = b%e;
ll ans1 = a1*a2*e + a1*b2 + a2*b1 + (b1*b2+2LL)/e;
ll ans2 = (b1*b2+2LL)%e;
int res = ans2;
printf("Case #%d: ",cs++);
if(ans1 != )
{
printf("%I64d",ans1);
printf("%08d\n",res);
}
else
printf("%I64d\n",ans2);
}
return ;
}