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 = 1; 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 != 0) { printf("%I64d",ans1); printf("%08d ",res); } else printf("%I64d ",ans2); } return 0; }