HDU 5047 推公式+别样输出

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\n",res);

        }

        else

            printf("%I64d\n",ans2);

    }

    return 0;

}

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