HDU 5047 Sawtooth 找规律+拆分乘

 

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 ?

InputThe first line of the input is T (1 ≤ T ≤ 100000), which stands for the number of test cases you need to solve.

Each case contains one single non-negative integer, indicating number of “M”s. (0 ≤ N ≤ 10
¹²)OutputFor each test case, print a line “Case #t: ”(without quotes,
t means the index of the test case) at the beginning. Then an integer
that is the maximum number of regions N the “M” figures can divide.Sample Input

2
1
2

Sample Output

Case #1: 2
Case #2: 19
 
题意：问用"M"形的线条去切割一个平面，问当有n个"M"时能把平面切割成几份？
用1条直线切割平面可以把平面分成2份
用2条直线切割平面可以把平面分成4份
用3条直线切割平面可以把平面分成7份
用n条直线切割平面可以把平面分成2+2+3+4+....+n份
假设"M"的线条可以无限延长成直线，那么一个M相当于4条直线，但是考虑到"M"实际上并没有延长成直线，那么一个"M"的切割份数一定小于4条直线的切割分数，
猜想二者可能满足某种关系。
直线:
n=4时，ans=11
n=8时，ans=37
“M”：
n=1时，ans=2
n=2时，ans=19
发现
11-9=2
37-2*9=19
设“M”个数为n，那么有4n*(4n+1)/2+1-9n  --->  n*(8n-7)+1
直接输出答案会爆ll。
可以用JAVA或者高精度算法。
这里用的是拆分乘。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll mod=1e9;
int main()
{
    //freopen("input.txt","r",stdin);
    ll T,kase=;
    scanf("%lld",&T);
    while(T--)
    {
        ll n;
        scanf("%lld",&n);
        ll a0=(*n-)%mod;
        ll a1=(*n-)/mod;
        ll b0=n%mod;
        ll b1=n/mod;
        ll num0=(a0*b0+);
        ll num1=(b0*a1+b1*a0+num0/mod);
        ll num2=(a1*b1+num1/mod);
        num0%=mod;
        num1%=mod;
        printf("Case #%lld: ",kase++);
        if(num2)    printf("%lld%09lld%09lld\n",num2,num1,num0);
        else if(num1)    printf("%lld%09lld\n",num1,num0);
        else    printf("%lld\n",num0);
    }
    return ;
}