HDU 5047 sawtooth

Sawtooth Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/65536 K (Java/Others)
Total submission (s): 258 accepted submission (s): 78


Problem descriptionthink 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 ≤0 ≤ 1012)
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

212

 
Sample output

Case #1: 2Case #2: 19

 
Source2014 ACM/ICPC Asia Regional Shanghai Online

Question and code:

The formula for derivation is simple: first, we can see that any two m can have 16 intersections, and then label M 1 -- N, the total intersection points are Σ 16 * I (1 = <I <n ),

Then, based on the number of edges + points + 1 = number of divided areas, the formula is 8 * n ^ 2-7 * n + 1. Because the number is relatively large, we will think of using Java. The result is that the kuangbin is very tight. orz ....

Then use the C ++ large number template, through observation we found: the above formula can be converted to N * (8 * N-7) + 1, here 8 * N can be calculated using bitwise operations, we let M = 8 * N-7, m maximum also

N is not 10 ^ 13, but n is 10 ^ 12 at the maximum. In this way, we can use the idea of large numbers to divide m into two parts and multiply them by N, finally, simply process the output.

Time: 187 Ms

#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