hdu4466 Triangle

最新推荐文章于 2019-03-26 21:34:42 发布

相知无悔

最新推荐文章于 2019-03-26 21:34:42 发布

阅读量1.1k

点赞数

CC 4.0 BY-SA版权

分类专栏： hdu 数论文章标签： hdu4466 Triangle

本文链接：https://blog.youkuaiyun.com/mengzhengnan/article/details/9567273

hdu 同时被 2 个专栏收录

188 篇文章

订阅专栏

数论

44 篇文章

订阅专栏

Triangle

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 236 Accepted Submission(s): 121

Problem Description

You have a piece of iron wire with length of n unit. Now you decide to cut it into several ordered pieces and fold each piece into a triangle satisfying:
*All triangles are integral.
* All triangles are pairwise similar.
You should count the number of different approaches to form triangles. Two approaches are considered different if either of the following conditions is satisfied:
*They produce different numbers of triangles.
* There exists i that the i^th (again, pieces are ordered) triangle in one approaches is not congruent to i^th triangle in another plan.
The following information can be helpful in understanding this problem.
* A triangle is integral when all sides are integer.
*Two triangles are congruent when all corresponding sides and interior angles are equal.
* Two triangles are similar if they have the same shape, but can be different sizes.
*For n = 9 you have 6 different approaches to do so, namely
(1, 1, 1) (1, 1, 1) (1, 1, 1)
(1, 1, 1) (2, 2, 2)
(2, 2, 2) (1, 1, 1)
(1, 4, 4)
(2, 3, 4)
(3, 3, 3)
where (a, b, c) represents a triangle with three sides a, b, c.

Input

There are several test cases.
For each test case there is a single line containing one integer n (1 ≤ n ≤ 5 * 10⁶).
Input is terminated by EOF.

Output

For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) and Y is the number of approaches, moduled by 10⁹ + 7.

Sample Input

Sample Output

Case 1: 0
Case 2: 0
Case 3: 1
Case 4: 0
Case 5: 1
Case 6: 2
Case 7: 1
Case 8: 6
Case 9: 3
Case 10: 4
Case 11: 10
Case 12: 25
Case 13: 10
Case 14: 16
Case 15: 525236
Case 16: 523080925

#include <stdio.h>
#include <iostream>
#include <math.h>
#define maxmod 1000000007
#define N 5000001
using namespace std;
int dp[N],fact[N];
void init()
{
    int i,j;
    dp[3]=1;
    for(i=4;i<N;i++)
    {
        dp[i]=dp[i-1]+floor((i-1)/2.0)-ceil(i/3.0)+1;
        if(!(i&1))//如果是偶数的话
        {
            dp[i]-=floor((i/2)/2.0);
            if(dp[i]<0)
            {
                dp[i]+=maxmod;

            }
            if(dp[i]>=maxmod)
            {
                dp[i]-=maxmod;
            }
        }
    }
    fact[1]=1;fact[2]=2;
    for(i=3;i<N;i++)
    {
      fact[i]=fact[i-1]*2;//记录２的阶乘
      if(fact[i]>=maxmod)
      {
          fact[i]-=maxmod;
      }
      for(j=2;i*j<N;j++)
      {
          dp[i*j]-=dp[i];//去掉所有的非质三角形,一定要从小去，
          if(dp[i*j]<0)
          {
              dp[i*j]+=maxmod;
          }
      }
    }


}
int main ()
{
    int i,n,tcase;
    __int64 result;
    tcase=1;
    init();
    while(scanf("%d",&n)!=EOF)
    {
        result=0;
        for(i=1;i*i<=n;i++)
        {
            if(n%i==0)//可以分成n/i个小三角形
            {
                result=(result+(__int64)dp[i]*fact[n/i]%maxmod)%maxmod;
                if(i*i!=n)//i与n/i不相等，就要加两个
                {
                result=(result+(__int64)dp[n/i]*fact[i]%maxmod)%maxmod;
                }
            }

        }

        printf("Case %d: %I64d\n",tcase++,result);
    }
    return 0;
}