2015
07-16

# hdu 4466-triangle-动态规划-[解题报告]hoj

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 * 10 6).
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 10 9 + 7.

Sample Input

 1
2
3
4
5
6
8
9
10
11
12
15
19
20
100
1000 

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 

1、本质三角形的定义：设a、b、c为三角形的三边，则gcd(a，b，c)=1。

2、先不考虑边互不互质，算出周长为i的三角形个数（动规实现，其实暴力应该也可以）。然后算周长为i，即

i个单位长度作边长的三角形的组合数。

a)你用隔板法考虑就是i个点，i-1个空格插隔板，djw[i]=2^(i-1)种；

b)你也可以按选与不选考虑。

注：放在一个格子里的三角形边长合并组成大边长的三角形。

最后就是n/i个周长为i的三角形组合数即为所求。

3、说一下用动态规划求dp[i]的过程：

函数dp(x),表示周长为 x的不同三角形(a,b,c)的数量,我们假设(
a <= b <= c )

则我们通过枚举最大周长 x, 假设其最大边为 c.

则问题可以划分为两类独立: 1: b = c  2:b != c

第一种情况: b = c, 周长为x的三角形 (a,c,c) 的方案数为:

因为 a+c+c = x ,  a <= c 那么

c最大取 A=floor((x-1)/2 ), c最小取 B=ceil( x/3 ), 此时三角形种类:
A-B+1

第二种情况: b != c, 周长为x的三角形 (a,b,c) 的方案数为:

因为 a+b+c = x,  b <= c-1,a+b > c,

这里我们转而考虑 , 形式如 (
a, b, c-1 )

> c ,

问题就是如何计算出这个M, 我们继续考虑.

a+b+(c-1) = x-1,

=>a+b+c = x, 又因为 a+b = c. 可以得到

=> c+c = x , 2*c = x, 所以 c = x/2, 因为c为整数,所以我们知道,只有当x为偶数时才会出现这个M.

又　a + b = c = x/2,  a <= b , 此时 a 的取值为 [
1, floor( (x/2)/2 ) ]

所以 x为偶数时, M = floor( (x/2)/2 )

当 x为奇数时, M = 0;

dp[i]表示周长为i，那么就有n / i 个这样的三角形。用隔板板，将三角形合并成大的，保证相似就是2
^ (n / i – 1)种情况

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
typedef long long ll;
using namespace std;
const int maxn = 5000005;
const int mod = 1000000007;

int dp[maxn], num[maxn];

void init() {
dp[3] = 1;
for (int i = 4; i < maxn; i++) {
dp[i] = dp[i-1] + floor((i-1)/2.0) - ceil(i/3.0) + 1;
if (!(i & 1))
dp[i] -= i / 4;
dp[i] %= mod;
if (dp[i] < 0) dp[i] += mod;
}

num[1] = 1;
for (int i = 2; i < maxn; i++) {
num[i] = (num[i-1] << 1) % mod;
for (int j = 2; i * j < maxn; j++) {
dp[i * j] -= dp[i];
if (dp[i * j] < 0)
dp[i * j] += mod;
}
}
}

int main() {
int n, t, cas = 1;
init();
while (scanf("%d", &n) != EOF) {
ll ans = 0;
for (int i = 1; i * i <= n; i++) {
if (n % i) continue;
ans = (ans + (1ll*dp[i]*num[n/i])) % mod;
if (i * i != n)
ans = (ans + (1ll*dp[n/i] * num[i])) % mod;
}
printf("Case %d: %lld\n", cas++, ans);
}
return 0;
}