HDU 4466 triangle (Chengdu, 12 years)

Reprint please indicate the source, thank you http://blog.csdn.net/ACM_cxlove? Viewmode = Contents
By --- cxlove

Question: a wire with a length of N is given. The wire is divided into several parts. Each part is folded into a triangle, which must be similar to all triangles.

If the three sides of a triangle are equal, the order of the triangle is different.

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4466

First, consider the number of triangles with a circumference of M.

Let's assume a <= B <= C

There are two scenarios: B = C

So for the upper bound of C, C maximum floor (x-1)/2), because a is at least 1

For the lower bound of C, the minimum value of C is Ceil (X/3)

Then all cases are floor (x-1)/2)-Ceil (X/3) + 1

Case 2: B! = C

We consider that (A, B, C), because a + B> C, B <= C-1, so (a, B, C-1) must also be a triangle.

In addition, you need to remove a + B = C + 1, where A + B + C = X, you can find that X must be an odd number, the number of floor (X-(x-1) /2)/2)

The recursive formula in the Code is obtained.

We also need to convert all triangles into persistent ones. Linear filtering can be used here.

Explain the purpose.

Next, we divide N into several parts, each of which is of the same length and can form several consecutive triangles. Because the questions require mutual quality.

We combine several of them into a large triangle similar to an essential triangle.

Suppose I have an essential triangle, the question requires different order as different, Plug board can be solved, I-1 empty, each can choose to separate or not separated

2 ^ (I-1) cases, together constitute a large triangle

#include<iostream>#include<cstdio>#include<map>#include<cstring>#include<cmath>#include<vector>#include<algorithm>#include<set>#include<string>#include<queue>#define inf 100000005#define M 20005#define N 5000005#define maxn 300005#define eps 1e-10#define zero(a) fabs(a)<eps#define Min(a,b) ((a)<(b)?(a):(b))#define Max(a,b) ((a)>(b)?(a):(b))#define pb(a) push_back(a)#define mp(a,b) make_pair(a,b)#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define MOD 1000000007#define lson step<<1#define rson step<<1|1#define sqr(a) ((a)*(a))#define Key_value ch[ch[root][1]][0]//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;int dp[N],fac[N];void Init(){    dp[3]=1;    for(int i=4;i<N;i++)    {        dp[i]=dp[i-1]-ceil(i/3.0)+floor((i-1)/2.0)+1;        if(!(i&1)) dp[i]-=floor((i-(i-1)/2.0)/2.0);        if(dp[i]>=MOD) dp[i]-=MOD;        if(dp[i]<0) dp[i]+=MOD;    }    fac[1]=1;fac[2]=2;    for(int i=3;i<N;i++)    {        fac[i]=fac[i-1]*2;        if(fac[i]>=MOD) fac[i]-=MOD;        for(int j=2;i*j<N;j++)        {            dp[j*i]-=dp[i];            if(dp[j*i]<0) dp[j*i]+=MOD;        }    }}int main(){    Init();    int cas=0,n;    while(cin>>n)    {        LL ans=0;        for(int i=1;i*i<=n;i++)        {            if(n%i) continue;            ans=(ans+(LL)dp[i]*fac[n/i])%MOD;            if(i*i!=n) ans=(ans+(LL)dp[n/i]*fac[i])%MOD;        }        cout<<"Case "<<++cas<<": "<<ans<<endl;    }    return 0;}