UVa11401

11401 - Triangle Counting

Problem G Triangle Counting

Input: Standard Input

Output: Standard Output

 

 

You are given n rods of length 1, 2…, n. You have to pick any 3 of them & build a triangle. How many distinct triangles can you make? Note that, two triangles will be considered different if they have at least 1 pair of arms with different length.

Input

The input for each case will have only a single positive integer n (3<=n<=1000000). The end of input will be indicated by a case with n<3. This case should not be processed.

Output

 

For each test case, print the number of distinct triangles you can make.

Sample Input                                                  Output for Sample Input

5 8 0

3 22

---

Problemsetter: Mohammad Mahmudur Rahman

法一！！
/*
先打表！！！后推公式！！！

i     3   4    5    6     7      8      9      10     11    12     13     14     15     16     17     18
 
f[i]  0   1    3    7     13     22     34     50     70    95     125    161    203    252    308    372

做差    1    2    4    6      9      12      16     20    25    30      36     42     49     56     64

再做差    1    2    2     3      3       4       4      5     5     6        6      7      7     8

在奇数项做差和偶数项做差即可！！！     
*/


/*//一开始TLE！！后来就把它用来打表了，推出了通项公式！！！

记得一道做过fft的题，打这个表轻轻松松！！！
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<set>
#include<vector>
#include<bitset>
using namespace std;
typedef long long ll;

const int M=1000010;
ll s[M],f[M];
int get(){
    char c;
    int res=0;
    while(c=getchar(),!isdigit(c));
    do{
        res=(res<<3)+(res<<1)+(c-'0');
    }while(c=getchar(),isdigit(c));
    return res;
}

*/


int main()
{
    ll i,ans1,ans2,ans3,ans,n,m;
    while(~scanf("%lld",&n))
    {
        if(n<3)break;
        ll k=0;f[n+1]=n;
        while((++k)<=n)f[n+1-k]=f[n+1+k]=n-k;
        f[1]=f[0]=0;
        //for(i=0;i<=2*n;i++)printf("%lld**",f[i]);
        for(i=1;i<=n;i++)f[i*2]--;
        for(i=1;i<=2*n;i++)f[i]/=2;
        //for(i=0;i<=2*n;i++)printf("%lld**",f[i]);
        s[1]=0;
        ll ans=0;
        for(i=2;i<=2*n;i++)s[i]=s[i-1]+f[i];
        for(i=1;i<=n;i++)
        {
            ans+=s[2*n]-s[i];
            ans-=(i-1)*(n-i);
            ans-=(n-1);
            ans-=(n-i)*(n-i-1)/2;
        }
        printf("%lld**
",ans);
    }
    return 0;
}






#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<set>
#include<vector>
#include<bitset>
using namespace std;
typedef long long ll;

int main()
{
    ll ans,k,n;
    while(1)
    {
        n=get();
        if(n<3)break;
        if(n%2==0){k=n/2-1;ans=k*(4*k*k+3*k-1)/6;printf("%lld
",ans);continue;}
        k=(n-1)/2-1;ans=k*(4*k*k+3*k-1)/6+(n/2-1)*(n/2);
        printf("%lld
",ans);
    }
    return 0;
}





法2！！





/*
递推：
设最大边为x的三角形有c[x]个，y+z>x；

y的值  z的个数 
1       0
2       1
3       3
。。。。。
x-1     x-2


除去y=z的情况有（x-1）/2种！！

剩下的再除以2即可！！

C[x]= ((x-1)*(x-2)/2-(x-1)/2)/2;

*/ 
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<set>
#include<vector>
#include<bitset>
using namespace std;
typedef long long ll;

const int M=1000010;
ll f[M];

int get(){
    char c;
    int res=0;
    while(c=getchar(),!isdigit(c));
    do{
        res=(res<<3)+(res<<1)+(c-'0');
    }while(c=getchar(),isdigit(c));
    return res;
}


int main()
{
    ll i;
    int n;
    f[3]=0;
    for(i=4;i<M;i++)f[i]=f[i-1]+((i-1)*(i-2)/2-(i-1)/2)/2;
    while(1)
    {
        n=get();
        if(n<3)break;
        printf("%lld
",f[n]);
    }
    return 0;
}