You are given n rods of length 1,2, . . . , n. You have to pick any 3 of them and build a triangle. How
many distinct triangles can you make? Note that, two triangles will be considered di?erent if they have
at least 1 pair of arms with di?erent 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
5
8
0
Sample Output
3
22
题意:问你从1......n中任意选择三个数,能形成三角形的方案数
题解:我是递推 假设已经找出 从前i个数找到的方案数, 那么dp[i] = dp[i-1] + i与前面的数形成的方案
根据排列组合计算可以得到
//meek///#include<bits/stdc++.h> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include<iostream> #include<bitset> #include<vector> #include <queue> #include <map> #include <set> #include <stack> using namespace std ; #define mem(a) memset(a,0,sizeof(a)) #define pb push_back #define fi first #define se second #define MP make_pair typedef long long ll; const int N = 1000000+100; const int M = 1000001; const int inf = 0x3f3f3f3f; const int MOD = 1000000007; ll dp[N],c[N][3]; ll n,tmp,tmpp; void init() { c[1][1] = 1;c[1][2] = 0; for(int i=2;i<N;i++) { c[i][1] = i; c[i][2] = c[i-1][2] + c[i-1][1]; } dp[3] = 0; for(int i=4;i<N;i++) { tmp = i%2?i/2+1:i/2; tmpp = (tmp-1)*(tmp-2)/2; dp[i] = dp[i-1] + c[i - tmp][2] + tmpp; } } int main() { init(); while(~scanf("%lld",&n)!=EOF) { if(n<3) break; printf("%lld\n",dp[n]); } return 0; }
代码
时间: 2024-10-08 10:13:51