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

?

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Problemsetter: Mohammad Mahmudur Rahman

解題思路，分段法，f [n] 記錄最長的那條邊不超過n的放法數， 假設c[n] 為最長邊為n的放法數 ?f [n] = f [n-1] + c[n] 。

假設最長邊為z，其它為 x,y;

則 ? z-x < y < z

當 ?x=1 時， z-1<y<z ?0 種方案，

當 ?x=2 時， z-2<y<z ?1 種方案，

......................................................

當 ?x=z-1時， 1<y<z ?z-2種方案

所以總計 (z-1)*(z-2)/2 種方案

但是其中包含了x與y想等的情況，因為 x取值為 [ z/2+1 , z-1 ] 區間內可能 x,y相等，共 (z-1)- (z/2+1)+1=z/2-1 種方案

而且x,y與 y,x認為為兩個，重復計算了，所以除以2

所以 c[n]= [?(z-1)*(z-2)/2 -(z/2-1) ] / 2

#include <iostream> using namespace std;const int maxn=1000000; unsigned long long f[maxn+10]; int n;void ini(){f[3]=0;for(long long z=4;z<=maxn;z++){f[z]=f[z-1] + ( (z-1)*(z-2)/2 - (z/2 -1) )/2;} }int main(){ini();while(cin>>n && n>=3){cout<<f[n]<<endl;}return 0; }

轉載于:https://www.cnblogs.com/toyking/p/3797375.html

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