hdu4277 ? 暴力

USACO ORZ

Time Limit : 5000/1500ms (Java/Other)???Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 4???Accepted Submission(s) : 2

Problem Description Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. The old rectangular shapes are out of favor; new geometries are the favorite.
I. M. Hei, the lead cow pasture architect, is in charge of creating a triangular pasture surrounded by nice white fence rails. She is supplied with N fence segments and must arrange them into a triangular pasture. Ms. Hei must use all the rails to create three sides of non-zero length. Calculating the number of different kinds of pastures, she can build that enclosed with all fence segments.?
Two pastures look different if at least one side of both pastures has different lengths, and each pasture should not be degeneration.

?

Input The first line is an integer T(T<=15) indicating the number of test cases. The first line of each test case contains an integer N. (1 <= N <= 15) The next line contains N integers li indicating the length of each fence segment. (1 <= li <= 10000)

?

Output For each test case, output one integer indicating the number of different pastures.

?

Sample Input 1 3 2 3 4 Sample Output 1 Source 2012 ACM/ICPC Asia Regional Changchun Online

?

#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<set>using namespace std;const int maxn=1000100;typedef long long ll; int T; int a,b,c; int l[maxn],n; set<ll> s;void dfs(int i) {if(i==n+1){if(a&&a<=b&&b<=c&&a+b>c){if(s.find((a<<16)|b)==s.end()) s.insert((a<<16)|b);}return;}a+=l[i];dfs(i+1);a-=l[i];b+=l[i];dfs(i+1);b-=l[i];c+=l[i];dfs(i+1);c-=l[i]; }int main() {cin>>T;while(T--){cin>>n;for(int i=1;i<=n;i++) scanf("%d",&l[i]);s.clear();a=b=c=0;dfs(1);cout<<(int)s.size()<<endl;}return 0; } View Code

?

轉載于:https://www.cnblogs.com/--560/p/4782233.html

總結

以上是生活随笔為你收集整理的hdu4277 暴力的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。