(HDU 5753)2016 Multi-University Training Contest 3 Permutation Bo (水)
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思路
两边的数是0和0,对于第1个数和第n个数,他们的大小只有两种可能。
以1为例:0-A-B或0-B-A。 能够让f=1的有
对于中间的数,一共六种情况:
A-B-C、A-C-B、B-C-A、B-A-C、C-A-B、C-B-A
能够让f=1的有
所有的可能性有
代码
#include <bits/stdc++.h>#define mem(a,b) memset(a,b,sizeof(a))#define rep(i,a,b) for(int i=a;i<b;i++)#define debug(a) printf("a =: %d\n",a);const int INF=0x3f3f3f3f;const int maxn=1e3+50;const int Mod=1e9+7;const double PI=acos(-1);typedef long long ll;using namespace std;double mid;double lr;int n;int w[maxn];void init(){ mid=2/6.0; lr=3/6.0;}int main(){ #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif init(); while(~scanf("%d",&n)){ for(int i=1;i<=n;i++) scanf("%d",w+i); double sum=0; if (n==1) sum=w[1]; else{ for(int i=1;i<=n;i++){ if (i>1 && i<n) sum+=w[i]*mid; else sum+=w[i]*lr; } } printf("%.9lf\n",sum); } return 0;}
0 0
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