Problem:
Analyse:
dp[i]为i开始走到结尾的价值,
那么dp[i]是从后面的6个转移过来的.
这样我们就倒着递推就好了(后面的要先算好).
注意后面不足六个的时候的处理情况.
/**********************jibancanyang**************************
*Author* :jibancanyang
*Created Time* : 一 5/ 9 20:38:59 2016
*File Name* : .cpp
**Code**:
***********************[email protected]**********************/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <stack>
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
vector<int> vi;
#define pr(x) cout << #x << ": " << x << " "
#define pl(x) cout << #x << ": " << x << endl;
#define pri(a) printf("%d\n",(a));
#define xx first
#define yy second
#define sa(n) scanf("%d", &(n))
#define sal(n) scanf("%lld", &(n))
#define sai(n) scanf("%I64d", &(n))
#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++)
const int mod = int(1e9) + 7, INF = 0x3fffffff;
const int maxn = 1e5 + 13;
int T, n;
double dp[111];
int a[111];
int main(void)
{
#ifdef LOCAL
freopen("/Users/zhaoyang/in.txt", "r", stdin);
//freopen("/Users/zhaoyang/out.txt", "w", stdout);
#endif
//ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
cin >> T;
int cas = 1;
while (T--) {
cin >> n;
for (int i = 1; i <= n; i++)
cin >> a[i];
dp[n] = (double)a[n];
for (int i = n - 1; i >= 1; i--) {
dp[i] = a[i];
int m = min(n, i + 6);
for (int j = i + 1; j <= m; j++) {
dp[i] += dp[j] / (m - i) ;
}
}
cout << "Case " << cas++ << ": " ;
printf("%.14f\n", dp[1]);
}
return 0;
}
时间: 2024-10-09 19:04:23