LightOJ 1030 Discovering Gold(期望DP)

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题意：n个位置，每处有一定价值。从1开始，扔骰子，扔几走几，走到哪就得到那个点的价值，直到走到n。如果掷出的点数超过了n就重新掷。问到达n得到金子价值的期望。

思路：dp[i] = (dp[i+1] + ... + dp[i+j]) / tot，其中1 <= j <= tot(tot为最大可以掷出的点数)，显然 tot = min(n - i, 6)。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 1e2 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;int v[N];double dp[N];int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//freopen("/Users/apple/out.txt", "w", stdout);#endif    si(T);    while(T --) {    si(n);    for(int i = 1; i <= n; i ++) si(v[i]), dp[i] = v[i] * 1.0;    for(int i = n - 1; i > 0; i --) {    int tot = min(n - i, 6);    for(int j = 1; j <= tot; j ++) dp[i] += dp[i+j] / tot;    }    printf("Case %d: %.6f\n", ++ cas, dp[1]);    }        return 0;}

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