LightOJ 1030-Discovering Gold

题意：

题意就是说给你一个长度为n的序列，初始在1位置，每次可以掷骰子，1-6，如果没有越界，就可以往前走，并且得到目标格子的值。走到最后一个格子结束。问最后获得分数的期望。一般来说期望都是倒着推。设dp[i]表示从i号格子出去的期望，那么dp[i]+=dp[i+step]*1.0/(6,n-i).

代码：

////  Created by  CQU_CST_WuErli//  Copyright (c) 2015 CQU_CST_WuErli. All rights reserved.//// #include<bits/stdc++.h>#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <string>#include <vector>#include <list>#include <map>#include <queue>#include <stack>#include <set>#include <algorithm>#include <sstream>#define CLR(x) memset(x,0,sizeof(x))#define OFF(x) memset(x,-1,sizeof(x))#define MEM(x,a) memset((x),(a),sizeof(x))#define ALL(x) x.begin(),x.end()#define AT(i,v) for (auto &i:v)#define For_UVa if (kase!=1) cout << endl#define BUG cout << "I am here" << endl#define lookln(x) cout << #x << "=" << x << endl#define look(x) cout << #x << "=" << x#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 Lson l,mid,rt<<1#define Rson mid+1,r,rt<<1|1#define Root 1,n,1#define BigInteger bigntemplate <typename T> T max(T& a,T& b) {return a>b?a:b;}template <typename T> T min(T& a,T& b) {return a<b?a:b;}int gcd(int a,int b) {return b==0?a:gcd(b,a%b);}long long gcd (long long a,long long b) {return b==0LL?a:gcd(b,a%b);}const int MAX_L=2005;// For BigIntegerconst int INF_INT=0x3f3f3f3f;const long long INF_LL=0x7fffffff;const int MOD=1e9+7;const double eps=1e-9;const double pi=acos(-1);typedef long long  ll;using namespace std;const int N=110;int n;double w[N];double dp[N];int main(){#ifdef LOCAL    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);//  freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);#endif    int T_T;    for (int kase=scanf("%d",&T_T);kase<=T_T;kase++) {        printf("Case %d: ",kase);        cin >> n;        for (int i=1;i<=n;i++) cin >> w[i];        CLR(dp);        dp[n]=w[n];        for (int i=n-1;i>=1;i--) {            int step=min(6,n-i);            dp[i]+=w[i];            for (int j=1;j<=step;j++) {                dp[i]+=1.0/step*dp[i+j];             }        }        printf("%.6lf\n",dp[1]);        }    return 0;}