loj 1030概率dp
题目链接:http://lightoj.com/volume_showproblem.php?problem=1030
思路:一直以来对这种概率题都挺感冒的=.=......还是说一下思路吧,dp[i]表示前i个位置所能能到的期望值,然后我们可以从后往前递推。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 using namespace std; 7 8 const int MAXN = (100 + 10); 9 double dp[MAXN], num[MAXN]; 10 int n; 11 12 int main() 13 { 14 int _case, t = 1; 15 scanf("%d", &_case); 16 while (_case--) { 17 scanf("%d", &n); 18 for (int i = 1; i <= n; i++) { 19 scanf("%lf", &num[i]); 20 } 21 memset(dp, 0, sizeof(dp)); 22 dp[n] = num[n]; 23 for (int i = n - 1; i >= 1; i--) { 24 int d = min(6, n-i); 25 dp[i] = num[i]; 26 for (int j = 1; j <= d; j++) { 27 dp[i] += dp[i+j]/d; 28 } 29 } 30 printf("Case %d: %.7lf\n", t++, dp[1]); 31 } 32 return 0; 33 }