LightOJ 1030 Discovering Gold(概率dp)
阿新 • • 發佈:2018-12-14
題解:
從後面往前加上本金乘上概率即可,控制好多少個可以走到這裡。
#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #include <string> #include <vector> #include <bitset> #include <stack> #include <cmath> #include <deque> #include <queue> #include <list> #include <set> #include <map> #pragma comment(linker, "/STACK:1024000000,1024000000") #define line printf("---------------------------\n") #define mem(a, b) memset(a, b, sizeof(a)) #define pi acos(-1) using namespace std; typedef long long ll; const double eps = 1e-9; const int inf = 0x3f3f3f3f; const int mod = 1e9+7; const int maxn = 100+10; double dp[maxn]; int main(){ int t, cas = 1; scanf("%d", &t); while(t--){ mem(dp, 0); int n; scanf("%d", &n); for(int i = 0; i < n; i++){ scanf("%lf", &dp[i]); } int cnt = 1; for(int i = n-2; i >= 0; i--){ cnt = cnt > 6 ? 6 : cnt; for(int j = 1; j <= cnt; j++){ dp[i] += dp[i+j]/cnt; } cnt++; } printf("Case %d: %.6lf\n", cas++, dp[0]); } }