lightoj - 1027 - A Dangerous Maze - dp / 概率dp

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题意：给n个门，正数表示xi分钟后出去，负数表示-xi分钟后回到原地，选每个门的概率相等，求出去的时间的期望。

题解：设E为所求期望，P为选正数的概率，S1为正数平均值，S2为负数平均值的绝对值，则E = P * S1 + (1 - P) * (S2 + S1)，进一步化简，可得E = sum(abs(x[i])) / 正数的个数。

#include <bits/stdc++.h>//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define ll long long#define SZ(x) ((int)(x).size()) #define ALL(v) (v).begin(), (v).end()#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)#define reveach(i, v) for (__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++ i) #define REP(i,a,n) for ( int i=a; i<int(n); i++ )#define FOR(i,a,n) for ( int i=n-1; i>= int(a);i-- )#define lson rt<<1, L, m#define rson rt<<1|1, m, Rtypedef pair<int, int> pii;typedef pair<ll, ll> pll;#define mp(x, y) make_pair(x, y)#define pb(x) push_back(x)#define fi first#define se second#define CLR(a, b) memset(a, b, sizeof(a))#define Min(a, b) a = min(a, b)#define Max(a, b) a = max(a, b)const int maxn = 1e2 + 7;int T;int kase;int n;int a[maxn];int main(){#ifdef ac    freopen("in.txt","r",stdin);#endif    //freopen("out.txt","w",stdout);    scanf("%d", &T);    while(T--){        scanf("%d", &n);        REP(i, 0, n) scanf("%d", &a[i]);        int up = 0, down = 0;        REP(i, 0, n){            up += abs(a[i]);            if(a[i] > 0) down ++;        }        if(down == 0){            printf("Case %d: inf\n", ++ kase);            continue;        }        int t = __gcd(up, down);        printf("Case %d: %d/%d\n", ++ kase, up / t, down / t);    }    return 0;}

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