微软2017年预科生计划在线编程笔试

Posted 2020-09-07 Dirge

Legendary Items

答案是每一件物品需要的期望步数和

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define st first
 5 #define nd second
 6 #define pii pair<int, int>
 7 #define pil pair<int, ll>
 8 #define pli pair<ll, int>
 9 #define pll pair<ll, ll>
10 #define pw(x) ((1LL)<<(x))
11 #define lson l, m, rt<<1
12 #define rson m+1, r, rt<<1|1
13 #define FIN freopen("input.txt","r",stdin);
14 #define FOUT freopen("output.txt","w+",stdout);
15 using namespace std;
16 /***********/
17 template <class T>
18 bool scan (T &ret) {
19     char c;
20     int sgn;
21     if (c = getchar(), c == EOF) return 0; //EOF
22     while (c != ‘-‘ && (c < ‘0‘ || c > ‘9‘) ) c = getchar();
23     sgn = (c == ‘-‘) ? -1 : 1;
24     ret = (c == ‘-‘) ? 0 : (c - ‘0‘);
25     while (c = getchar(), c >= ‘0‘ && c <= ‘9‘) ret = ret * 10 + (c - ‘0‘);
26     ret *= sgn;
27     return 1;
28 }
29 template<typename N,typename PN>inline N flo(N a,PN b){return a>=0?a/b:-((-a-1)/b)-1;}
30 template<typename N,typename PN>inline N cei(N a,PN b){return a>0?(a-1)/b+1:-(-a/b);}
31 template<typename T>
32 inline int sgn(T a) {return a>0?1:(a<0?-1:0);}
33 template <class T1, class T2>
34 bool gmax(T1 &a, const T2 &b) { return a < b? a = b, 1:0;}
35 template <class T1, class T2>
36 bool gmin(T1 &a, const T2 &b) { return a > b? a = b, 1:0;}
37 
38 template<class A, class B, class C>
39 struct Triple {
40     A st; B nd; C rd;
41     bool operator <(const Triple &rhs) const {
42         return st == rhs.st? (nd == rhs.nd? rd < rhs.rd: nd < rhs.nd): st < rhs.st;
43     }
44 };
45 typedef Triple<int, int, int> tiii;
46 
47 template<class T1, class T2>
48 ostream& operator <<(ostream &out, pair<T1, T2> p) {
49     return out << "(" << p.st << ", " << p.nd << ")";
50 }
51 template<class A, class B, class C>
52 ostream& operator <<(ostream &out, Triple<A, B, C> t) {
53     return out << "(" << t.st << ", " << t.nd << ", " << t.rd << ")";
54 }
55 template<class T>
56 ostream& operator <<(ostream &out, vector<T> vec) {
57     out << "("; for(auto &x: vec) out << x << ", "; return out << ")";
58 }
59 const int inf = 0x3f3f3f3f;
60 const ll INF = 1e18;
61 const ll mod = 1e9+7;
62 const double eps = 1e-5;
63 const int N = 2e5+10;
64 /***********/
65 
66 double e(int p, int q){//当前概率p, 每次增长q
67     double ans = 0, ret = 1;
68     for(int i = 1; ; p += q, i++){
69         if(p >= 100){
70             ans += i*ret;
71             return ans;
72         }
73         ans += i*ret*p*0.01;
74         ret *= 1-p*0.01;
75     }
76 }
77 double f[101][101];
78 int main(){
79     for(int i = 0; i <= 100; i++)
80         for(int j = 1; j <= 100; j++)
81             f[i][j] = e(i, j);
82     int n, p, q;
83     while(~scanf("%d%d%d", &p, &q, &n)){
84         double ans = 0;
85         for(int i = 0; i < n; i++)
86             if(p >> i)
87                 ans += f[p>>i][q];
88             else {
89                 //(n-i)个f[0][q]累加精度损失过大, WA
90                 ans += f[0][q]*(n-i);
91                 break;
92             }
93         printf("%.2f\n", ans);
94     }
95     return 0;
96 }

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