[HDOJ4788]Hard Disk Drive（水题）

Posted 2020-08-09

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4788

题意：1KB有的时候认为是1000B，有的时候是1024B。问两种方法之间的损失。

YY个公式水过。

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <iomanip>
 4 #include <cstring>
 5 #include <climits>
 6 #include <complex>
 7 #include <cassert>
 8 #include <cstdio>
 9 #include <bitset>
10 #include <vector>
11 #include <deque>
12 #include <queue>
13 #include <stack>
14 #include <ctime>
15 #include <set>
16 #include <map>
17 #include <cmath>
18 using namespace std;
19 #define fr first
20 #define sc second
21 #define cl clear
22 #define BUG puts("here!!!")
23 #define W(a) while(a--)
24 #define pb(a) push_back(a)
25 #define Rint(a) scanf("%d", &a)
26 #define Rll(a) scanf("%I64d", &a)
27 #define Rs(a) scanf("%s", a)
28 #define Cin(a) cin >> a
29 #define FRead() freopen("in", "r", stdin)
30 #define FWrite() freopen("out", "w", stdout)
31 #define Rep(i, len) for(int i = 0; i < (len); i++)
32 #define For(i, a, len) for(int i = (a); i < (len); i++)
33 #define Cls(a) memset((a), 0, sizeof(a))
34 #define Clr(a, x) memset((a), (x), sizeof(a))
35 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
36 #define lrt rt << 1
37 #define rrt rt << 1 | 1
38 #define pi 3.14159265359
39 #define RT return
40 #define lowbit(x) x & (-x)
41 #define onecnt(x) __builtin_popcount(x)
42 typedef long long LL;
43 typedef long double LD;
44 typedef unsigned long long ULL;
45 typedef pair<int, int> pii;
46 typedef pair<string, int> psi;
47 typedef pair<LL, LL> pll;
48 typedef map<string, int> msi;
49 typedef vector<int> vi;
50 typedef vector<LL> vl;
51 typedef vector<vl> vvl;
52 typedef vector<bool> vb;
53 
54 const int maxn = 1010;
55 double tmp;
56 char s[maxn];
57 map<char, int> q;
58 
59 signed main() {
60     // FRead();
61     int T, _ = 1;
62     Rint(T);
63     q[‘B‘] = 1;q[‘K‘] = 2;q[‘M‘] = 3;q[‘G‘] = 4;
64     q[‘T‘] = 5;q[‘P‘] = 6;q[‘E‘] = 7;q[‘Z‘] = 8;
65     q[‘Y‘] = 9;
66     W(T) {
67         scanf("%lf[%s", &tmp, s);
68         printf("Case #%d: ", _++);
69         int t = q[s[0]] - 1;
70         double ret = pow(1000.0, t) / pow(1024.0, t);
71         printf("%.2lf%%\n", (double)(1.0 - ret) * 100.0);
72     }
73     RT 0;
74 }