Codeforces Hello 2018

A. Modular Exponentiation

$2^n$很大的时候直接输出$m$， 不然就把$2^n$算出来。

B. Christmas Spruce

没什么可说的。

C. Party Lemonade

感觉有点厉害。如果$2c_{i-1}<c_i$ ，那么咱与其买$i$，不如买两个$i-1$。这样循环一遍，更新所有的$c_i$。

然后在从$0$到$31$循环，相当于枚举$l$的二进制位。如果二进制位为$1$就买，如果之前买的能用当前的替换，就把$ans$重新赋值为$c_i$。

 1 #include<map>
 2 #include<cstdio>
 3 #include<iostream>
 4 using namespace std;
 5 typedef long long ll;
 6 int n, m;
 7 ll ans, a[32];
 8 inline void gmin(ll &x, ll y) {
 9     if (x > y) x = y;
10 }
11 int main() {
12     cin >> n >> m;
13     for (int i = 0; i < n; ++i)
14         cin >> a[i];
15     for (int i = 1; i < n; ++i)
16         a[i] = min(a[i], 2 * a[i-1]);
17     for (int i = n; i < 32; ++i)
18         a[i] = a[i-1] * 2;
19     for (int i = 0; i < 32; ++i) {
20         if (ans > a[i]) ans = a[i];
21         if (m & (1 << i)) ans += a[i];
22     }
23     printf("%lld\n", ans);
24     return 0;
25 }

D. Too Easy Problems

二分答案。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 using namespace std;
 5 int n, T;
 6 struct P {
 7     int a, t, id;
 8     inline void init(int i) {
 9         cin >> a >> t; id = i;
10     }
11     inline bool operator<(const P &p) const {
12         return t < p.t;
13     }
14 } a[200005];
15 bool check(int k) {
16     int sum = 0, cnt = 0;
17     for (int i = 1; i <= n && cnt < k; ++i) {
18         if (a[i].a >= k) {
19             sum += a[i].t; ++cnt;
20         }
21     } return sum <= T && cnt == k;
22 }
23 inline void pt(int k) {
24     int cnt = 0;
25     for (int i = 1; i <= n && cnt < k; ++i)
26         if (a[i].a >= k) printf("%d ", a[i].id), ++cnt;
27 }
28 int main() {
29     cin >> n >> T;
30     for (int i = 1; i <= n; ++i) a[i].init(i);
31     sort(a + 1, a + 1 + n);
32     int l = 0, r = n;
33     while (l + 1 < r) {
34         int m = (l + r) >> 1;
35         if (check(m)) l = m; else r = m;
36     }
37     while (check(l+1)) ++l;
38     printf("%d\n", l);
39     printf("%d\n", l);
40     pt(l);
41     return 0;
42 }

原文地址：https://www.cnblogs.com/p0ny/p/8253477.html