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$c_i$。

#include<map>
#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
int n, m;
ll ans, a[32];
inline void gmin(ll &x, ll y) {
    if (x > y) x = y;    
}
int main() {
    cin >> n >> m;
    for (int i = 0; i < n; ++i)
        cin >> a[i];
    for (int i = 1; i < n; ++i)
        a[i] = min(a[i], 2 * a[i-1]);
    for (int i = n; i < 32; ++i)
        a[i] = a[i-1] * 2;
    for (int i = 0; i < 32; ++i) {
        if (ans > a[i]) ans = a[i];
        if (m & (1 << i)) ans += a[i];
    }
    printf("%lld\n", ans);
    return 0;    
}

D. Too Easy Problems

二分答案。

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int n, T;
struct P {
    int a, t, id;
    inline void init(int i) {
        cin >> a >> t; id = i;
    }
    inline bool operator<(const P &p) const {
        return t < p.t;    
    }
} a[200005];
bool check(int k) {
    int sum = 0, cnt = 0;
    for (int i = 1; i <= n && cnt < k; ++i) {
        if (a[i].a >= k) {
            sum += a[i].t; ++cnt;
        }
    } return sum <= T && cnt == k;
}
inline void pt(int k) {
    int cnt = 0;
    for (int i = 1; i <= n && cnt < k; ++i)
        if (a[i].a >= k) printf("%d ", a[i].id), ++cnt;
}
int main() {
    cin >> n >> T;
    for (int i = 1; i <= n; ++i) a[i].init(i);
    sort(a + 1, a + 1 + n);
    int l = 0, r = n;
    while (l + 1 < r) {
        int m = (l + r) >> 1;
        if (check(m)) l = m; else r = m;
    } 
    while (check(l+1)) ++l;
    printf("%d\n", l);
    printf("%d\n", l);
    pt(l);
    return 0;    
}