Hello 2018

A

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
LL solve(LL n, LL m) {
LL ans = 1;
for (int i = 1; i <= n; ++i) {
ans = 2LL * ans;
if (ans > m) return m;
}
return m % ans;
}

int main(){
LL n, m; scanf("%lld%lld", &n, &m);
printf("%lld\n", solve(n, m));
return 0;
}
B

注意判断的条件

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
vector<int> G[qq];
int num[qq];
void dfs(int u, int fa) {
if (G[u].size() == 0) return;
int sz = G[u].size();
for (int i = 0; i < sz; ++i) {
int v = G[u][i];
if (v == fa) continue;
dfs(v, u);
if (num[v] == 0) num[u]++;
}
}

int main(){
int n; scanf("%d", &n);
for (int x, i = 2; i <= n; ++i) {
scanf("%d", &x);
G[x].pb(i);
}
dfs(1, -1);
bool f = true;
for (int i = 1; i <= n; ++i) {
if (G[i].size() != 0) {
if (num[i] < 3) f = false;
}
// if (!(num[i] == 0 || num[i] >= 3)) f = false;
}
if (f) puts("Yes");
else puts("No");
return 0;
}

C

题意：n个容器，第i个容器可以装2^(i - 1)的水，容器价格是ci，现在要装下L的水，问最少的价格

思路：这题我开始具体也不知道怎么写，看到过了人很多，就瞎写了个做法，我是dp[i]代表恰好装下2^i的最小花费，然后一步一步去贪心，对与L的在二进制中的第i为为1和为0进行讨论

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
LL dp[50];
LL num[50];
LL quick(LL n) {
LL ans = 1;
for (int i = 0; i < n; ++i) {
ans = ans * 2LL;
}
return ans;
}

int main(){
LL n, L; scanf("%lld%lld", &n, &L);
for (int i = 0; i < n; ++i) {
scanf("%lld", num + i);
dp[i] = num[i];
for (int j = 0; j < i; ++j) {
dp[i] = min(dp[i], dp[j] * quick(i - j));
}
}
for (int i = n; i < 32; ++i) {
dp[i] = 1e18;
for (int j = 0; j < i; ++j) {
dp[i] = min(dp[i], dp[j] * quick(i - j));
}
}
/*for (int i = 0; i < 32; ++i) {
printf("%lld\n", dp[i]);
}*/
LL tmp = 0;
LL minx = 1e18;;
for (int i = 31; i >= 0; --i) {
// printf("%lld\n", dp[i]);
LL base = quick(i);
if (base >= L) {
minx = min(minx, dp[i]);
} else {
if (L & (1 << i)) {
LL x = (L & ((1 << i) - 1));
if (x != 0) {
minx = min(minx, tmp + 2LL * dp[i]);
} else {
minx = min(minx, tmp + dp[i]);
}
tmp += dp[i];
} else {
LL x = (L & ((1 << i) - 1));
if (x != 0) minx = min(minx, tmp + dp[i]);
}
}
// printf("%lld\n", minx);
}
printf("%lld\n", minx);
return 0;
}

D
题意：给出n个问题和一个可用时间T，然后给出n个问题的ai与ti，现在你需要从中选出一些问题，假设问题的索引是pj，总共选了k个问题，那么你的得分就是
k ≤ apj 

条件满足的个数，现在要求得分最大化并输出你选的问题

思路：这题目只要意识到如果存在某个答案ans满足，那么你一定可以选出ans个ai>=k的问题来，所以就可以直接二分去做了，在选择时间方面可以用一个最小堆来做，好像可以不用二分，听大佬们说nlogn可以直接做

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 2e5 + 10;
const int INF = 1e9 + 10;
struct Node {
LL ai, ti, id;
bool operator < (const Node &d) {
if (ai == d.ai) return ti < d.ti;
return ai < d.ai;
}
}p[qq];
LL n, T;
bool check(int x) {
priority_queue<int, vector<int>, greater<int> > Q;
for (int i = 0; i < n; ++i) {
if (p[i].ai >= x) Q.push(p[i].ti);
}
// printf("%d\n", Q.top());
if (Q.size() < x) return false;
LL tmp = 0;
while (x--) {
tmp += Q.top();
Q.pop();
}
if (tmp > T) return false;
return true;
}
struct List {
int ti, id;
bool operator < (const List &d) {
return ti < d.ti;
}
}num[qq];

int main(){
scanf("%lld%lld", &n, &T);
for (int i = 0; i < n; ++i) {
scanf("%lld%lld", &p[i].ai, &p[i].ti);
p[i].id = i + 1;
}
sort(p, p + n);
int l = 0, r = n;
int ans;
while (l <= r) {
int mid = (l + r) >> 1;
if (check(mid)) {
ans = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
printf("%d\n", ans);
if (ans == 0) {
printf("0\n");
return 0;
}
int cnt = 0;
for (int i = 0; i < n; ++i) {
if (p[i].ai >= ans) num[cnt].ti = p[i].ti, num[cnt++].id = p[i].id;
}
sort(num, num + cnt);
/*for (int i = 0; i < cnt; ++i) {
printf("%d ", num[i]);
}
puts("");*/
printf("%d\n", ans);
for (int i = 0; i < ans; ++i) {
printf("%d%c", num[i].id, i == ans - 1 ? '\n' : ' ');
}
return 0;
}