Codeforces Round #413(Div. 1 + Div. 2, combined)——ABCD
题目在这里
A.Carrot Cakes
乱七八糟算出两个时间比较一下就行了
又臭又长仅供参考
#include <bits/stdc++.h> #define rep(i, j, k) for(int i = j;i <= k;i ++) #define rev(i, j, k) for(int i = j;i >= k;i --) using namespace std; typedef long long ll; int main() {
ios::sync_with_stdio(false);
int n, t, k, d;
cin >> n >> t >> k >> d;
int t1 = (n + k - ) / k * t;
int t2 = ;
int t3 = (n + k - ) / k;
int t4 = (d + t - ) / t;
int t5 = t3 - t4;
if(t5 & ) t2 = d + (t5 + ) / * t;
else t2 = t4 * t + t5 / * t;
if(t2 < t1) puts("YES");
else puts("NO");
return ;
}
B.T-shirt buying
颜色数只有三个,开三个优先队列就行了
然后同一件衣服不要重复卖
#include <bits/stdc++.h> #define rep(i, j, k) for(int i = j;i <= k;i ++) #define rev(i, j, k) for(int i = j;i >= k;i --) using namespace std; const int maxn = ; typedef long long ll; struct node {
int x, y; bool operator < (const node &a) const {
return x > a.x;
}
}; priority_queue <node> q[]; int n, m, v[maxn], a, b, p[maxn]; int main() {
ios::sync_with_stdio(false);
cin >> n;
rep(i, , n) cin >> p[i];
rep(i, , n) cin >> a, q[a].push((node){p[i], i});
rep(i, , n) cin >> b, q[b].push((node){p[i], i});
cin >> m;
rep(i, , m) {
cin >> a;
while(!q[a].empty() && v[q[a].top().y]) q[a].pop();
if(!q[a].empty()) cout << q[a].top().x << " ", v[q[a].top().y] = , q[a].pop();
else cout << "-1 ";
}
return ;
}
C.Fountains
三种情况,都用c或d买,一个c一个d
第三种肥肠简单的
前面两种的话
对于新插入的物品,如果价格为x
那么查询一下价格<=(c-x)的物品中的最大beauty
然后两个物品组合一下就行了
查询的话,直接树状数组就可以了,O(nlogn)
需要注意保证是两个不同物品,不能只有一个
#include <bits/stdc++.h> #define lb(x) (x & (-x)) #define rep(i, j, k) for(int i = j;i <= k;i ++) #define rev(i, j, k) for(int i = j;i >= k;i --) using namespace std; typedef long long ll; struct node{
int x, y;
char str[];
void init() {
scanf("%d %d %s", &x, &y, str);
}
}a[]; int n, c, d, e, c1[], c2[]; int ask(int *C, int i) {
int ret = ;
while(i > ) ret = max(ret, C[i]), i -= lb(i);
return ret;
} void ins(int *C, int i, int x) {
while(i <= e) C[i] = max(C[i], x), i += lb(i);
} int main() {
cin >> n >> c >> d, e = max(c, d);
int t1 = , t2 = , t3, t4 = , t5 = , ans = ;
rep(i, , n) {
a[i].init();
if(a[i].str[] == 'C' && a[i].y <= c) {
if(a[i].x > t1) t1 = a[i].x;
t3 = ask(c1, c - a[i].y);
if(t3 != && a[i].x + t3 > t4) t4 = a[i].x + t3;
ins(c1, a[i].y, a[i].x);
}
if(a[i].str[] == 'D' && a[i].y <= d) {
if(a[i].x > t2) t2 = a[i].x;
t3 = ask(c2, d - a[i].y);
if(t3 != && a[i].x + t3 > t5) t5 = a[i].x + t3;
ins(c2, a[i].y, a[i].x);
}
}
ans = max(t4, t5);
if(t1 != && t2 != && t1 + t2 > ans) ans = t1 + t2;
cout << ans;
return ;
}
我还肥肠脑残地,取消了cin与stdio的同步后
同时用cin和scanf,然后WA了一发...
D.Field expansion
显然ans最大就是loga级别的
所以我们直接暴搜就行了
然后考虑到很骚的情况
a = b = 10W h = w = 1
ai = 2 n = 10W
这样的话我们用map防重就好了
很重要的一件事情 :
注意到题意说能放下
所以并没有h一定对应a,w一定对应b
也可以试试h怼b,w怼a!
#include <bits/stdc++.h> #define rep(i, j, k) for(int i = j;i <= k;i ++) #define rev(i, j, k) for(int i = j;i >= k;i --) using namespace std; typedef long long ll; int a, b, h, w, n, c[]; bool cmp(int x, int y) {
return x > y;
} typedef pair<ll, ll> node; vector <node> e; map <node,bool> p; int main() {
ios::sync_with_stdio(false);
cin >> a >> b >> h >> w >> n;
rep(i, , n) cin >> c[i];
sort(c + , c + n + , cmp);
node tmp;
if(h >= a && w >= b || w >= a && h >= b) {puts("");return ;}
e.push_back(make_pair(h, w));
for(int i = ;i <= n;i ++) {
int t = e.size();
for(int j = ;j < t;j ++) {
if(e[j].first < a) {
tmp = make_pair(e[j].first * c[i], e[j].second);
if(!p[tmp]) e.push_back(tmp), p[tmp] = ;
}
if(e[j].second < b) {
tmp = make_pair(e[j].first, e[j].second * c[i]);
if(!p[tmp]) e.push_back(tmp), p[tmp] = ;
}
}
for(int j = ;j < e.size();j ++) {
if(e[j].first >= a && e[j].second >= b || e[j].first >= b && e[j].second >= a) {
cout << i;
return ;
}
}
}
puts("-1");
return ;
}
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