多校第九场 HDU6682 HDU6685
HDU6682 Rikka with Mista
复制代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62#include <bits/stdc++.h> #define ll long long using namespace std; int w[50]; struct D{ ll l, r; }; vector<D> x, y, A[20], B[20]; void Init(int l, int r, vector<D>& x, ll sum){ if(l > r){x.push_back({sum, 0}); return; } Init(l + 1, r, x, sum); Init(l + 1, r, x, sum + w[l]); } ll get(vector<D>& A, vector<D>& B, ll lim, int opt){//opt表示是否进位 ll ans = 0; if(opt == 0){ int now = B.size() - 1; for(int i = 0; i < A.size(); i++){ while(now >= 0 && B[now].r + A[i].r >= lim) now--; ans += now + 1; } }else{ int now = 0; for(int i = A.size() - 1; i >= 0; i--){ while(now < B.size() && B[now].r + A[i].r < lim) now++; ans += B.size() - now; } } return ans; } void solve(){ int n; scanf("%d", &n); for(int i = 1; i <= n; i++) scanf("%d", &w[i]); int mid = n>>1; x.clear(); y.clear(); Init(1, mid, x, 0); Init(mid + 1, n, y, 0); ll ans = 0, base = 1; int now1, now2, k0, k1; for(int t = 1; t <= 12; t++){ for(int i = 0; i < 10; i++) A[i].clear(), B[i].clear(); for(int i = 0; i < x.size(); i++) A[x[i].l % 10].push_back({x[i].l / 10, x[i].r}); for(int i = 0; i < y.size(); i++) B[y[i].l % 10].push_back({y[i].l / 10, y[i].r}); for(int i = 0; i < 10; i++){ k0 = (14 - i) % 10, k1 = (13 - i) % 10; ans += get(A[i], B[k0], base, 0) + get(A[i], B[k1], base, 1); } now1 = now2 = 0; for(int i = 0; i < 10; i++){ ll ex = i * base; for(int j = 0; j < A[i].size(); j++) x[now1++] = {A[i][j].l, A[i][j].r + ex}; for(int j = 0; j < B[i].size(); j++) y[now2++] = {B[i][j].l, B[i][j].r + ex}; } base *= 10; } printf("%lldn", ans); } int main(){ int T; scanf("%d", &T); while(T--) solve(); }
HDU6685 Rikka with Coin
50最多一张
20最多三张
10最多1张
100可能少一张
然后暴力就ok了
复制代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40#include <bits/stdc++.h> #define ll long long #define INF 0x3f3f3f3f using namespace std; int a[110]; void solve(){ int n, ans = INF; scanf("%d", &n); bool flag = true; for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); if(a[i] % 10) flag = false; a[i] /= 10; } sort(a + 1, a + 1 + n); int Maxdollar = max(1, a[n] / 10); if(!flag){ printf("-1n"); return; } for(int i = 0; i <= 1; i++) for(int j = 0; j <= 3; j++) for(int k = 0; k <= 1; k++) for(int h = Maxdollar - 1; h <= Maxdollar; h++){ bool OK = true; for(int cnt = 1; cnt <= n; cnt++){ bool ok = false; for(int x = 0; x <= i; x++) for(int y = 0; y <= j; y++) for(int z = 0; z <= k; z++) { int r = x * 5 + y * 2 + z; if (h * 10 + r >= a[cnt] && a[cnt] >= r && (h * 10 + r) % 10 == a[cnt] % 10) ok = true; } if(!ok){OK = false; break;} } if(OK) ans = min(ans, i + j + k + h); } printf("%dn", ans); } int main(){ int T; scanf("%d", &T); while(T--) solve(); }
最后
以上就是高挑大地最近收集整理的关于多校第九场 HDU6682 HDU6685的全部内容,更多相关多校第九场内容请搜索靠谱客的其他文章。
本图文内容来源于网友提供,作为学习参考使用,或来自网络收集整理,版权属于原作者所有。
发表评论 取消回复