Wannafly挑战赛15

A.小化价格

贪心

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 int a[maxn], b[maxn], c[maxn], id[maxn];
 5 multiset<int> S;
 6  
 7 bool cmp(int i, int j){
 8     return b[i] > b[j];
 9 }
10  
11 int main(){
12     int n, m, p = 0;
13     scanf("%d %d", &n, &m);
14     for(int i = 1; i <= n; ++i) scanf("%d", a + i);
15     for(int i = 1; i <= m; ++i){
16         scanf("%d %d", b + i, c + i);
17         id[i] = i;
18     }
19     sort(a + 1, a + 1 + n);
20     sort(id + 1, id + 1 + m, cmp);
21     long long ans = 0;
22     for(int i = n; i >= 1; i--){
23         while(p < m && b[id[p + 1]] >= a[i]){
24             ++p;
25             S.insert(c[id[p]]);
26         }
27         if(S.empty()){ans = -1; break;}
28         ans += *S.begin();
29         S.erase(S.begin());
30     }
31     cout << ans << endl;
32     return 0;
33 }
Aguin

B.辆安排

贪心

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 int cnt[6];
 5  
 6 int main(){
 7     int n, x;
 8     scanf("%d", &n);
 9     for(int i = 1; i <= n; ++i){
10         scanf("%d", &x);
11         cnt[x]++;
12     }
13     int ans = cnt[5];
14     while(cnt[1] + cnt[2] + cnt[3] + cnt[4]){
15         int r = 5;
16         ans++;
17         while(r >= 4 && cnt[4]) r -= 4, cnt[4]--;
18         while(r >= 3 && cnt[3]) r -= 3, cnt[3]--;
19         while(r >= 2 && cnt[2]) r -= 2, cnt[2]--;
20         while(r >= 1 && cnt[1]) r -= 1, cnt[1]--;
21     }
22     cout << ans << endl;
23     return 0;
24 }
Aguin

C.

出题人骗我

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4  
 5 int main(){
 6     LL n, q, x;
 7     scanf("%lld %lld", &n, &q);
 8     while(q--){
 9         scanf("%lld", &x);
10         LL y = n, ans = 1;
11         while(1){
12             if(x % 2) {ans += x / 2; break;}
13             x /= 2;
14             if(y % 2) x++;
15             ans += y / 2;
16             y -= y / 2;
17         }
18         printf("%lld\n", ans);
19     }
20     return 0;
21 }
Aguin

D.字串

随便计数

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 typedef long long LL;
 5  
 6 int c[10][maxn];
 7 int lowbit(int s) {
 8     return s & (-s);
 9 }
10 void modify(int o, int x, int v) {
11     for (int i = x; i < maxn; i += lowbit(i)) c[o][i] += v;
12     return;
13 }
14 int query(int o, int x) {
15     int ret = 0;
16     for (int i = x; i > 0; i -= lowbit(i)) ret += c[o][i];
17     return ret;
18 }
19  
20 char s[maxn];
21 int main(){
22     int q, L, R;
23     scanf("%s %d %d %d", s + 1, &q, &L, &R);
24     int l = strlen(s + 1);
25     LL ans = 0;
26     for(int i = 1; i <= l; ++i){
27         int x = s[i] - '0';
28         modify(x, i, 1);
29         for(int j = x + 1; j <= 9; ++j) ans += query(j, i - L + 1) - query(j, i - R);
30     }
31     while(q--){
32         int i, x;
33         scanf("%d %d", &i, &x);
34         int y = s[i] - '0';
35         modify(y, i, -1);
36         for(int j = 0; j <= 9; ++j) {
37             if (j < x) ans += query(j, min(l, i + R - 1)) - query(j, min(l, i + L - 2));
38             else if (j > x) ans += query(j, i - L + 1) - query(j, i - R);
39             if (j < y) ans -= query(j, min(l, i + R - 1)) - query(j, min(l, i + L - 2));
40             else if (j > y) ans -= query(j, i - L + 1) - query(j, i - R);
41         }
42         modify(x, i, 1);
43         s[i] = '0' + x;
44         printf("%lld\n", ans);
45     }
46     return 0;
47 }
Aguin

E.W的斜率

显然斜着扫

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 1e6 + 10;
 5 int x[maxn], y[maxn], id[maxn], P, Q;
 6  
 7 bool cmp(int i, int j){
 8     return (double) x[i] * P + (double) y[i] * Q < (double) x[j] * P + (double) y[j] * Q;
 9 }
10  
11 double cal(int i){
12     double k1 = x[id[i]+1] == x[id[i]] ? 1e18 : 1.0 * (y[id[i]+1] - y[id[i]]) / (x[id[i]+1] - x[id[i]]);
13     double k2 = Q == 0 ? 1e18 : 1.0 * P / Q;
14     return fabs(k1 - k2);
15 }
16  
17 LL gcd(LL a, LL b){
18     if(a == 0) return b;
19     if(b == 0) return a;
20     return a % b ? gcd(b, a % b) : b;
21 }
22  
23 int main(){
24     int n;
25     scanf("%d %d %d", &n, &P, &Q);
26     for(int i = 1; i <= n; ++i){
27         scanf("%d %d", x + i, y + i);
28         id[i] = i;
29     }
30     sort(id + 1, id + 1 + n, cmp);
31     int p = 1;
32     double m = cal(1);
33     for(int i = 2; i < n; ++i)
34         if(cal(i) < m) m = cal(i), p = i;
35     LL dy = y[id[p]+1] - y[id[p]], dx = x[id[p]+1] - x[id[p]];
36     LL g = gcd(abs(dx), abs(dy));
37     if(dx < 0 && dy < 0) dx = -dx, dy = -dy;
38     dx /= g, dy /= g;
39     printf("%lld/%lld\n", dy, dx);
40     return 0;
41 }
Aguin

F.

优秀分块

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 111111;
 4 int sg[maxn], cnt[maxn], b[maxn];
 5 vector<int> G[maxn];
 6  
 7 struct query {
 8     int l, r;
 9 } q[maxn];
10 int block;
11  
12 bool cmp(query A, query B) {
13     int a = A.l / block, b = B.l / block;
14     if (a != b) return a < b;
15     return A.r < B.r;
16 }
17  
18 inline void Add(int x) {
19     cnt[x]++;
20     if (cnt[x] == 1) b[x / block]++;
21 }
22  
23 inline void Minus(int x) {
24     cnt[x]--;
25     if (cnt[x] == 0) b[x / block]--;
26 }
27  
28 int get() {
29     for (int j = 0; ; ++j)
30         if (b[j] != block)
31             for (int k = j * block; ; ++k)
32                 if (!cnt[k]) return k;
33 }
34  
35 int main() {
36     int n, m, k;
37     scanf("%d %d %d", &n, &m, &k);
38     assert(1 <= n && n <= 100000 && 1 <= m && m <= 200000 && 1 <= k && k <= 100000);
39     block = sqrt(n + 0.5);
40     for (int i = 1; i <= m; ++i) {
41         int u, v;
42         scanf("%d %d", &u, &v);
43         assert(1 <= v && v < u && u <= n);
44         G[u].push_back(v);
45     }
46     for (int i = 1; i <= n; ++i) {
47         for (int j = 0; j < G[i].size(); ++j) Minus(sg[G[i][j]]);
48         sg[i] = get();
49         for (int j = 0; j < G[i].size(); ++j) Add(sg[G[i][j]]);
50         Add(sg[i]);
51     }
52     for (int i = 1; i <= k; ++i) scanf("%d %d", &q[i].l, &q[i].r);
53     sort(q + 1, q + 1 + k, cmp);
54     memset(cnt, 0, sizeof(cnt));
55     memset(b, 0, sizeof(b));
56     int r = 0, l = 1, ans = 0;
57     for (int i = 1; i <= k; ++i) {
58         while (l < q[i].l) Minus(sg[l++]);
59         while (l > q[i].l) Add(sg[--l]);
60         while (r < q[i].r) Add(sg[++r]);
61         while (r > q[i].r) Minus(sg[r--]);
62         ans ^= get();
63     }
64     puts(ans ? "Alice" : "Bob");
65     return 0;
66 }
Aguin

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转载自www.cnblogs.com/Aguin/p/9028835.html