牛客练习赛17

Posted 2020-11-02 Algorithms Crush Me

长方体

1 #include <bits/stdc++.h>
2 using namespace std;
3  
4 int main(){
5     int X1, X2, X3;
6     scanf("%d %d %d", &X1, &X2, &X3);
7     printf("%d\n", 4 * (int)(sqrt(1.0 * X1 * X2 / X3 + 0.5) + sqrt(1.0 * X1 * X3 / X2 + 0.5) + sqrt(1.0 * X2 * X3 / X1 + 0.5)));
8     return 0;
9 }

Aguin

 

好位置

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 2e5 + 10;
 4 char s[maxn], t[maxn];
 5 int a[maxn], r[66];
 6  
 7 int main(){
 8     scanf("%s %s", s + 1, t + 1);
 9     int ls = strlen(s + 1), lt = strlen(t + 1);
10     int p = 1, ok = 1;
11     for(int i = 1; i <= ls; ++i) {
12         if (p <= lt && s[i] == t[p]) ++p;
13         a[i] = p - 1;
14     }
15     p = lt;
16     for(int i = ls; i >= 1; --i){
17         if(p >= 1 && s[i] == t[p]){
18             r[s[i]-‘a‘] = p;
19             p--;
20         }
21         if(r[s[i]-‘a‘] && r[s[i]-‘a‘] <= a[i-1] + 1) continue;
22         ok = 0;
23         break;
24     }
25     puts(ok ? "Yes" : "No");
26     return 0;
27 }

Aguin

 

操作数

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const LL mod = 1e9 + 7;
 5 LL a[2222];
 6  
 7 LL qpow(LL a, LL b) {
 8     LL ret = 1LL;
 9     while (b) {
10         if (b & 1) ret = ret * a % mod;
11         a = a * a % mod;
12         b >>= 1;
13     }
14     return ret;
15 }
16  
17 LL inv(LL x) {
18     return qpow(x, mod - 2);
19 }
20  
21 int main() {
22     int n, k;
23     scanf("%d %d", &n, &k);
24     for (int i = 1; i <= n; ++i) {
25         scanf("%lld", a + i);
26         LL ans = 0, tmp = 1;
27         for (int j = i; j >= 1; --j) {
28             ans = (ans + a[j] * tmp) % mod;
29             tmp = tmp * (k - j + i) % mod * inv(i - j + 1) % mod;
30         }
31         printf("%lld%c", k ? ans : a[i], i == n ? ‘\n‘ : ‘ ‘);
32     }
33     return 0;
34 }

Aguin

 

经纬度

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const double pi = acos(-1);
 4  
 5 int main(){
 6     int T;
 7     scanf("%d", &T);
 8     while(T--){
 9         double lat1, lng1, lat2, lng2, r = 6371009;
10         scanf("%lf %lf %lf %lf", &lat1, &lng1, &lat2, &lng2);
11         lat1 = lat1 / 180 * pi;
12         lat2 = lat2 / 180 * pi;
13         lng1 = lng1 / 180 * pi;
14         lng2 = lng2 / 180 * pi;
15         double x1 = r * cos(lat1) * sin(lng1), y1 = r * cos(lat1) * cos(lng1), z1 = r * sin(lat1);
16         double x2 = r * cos(lat2) * sin(lng2), y2 = r * cos(lat2) * cos(lng2), z2 = r * sin(lat2);
17         double x = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) + (z1 - z2) * (z1 - z2));
18         double c = acos((2 * r * r - x * x) / 2 / r / r);
19         double d = c * r - x;
20         printf("%.0f\n", d);
21     }
22     return 0;
23 }

Aguin

 

求长度

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef pair<int, int> pii;
 4 const int maxn = 1e5 + 10;
 5 const int INF = 1e9;
 6 vector<pii> G[maxn];
 7 int n, p[11];
 8  
 9 int dis[maxn], vis[maxn];
10 void dijkstra(int s){
11     for(int i = 0; i <= n; ++i) dis[i] = INF, vis[i] = 0;
12     dis[s] = 0;
13     priority_queue<pii> pq;
14     pq.push(pii(0, s));
15     while(!pq.empty()){
16         int x = pq.top().second; pq.pop();
17         if(vis[x]) continue;
18         vis[x] = 1;
19         for(int i = 0; i < G[x].size(); ++i){
20             int to = G[x][i].first, d = G[x][i].second;
21             if(dis[to] > dis[x] + d) {
22                 dis[to] = dis[x] + d;
23                 pq.push(pii(-dis[to], to));
24             }
25         }
26     }
27 }
28  
29 int d[11][11], f[1<<10][10];
30 int main() {
31     int T;
32     scanf("%d", &T);
33     while (T--) {
34         int m, S;
35         scanf("%d %d", &n, &m);
36         for (int i = 0; i <= n; ++i) G[i].clear();
37         for (int i = 1; i <= m; ++i) {
38             int x, y, z;
39             scanf("%d %d %d", &x, &y, &z);
40             G[x].push_back(pii(y, z));
41             G[y].push_back(pii(x, z));
42         }
43         scanf("%d", &S);
44         for(int i = 0; i < (1 <<S); ++i)
45             for(int j = 0; j < S; ++j)
46                 f[i][j] = INF;
47         for (int i = 1; i <= S; ++i) scanf("%d", p + i);
48         for (int i = 1; i <= S; ++i) {
49             dijkstra(p[i]);
50             f[1<<(i-1)][i-1] = d[0][i] = d[i][0] = dis[0];
51             for (int j = 1; j <= i; ++j) {
52                 d[i][j] = d[j][i] = dis[p[j]];
53             }
54         }
55         for(int i = 0; i < (1 << S); ++i){
56             for(int j = 0; j < S; ++j){
57                 if(f[i][j] == INF) continue;
58                 for(int k = 0; k < S; ++k){
59                     if(i & (1 << k)) continue;
60                     if(d[j+1][k+1] == INF) continue;
61                     f[i^(1<<k)][k] = min(f[i^(1<<k)][k], f[i][j] + d[j+1][k+1]);
62                 }
63             }
64         }
65         int ans = INF;
66         for(int i = 0; i < S; ++i){
67             ans = min(ans, f[(1<<S)-1][i] + d[0][i+1]);
68         }
69         printf("%d\n", ans);
70     }
71     return 0;
72 }

Aguin

 

玩游戏

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 5e5 + 10;
 4 char s[maxn], t[maxn];
 5  
 6 bool check(int x, int lt){
 7     for(int i = 1; i <= lt; ++i)
 8         if(s[x+i] != t[i]) return false;
 9     return true;
10 }
11  
12 int main(){
13     int T;
14     scanf("%d", &T);
15     while(T--){
16         scanf("%s %s", s + 1, t + 1);
17         int ls = strlen(s + 1), lt = strlen(t + 1), ok = 0;
18         if(ls == lt){
19             if(check(0, lt)) ok = 1;
20         }
21         else if((ls - lt) % 2 == 1) {
22             if (check((ls - lt - 1) / 2, lt) && check((ls - lt - 1) / 2 + 1, lt)) ok = 1;
23         }
24         else {
25             if ((check((ls - lt - 1) / 2, lt) || check((ls - lt - 1) / 2 + 1, lt)) && (check((ls - lt - 1) / 2 + 1, lt) || check((ls - lt - 1) / 2 + 2, lt))) ok = 1;
26         }
27         puts(ok ? "Alice" : "Bob");
28     }
29     return 0;
30 }

Aguin