2018 ACM-ICPC 宁夏邀请赛

A 小甜甜

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cmath>
 5 #include <vector>
 6 #include <queue>
 7 #include <set>
 8 #include <map>
 9 #include <string>
10 #include <string.h>
11 #include <stdlib.h>
12 #include <time.h>
13 #include <climits>
14
15 using namespace std;
16
17 const int maxN=5e6+7;
18
19 unsigned int sta[maxN];
20 int to[maxN];
21 int top=0;
22
23 int n,p,q,m;
24 unsigned int SA,SB,SC;
25 long long ans=0;
26
27 unsigned int rng61(){
28     SA^=SA<<16;
29     SA^=SA>>5;
30     SA^=SA<<1;
31     unsigned int t=SA;
32     SA=SB;
33     SB=SC;
34     SC^=t^SA;
35     return SC;
36 }
37
38 void PUSH(unsigned int x,int i){
39     //cerr<<"PUSH"<<x<<endl;
40     top++;
41     sta[top]=x;
42
43     if (x>sta[to[top-1]]) to[top]=top;
44     else to[top]=to[top-1];
45
46     ans=ans^(1LL*sta[to[top]]*i);
47 }
48
49 void POP(int i){
50     //cerr<<"POP"<<endl;
51     if (top==0) return;
52     top--;
53     if (top==0) return;
54
55     ans=ans^(1LL*sta[to[top]]*i);
56 }
57
58 void gen(){
59     scanf("%d%d%d%d%u%u%u",&n,&p,&q,&m,&SA,&SB,&SC);
60     ans=top=0;
61
62     for (int i=1;i<=n;i++) {
63         if (rng61()%(p+q)<p) PUSH(rng61()%m+1,i);
64         else POP(i);
65     }
66
67     printf("%lld\n",ans);
68 }
69
70
71 int main(){
72     int T;
73     scanf("%d",&T);
74     int kcase=0;
75     while (T--){
76         printf("Case #%d: ",++kcase);
77         gen();
78     }
79     return 0;
80 }

B 小洛洛

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <cstdlib>
 6 #include <cmath>
 7 using namespace std;
 8
 9 typedef long long ll;
10
11 const int N = 55;
12 struct P_ {
13     int x, y;
14     P_ operator-(const P_ &r) const {
15         return (P_) { x - r.x, y - r.y };
16     }
17     double norm() const {
18         return hypot((double)x, (double)y);
19     }
20     double operator*(const P_ &r) const {
21         return (double)x * r.x + (double)y * r.y;
22     }
23 } pts[N];
24
25 int main() {
26     int T;
27     scanf("%d", &T);
28     for(int it = 1; it <= T; ++it) {
29         int n;
30         scanf("%d", &n);
31         for(int i = 0; i < n; ++i) {
32             scanf("%d%d", &pts[i].x, &pts[i].y);
33         }
34         pts[n] = pts[0];
35         pts[n + 1] = pts[1];
36         P_ p0;
37         scanf("%d%d", &p0.x, &p0.y);
38         double ans = 0.0;
39         for(int i = 0; i < n; ++i) {
40             double r = (p0 - pts[i + 1]).norm();
41             P_ a = pts[i + 1] - pts[i],
42                b = pts[i + 2] - pts[i + 1];
43             double th = acos(a * b / (a.norm() * b.norm()));
44             ans += r * th;
45         }
46         printf("Case #%d: %.3f\n", it, ans);
47     }
48     return 0;
49 }

C BPM136

计算出偏移量即可

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<cstdlib>
 6 #include<algorithm>
 7
 8 using namespace std;
 9
10 typedef long long ll;
11
12 const int N = 100;
13
14 char s1[N];
15 char s2[N];
16 int n,m;
17
18 char s[N];
19
20 int main() {
21     int T,kcas=0;
22     scanf("%d",&T);
23     while(T--) {
24         scanf("%d%d",&n,&m);
25         scanf("%s%s%s",s1,s2,s);
26         int ca=s1[0]-s2[0];
27
28         printf("Case #%d: ",++kcas);
29         for(int i=0;i<m;i++) putchar((s[i]-‘A‘+(ca)+26)%26+‘A‘);
30         puts("");
31     }
32     return 0;
33 }

D 小洛洛

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <cstdlib>
 6 using namespace std;
 7
 8 int main() {
 9     int T;
10     scanf("%d", &T);
11     for(int it = 1; it <= T; ++it) {
12         int n, m;
13         scanf("%d%d", &n, &m);
14         double ans1 = (n == 1 ? 1 : 0.5);
15         double ans2 = (m + 1.0) / (2.0 * m);
16         printf("Case #%d: %.6f %.6f\n", it, ans1, ans2);
17     }
18     return 0;
19 }

E 小洛洛

模拟即可，一脸蛋疼

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <cstdlib>
 6 using namespace std;
 7
 8 const int N = 500003;
 9 struct T_ {
10     int n;
11     int v[3];
12     T_ *ch[4];
13 } ss[N], *sp;
14
15 void dfs_(T_ *c) {
16     if(!c) return;
17     for(int i = 0; i < c->n; ++i) {
18         if(i != 0)
19             putchar(‘ ‘);
20         printf("%d", c->v[i]);
21     }
22     putchar(‘\n‘);
23     for(int i = 0; i <= c->n; ++i)
24         dfs_(c->ch[i]);
25 }
26
27
28 void fuxk_rt_(T_ *&c) {
29     if(c->n == 3) {
30         T_ *c2 = sp++, *nc = sp++;
31         *c2 = (T_){ 1, { c->v[2] }, { c->ch[2], c->ch[3] } };
32         *nc = (T_){ 1, { c->v[1] }, { c, c2 } };
33         c->n = 1;
34         c = nc;
35     }
36 }
37
38 bool insert_(T_ *&c, int val) {
39     if(!c) {
40         *(c = sp++) = (T_) { 1, { val }, { NULL, NULL } };
41         return true;
42     } else if(c->n == 3)
43         return false;
44     else if(c->ch[0]) {
45 rerun:
46         int i = 0;
47         while(i < c->n && c->v[i] < val) ++i;
48
49         T_ *sub = c->ch[i];
50         if(!insert_(sub, val)) {
51             for(int j = c->n; j > i; --j)
52                 c->v[j] = c->v[j - 1];
53             for(int j = ++c->n; j > i; --j)
54                 c->ch[j] = c->ch[j - 1];
55
56             T_ *c2 = sp++;
57             *c2 = (T_){ 1, { sub->v[2] }, { sub->ch[2], sub->ch[3] } };
58             sub->n = 1;
59             c->v[i] = sub->v[1];
60             c->ch[i] = c2;
61             swap(c->ch[i], c->ch[i + 1]);
62             goto rerun;
63         }
64         return true;
65     } else {
66         int i = 0;
67         while(i < c->n && c->v[i] < val) ++i;
68         for(int j = c->n; j > i; --j)
69             c->v[j] = c->v[j - 1];
70         ++c->n;
71         c->ch[c->n] = NULL;
72         c->v[i] = val;
73         return true;
74     }
75 }
76
77 int main() {
78     int T;
79     scanf("%d", &T);
80     for(int it = 1; it <= T; ++it) {
81         int n;
82         scanf("%d", &n);
83
84         sp = ss;
85         T_ *rt = NULL;
86         for(int i = 0; i < n; ++i) {
87             int x;
88             scanf("%d", &x);
89             while(!insert_(rt, x))
90                 fuxk_rt_(rt);
91         }
92         printf("Case #%d:\n", it);
93         dfs_(rt);
94     }
95     return 0;
96 }

F BPM136

考虑floyed，每次加进点更新就好了

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<cstdlib>
 6 #include<algorithm>
 7
 8 using namespace std;
 9 #define int long long
10
11 const int N = 205;
12 const int M = 20005;
13
14 int d[N][N];
15 int f[N][N];
16 int a[N];
17 int n,m;
18
19 struct ques {
20     int x,y,id;
21     int w;
22     int ans;
23 }q[M];
24
25 bool cmp_id(ques a,ques b) {
26     return a.id<b.id;
27 }
28 bool cmp_w(ques a,ques b) {
29     return a.w>b.w;
30 }
31
32 struct node {
33     int x;
34     int w;
35 }p[N];
36
37 bool cmp_node_w(node a,node b) {
38     return a.w<b.w;
39 }
40
41 bool vis[N];
42
43 main() {
44     int T,kcas=0;
45     scanf("%lld",&T);
46     while(T--) {
47         scanf("%lld%lld",&n,&m);
48         for(int i=1;i<=n;i++) {
49                scanf("%lld",&a[i]);
50             p[i].w=a[i];
51             p[i].x=i;
52         }
53         memset(vis,0,sizeof(vis));
54         for(int i=1;i<=n;i++) {
55             for(int j=1;j<=n;j++) {
56                 scanf("%lld",&d[i][j]);
57                 f[i][j]=d[i][j];
58             }
59         }
60
61         for(int i=1;i<=m;i++) {
62             scanf("%lld%lld%lld",&q[i].x,&q[i].y,&q[i].w);
63             q[i].id=i;
64         }
65
66         sort(q+1,q+m+1,cmp_w);
67         sort(p+1,p+n+1,cmp_node_w);
68
69         int l=1;
70         for(int o=m;o>=1;o--) {
71             while(l <= n && p[l].w<=q[o].w) {
72                 int k=p[l].x;
73                 vis[k]=1;
74                 for(int i=1;i<=n;i++) {
75                     for(int j=1;j<=n;j++) {
76                         if(f[i][k]+f[k][j]<f[i][j]) {
77                             f[i][j]=f[i][k]+f[k][j];
78                         }
79                     }
80                 }
81                 l++;
82             }
83
84             q[o].ans=f[q[o].x][q[o].y];
85         }
86         sort(q+1,q+m+1,cmp_id);
87         printf("Case #%lld:\n",++kcas);
88         for(int i=1;i<=m;i++) {
89             printf("%lld\n",q[i].ans);
90         }
91     }
92     return 0;
93 }

G BPM136

树形DP，fi，j表示在i的子树中选j个，在最优情况下的贡献是多少

转移min(size,m)之后复杂度为O（nk）

whx的证明：找个割出来，割以上子树大小大于k，割以下子树大小小于k。然后上面最多n/k个合并，下面每对在割上共一个点的点对被算一次，每人最多和k个被算

每条边的贡献为p(k-p)

  1 /* ***********************************************
  2 Author        :BPM136
  3 Created Time  :2018/7/17 21:39:31
  4 File Name     :G.cpp
  5 ************************************************ */
  6
  7 #include<iostream>
  8 #include<cstdio>
  9 #include<algorithm>
 10 #include<cstdlib>
 11 #include<cmath>
 12 #include<cstring>
 13 #include<vector>
 14 using namespace std;
 15
 16 typedef long long ll;
 17
 18 const int N = 100005;
 19 const int M = 105;
 20 const ll inf = 1e17;
 21
 22 struct edge {
 23     int y,w,next;
 24 }e[N<<1];
 25 int last[N],ne;
 26
 27 ll f[N][M];
 28 ll sum[N][M];
 29 int siz[N];
 30 int n,m;
 31
 32 void add(int x,int y,int w) {
 33     e[++ne].y=y; e[ne].w=w; e[ne].next=last[x]; last[x]=ne;
 34 }
 35 void add2(int x,int y,int w) {
 36     add(x,y,w);
 37     add(y,x,w);
 38 }
 39
 40 ll cur[M];
 41 ll sum_cur[M];
 42 int du[N];
 43 void dfs(int x,int pre) {
 44     siz[x]=0;
 45     f[x][0]=0; for(int i=1;i<=m;i++) f[x][i]=inf;
 46     sum[x][0]=0; for(int i=1;i<=m;i++) sum[x][i]=inf;
 47     int flag=1;
 48     for(int i=last[x];i!=0;i=e[i].next) {
 49         int y=e[i].y;
 50         if(y==pre) continue;
 51         flag=0;
 52         dfs(y,x);
 53
 54         for(int p=0;p<=m;p++) cur[p]=f[x][p],f[x][p]=inf;
 55         for(int p=0;p<=m;p++) sum_cur[p]=sum[x][p],sum[x][p]=inf;
 56         int bound0=min(m,siz[x]);
 57         int bound1=min(m,siz[y]);
 58         for(int p=0;p<=bound0;p++) {
 59             for(int q=0;q<=bound1 && p+q<=m;q++) {
 60                 if(f[x][p+q]>cur[p]+f[y][q]+(ll)e[i].w*(m-q)*q) {
 61                     f[x][p+q]=cur[p]+f[y][q]+(ll)e[i].w*(m-q)*q;
 62                 }
 63             }
 64         }
 65         siz[x]+=siz[y];
 66     }
 67     if(siz[x]==0 && flag==1) {
 68         siz[x]=1;
 69         f[x][1]=0;
 70         sum[x][1]=0;
 71     }
 72 }
 73
 74 int main() {
 75     int T,kcas=0;
 76     scanf("%d",&T);
 77     while(T--) {
 78         scanf("%d%d",&n,&m);
 79         memset(last,0,sizeof(last));
 80         memset(du,0,sizeof(du));
 81         ne=0;
 82         int tmp;
 83         for(int i=1;i<n;i++) {
 84             int x,y,w;
 85             scanf("%d%d%d",&x,&y,&w);
 86             tmp=w;
 87             add2(x,y,w);
 88             du[x]++; du[y]++;
 89         }
 90         if(n==2) {
 91             if(m==1) printf("Case #%d: 0\n",++kcas);
 92             if(m==2) printf("Case #%d: %d\n",++kcas,tmp);
 93             continue;
 94         }
 95
 96         int root=0;
 97         for(int i=1;i<=n;i++) if(du[i]!=1) {
 98             root=i;
 99             break;
100         }
101         dfs(root,-1);
102         printf("Case #%d: %lld\n",++kcas,f[root][m]);
103     }
104     return 0;
105 }

H 小甜甜+BPM136

考虑如果需要zi次才能打到对手，那么微扰法，只需要ATKj * Zi < ATKi * Zj 就是好的，于是sort即可

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cstdlib>
 7
 8 using namespace std;
 9
10 typedef long long ll;
11
12 const int N = 100005;
13
14 struct node {
15     ll ATK;
16     ll HP;
17     ll Z;
18     ll S;
19 }a[N];
20
21 bool cmp(node a,node b) {
22     return b.ATK*a.Z<b.Z*a.ATK;
23 }
24
25 int n;
26
27 ll EF(ll a) {
28     ll l=1,r=1e9;
29     ll ret=0;
30     while(l<=r) {
31         ll mid=(l+r)>>1;
32         if(mid*mid+mid-2*a>=0) {
33             ret=mid;
34             r=mid-1;
35         } else l=mid+1;
36     }
37     return ret;
38 }
39
40 int main() {
41     int T,kcas=0;
42     scanf("%d",&T);
43     while(T--) {
44         scanf("%d",&n);
45         for(int i=1;i<=n;i++) {
46             scanf("%lld%lld",&a[i].HP,&a[i].ATK);
47             a[i].Z=EF(a[i].HP);
48         }
49         sort(a+1,a+n+1,cmp);
50         for(int i=1;i<=n;i++) a[i].S=a[i-1].S+a[i].Z;
51         ll ans=0;
52         for(int i=1;i<=n;i++) ans+=a[i].S*a[i].ATK;
53         printf("Case #%d: %lld\n",++kcas,ans);
54     }
55     return 0;
56 }

I 待补

J 待补

K 小洛洛

状压暴力check，大力出奇迹

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <cstdlib>
 6 #include <map>
 7 using namespace std;
 8
 9 typedef long long ll;
10
11 const int N = 40;
12 int n;
13 ll MOD, w[N], nei[N];
14 bool tbl[N][N];
15
16 map<ll, int> mp;
17
18 ll f_(ll rest) {
19     if(!rest) return 1;
20     if(mp.find(rest) == mp.end()) {
21         int i = 0;
22         while(~rest & (1LL << i)) ++i;
23         ll ret = 0;
24         ll cnei = nei[i] & rest;
25         if(~cnei & (1LL << i)) { // not sel
26             ll ws = 1;
27             for(int j = 0; j < n; ++j)
28                 if(cnei & (1LL << j))
29                     ws = ws * w[j] % MOD;
30             ret = (ret + ws * f_(rest & ~(1LL << i | cnei))) % MOD;
31         }
32         ret = (ret + w[i] * f_(rest ^ (1LL << i))) % MOD;
33         //printf("<%lld: %lld>\n", rest, ret);
34         mp[rest] = ret;
35     }
36     return mp[rest];
37 }
38
39 bool dfvis[N];
40 ll dfs_(int c) {
41     ll ret = 1LL << c;
42     for(int i = 0; i < n; ++i)
43         if(!dfvis[i] && tbl[c][i]) {
44             dfvis[i] = true;
45             ret |= 1L << i;
46             ret |= dfs_(i);
47         }
48     return ret;
49 }
50
51 int main() {
52     int T;
53     scanf("%d", &T);
54     for(int it = 1; it <= T; ++it) {
55         memset(tbl, 0, sizeof tbl);
56         memset(dfvis, 0, sizeof dfvis);
57
58         int m;
59         scanf("%d%d%lld", &n, &m, &MOD);
60         for(int i = 0; i < n; ++i)
61             scanf("%lld", w + i);
62         for(int i = 0; i < m; ++i) {
63             int a, b;
64             scanf("%d%d", &a, &b);
65             tbl[--a][--b] = true;
66             tbl[b][a] = true;
67         }
68
69         for(int i = 0; i < n; ++i) {
70             nei[i] = 0;
71             for(int j = 0; j < n; ++j)
72                 if(i != j && tbl[i][j])
73                     nei[i] |= 1LL << j;
74         }
75
76         ll ans = 1;
77         for(int i = 0; i < n; ++i)
78             if(!dfvis[i]) {
79                 ll blk = dfs_(i);
80                 mp.clear();
81                 ans = ans * f_(blk) % MOD;
82             }
83         printf("Case #%d: %lld\n", it, ans);
84     }
85     return 0;
86 }

L 待补

M 待补

原文地址：https://www.cnblogs.com/MyGirlfriends/p/9326893.html

时间： 2024-10-09 10:43:39

2018 ACM-ICPC 宁夏邀请赛的相关文章

2017 ACM/ICPC 广西邀请赛题解

题目链接 Problems HDOJ上的题目顺序可能和现场比赛的题目顺序不一样, 我这里的是按照HDOJ的题目顺序来写的. Problem 1001 签到 #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) typedef long long LL; LL n, f[31]; int ans; int main(){ f[1] = 1LL; fo

2013 ACM/ICPC 杭州邀请赛(8.5)

A.dp 但是m次数太多,开不出那么大的数组,所以要用算完的值替代新的值,这样来回循环,滚动数组,我用的方法比较麻烦,看到有%2的,也有^1的我写的是每一次从0到n循环一边,能到的话概率相加还有一种思路是当前的概率至与上一次的概率有关,每一次都是dp[i][j] = dp[i-1][ (j-x+n)%n ] + dp[i][ (j+x)%n ] I题就是硬暴,开始想用dfs,发现不对,后来改bfs,都是想麻烦了,四重循环50的4次方,不会超时看到有降了复杂度的方法,开始把H,P堆到vec

2018 ACM/ICPC 南京 I题 Magic Potion

题解:最大流板题:增加两个源点,一个汇点.第一个源点到第二个源点连边,权为K,然后第一个源点再连其他点(英雄点)边权各为1,然后英雄和怪物之间按照所给连边(边权为1). 每个怪物连终点,边权为1: 参考代码: 1 #include<bits/stdc++.h> 2 using namespace std; 3 #define INF 0x3f3f3f3f 4 const int maxn = 2100; 5 int n,m,k,s,t,u,v,w,num,num1; 6 struct Edge

【转】ACM/ICPC生涯总结暨退役宣言—alpc55

转自:http://hi.baidu.com/accplaystation/item/ca4c2ec565fa0b7fced4f811 ACM/ICPC生涯总结暨退役宣言—alpc55 前言早就该写这篇文章了,但是也很不想去写.毕竟是为之奋斗了两年的目标,不是说舍得就舍得的.然而,自己毕竟是到了该退的时候了,与其扭扭捏捏,不如挥一挥衣袖,尚落得一份潇洒.回首这两年来,有很多是需要总结的.在这里不分巨细的记录下来,或许有点像流水账,但是更多的,是一份对过去的难忘. 童年我的ACM/ICPC的生

2016 ACM/ICPC亚洲区青岛站现场赛（部分题解）

摘要本文主要列举并求解了2016 ACM/ICPC亚洲区青岛站现场赛的部分真题,着重介绍了各个题目的解题思路,结合详细的AC代码,意在熟悉青岛赛区的出题策略,以备战2018青岛站现场赛. HDU 5984 Pocky 题意给出一根棒子(可以吃的)的长度x和切割过程中不能小于的长度d,每次随机的选取一个位置切开,吃掉左边的一半,对右边的棒子同样操作,直至剩余的长度不大于d时停止.现在给出x和d,问切割次数的数学期望是多少. 解题思路当看到第二个样例2 1时,结果是1.693147,联想到ln

《ACM/ICPC 算法训练教程》读书笔记一之数据结构（堆）

书籍简评:<ACM/ICPC 算法训练教程>这本书是余立功主编的,代码来自南京理工大学ACM集训队代码库,所以小编看过之后发现确实很实用,适合集训的时候刷题啊~~,当时是听了集训队final的意见买的,感觉还是不错滴. 相对于其他ACM书籍来说,当然如书名所言,这是一本算法训练书,有着大量的算法实战题目和代码,尽管小编还是发现了些许错误= =,有部分注释的语序习惯也有点不太合我的胃口.实战题目较多是比较水的题,但也正因此才能帮助不少新手入门,个人认为还是一本不错的算法书,当然自学还是需要下不少

2014 ACM/ICPC Asia Regional Guangzhou Online Wang Xifeng's Little Plot HDU5024

一道好枚举+模拟题目.转换思维视角这道题是我做的,规模不大N<=100,以为正常DFS搜索,于是傻乎乎的写了起来.各种条件限制模拟过程但仔细一分析发现对每个点进行全部八个方向的遍历100X100X100^8 .100X100个点,每个点在走的时候8中选择,TLE 于是改为另一个角度: 以符合要求的点为拐弯点,朝两个垂直的方向走,求出最远的距离.这样只要对每个点各个方向的长度知道,组合一下对应的就OK. 避免了每个点深搜. PS:搜索的时候x,y写反了,导致构图出现问题,以后用[dy][dx]

HDU 5014 Number Sequence(2014 ACM/ICPC Asia Regional Xi'an Online) 题解

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5014 Number Sequence Problem Description There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules: ● ai ∈ [0,n] ● ai ≠ aj( i ≠ j ) For sequence a and sequ

hdu6206 Apple 2017 ACM/ICPC Asia Regional Qingdao Online

地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=6206 题目: Apple Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 530 Accepted Submission(s): 172 Problem Description Apple is Taotao's favouri

ACM ICPC 2008–2009 NEERC MSC A, B, C, G, L

这套题是我上周日, 就是前天打得一场组队赛, 题目不太好找题目链接:http://codeforces.com/gym/100861 在virtual judge 上也可以提交哦! A ACM ICPC Rules: 题目大意: 有很多所高校参加预选赛, 并在预选赛取得了排名, 但是对于每所学校, 除了MSU有4个名额之外其他大学只有两个名额( 也就是说, 只有每个大学的前2名进决赛(MSU前四名)&& 最多有10个队伍进入决赛), 高中队伍不能进入决赛. 给出预选赛的排名, 输出可以进