首页 > 代码库 > 河南工业大学2017校赛题解

河南工业大学2017校赛题解

问题 A: 饶学妹的比赛

题意:

给你一场比赛每人提交的记录,计算最后的排名

题解:

模拟+排序

代码:

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <queue>
 5 #include <stack>
 6 #include <cstdio>
 7 #include <string>
 8 #include <vector>
 9 #include <cstdlib>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <functional>
15 using namespace std;
16 #define rep(i,a,n) for (int i=a;i<n;i++)
17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
18 #define all(x) (x).begin(),(x).end()
19 #define pb push_back
20 #define mp make_pair
21 #define lson l,m,rt<<1  
22 #define rson m+1,r,rt<<1|1 
23 typedef long long ll;
24 typedef vector<int> VI;
25 typedef pair<int, int> PII;
26 const ll MOD = 1e9 + 7;
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 1010;
29 // head
30 
31 struct Node {
32     int id;
33     string name;
34     int a[20] = { 0 };
35     int b[20] = { 0 };
36     int num = 0;
37     int time = 0;
38     const bool operator<(const Node &t) const {
39         if (num != t.num) return num > t.num;
40         else if (time != t.time) return time < t.time;
41         else return id < t.id;
42     }
43 }p[MAXN];
44 
45 int main() {
46     int n, m, k;
47     cin >> n >> m >> k;
48     rep(i, 1, n + 1) {
49         p[i].id = i;
50         cin >> p[i].name;
51     }
52     while (m--) {
53         int timei, idi, pidi;
54         string s;
55         cin >> timei >> idi >> pidi >> s;
56         if (s == "AC") {
57             if (!p[idi].b[pidi]) {
58                 p[idi].b[pidi] = 1;
59                 p[idi].a[pidi] += timei;
60             }
61         }
62         else {
63             if (!p[idi].b[pidi])
64                 p[idi].a[pidi] += 20;
65         }
66     }
67     rep(i, 1, n + 1) {
68         rep(j, 1, k + 1) if (p[i].b[j]) {
69             p[i].num++;
70             p[i].time += p[i].a[j];
71         }
72     }
73     sort(p + 1, p + 1 + n);
74     rep(i, 1, n + 1) 
75         cout << p[i].name <<   << p[i].num <<   << p[i].time << endl;
76     return 0;
77 }

问题 B: 地狱飞龙

题意:

有一头龙从(0,0)往y轴正方向跑,有一人从(x,0)往x轴负方向跑,这头龙每秒对人造成k/d2的伤害(d为距离)

求这个人必须有多少HP,才永远不会死

题解:

就是一道数值积分,直接套模板就行了,但是:

这道题数据有误!

这道题数据有误!!

这道题数据有误!!!

你想啊,如果积分值为2.354,那不得有2.36的血才行吗,所以是不能四舍五入的,不然你就挂了

但是答案是四舍五入啊啊啊,

我们比赛时wa了50+都没想到是出题人sb啊

“AC”代码如下

代码:

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <queue>
 5 #include <stack>
 6 #include <cstdio>
 7 #include <string>
 8 #include <vector>
 9 #include <cstdlib>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <functional>
15 using namespace std;
16 #define rep(i,a,n) for (int i=a;i<n;i++)
17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
18 #define all(x) (x).begin(),(x).end()
19 #define pb push_back
20 #define mp make_pair
21 #define lson l,m,rt<<1  
22 #define rson m+1,r,rt<<1|1 
23 typedef long long ll;
24 typedef vector<int> VI;
25 typedef pair<int, int> PII;
26 const ll MOD = 1e9 + 7;
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 2010;
29 // head
30  
31 const double EPS = 1e-6;
32 double v1, v2, x, k;
33  
34 double f(double t){
35     double d = v1*t*v1*t + (x - v2*t)*(x - v2*t);
36     return k / d;
37 }
38  
39 inline double getAppr(double l, double r) {
40     return (f(l) + 4.0*f((l + r) / 2) + f(r)) * (r - l) / 6.0;
41 }
42  
43 double Simpson(double l, double r) {
44     double sum = getAppr(l, r);
45     double mid = (l + r) / 2;
46     double suml = getAppr(l, mid);
47     double sumr = getAppr(mid, r);
48     return (fabs(sum - suml - sumr) < EPS) ? sum : Simpson(l, mid) + Simpson(mid, r);
49 }
50  
51 int main() {
52     int T;
53     cin >> T;
54     while (T--) {
55         scanf("%lf%lf%lf%lf", &v1, &v2, &x, &k);
56         double ans = Simpson(0, 20000);
57         printf("%.2lf\n", ans);
58     }
59     return 0;
60 }

问题 C: 魔法宝石

题意:

创造第i个宝石需要a[i]膜法,或者可以两个合成一个,不需要膜法,问你合成每个的最小花费

题解:

反正就是最短路吧,xjb改一下dijkstra就好了

代码:

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <queue>
 5 #include <stack>
 6 #include <cstdio>
 7 #include <string>
 8 #include <vector>
 9 #include <cstdlib>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <functional>
15 using namespace std;
16 #define rep(i,a,n) for (int i=a;i<n;i++)
17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
18 #define all(x) (x).begin(),(x).end()
19 #define pb push_back
20 #define mp make_pair
21 #define lson l,m,rt<<1  
22 #define rson m+1,r,rt<<1|1 
23 typedef long long ll;
24 typedef vector<int> VI;
25 typedef pair<int, int> PII;
26 const ll MOD = 1e9 + 7;
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 1e5 + 7;
29 // head
30 
31 vector<PII> C[MAXN];
32 int d[MAXN];
33 priority_queue<PII, vector<PII>, greater<PII> > que;
34 
35 void dijkstra() {
36     while (!que.empty()) {
37         PII p = que.top(); que.pop();
38         int v = p.second;
39         if (d[v] < p.first) continue;
40         d[v] = p.first;
41         rep(i, 0, C[v].size()) {
42             PII p = C[v][i];
43             int a = v;
44             int b = p.first;
45             int c = p.second;
46             if (d[c] > d[a] + d[b]) {
47                 d[c] = d[a] + d[b];
48                 que.push(mp(d[a] + d[b], c));
49             }
50         }
51     }
52 }
53 
54 int main() {
55     int T;
56     cin >> T;
57     while (T--) {
58         rep(i, 0, MAXN) C[i].clear();
59         int n, m;
60         cin >> n >> m;
61         rep(i, 1, n + 1) {
62             int a;
63             scanf("%d", &a);
64             d[i] = a;
65         }
66         while (m--) {
67             int a, b, c;
68             scanf("%d%d%d", &a, &b, &c);
69             que.push(mp(d[a]+d[b], c));
70             C[a].pb(mp(b, c));
71             C[b].pb(mp(a, c));
72         }
73         dijkstra();
74         rep(i, 1, n + 1) {
75             printf("%d", d[i]);
76             if (i == n) printf("\n");
77             else printf(" ");
78         }
79     }
80     return 0;
81 }

问题 D: rqy的键盘

题意:

为你给个字母按键的左右两边的按键是什么,没有就输出No letter.

题解:

模拟一下就好了

代码:

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <queue>
 5 #include <stack>
 6 #include <cstdio>
 7 #include <string>
 8 #include <vector>
 9 #include <cstdlib>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <functional>
15 using namespace std;
16 #define rep(i,a,n) for (int i=a;i<n;i++)
17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
18 #define all(x) (x).begin(),(x).end()
19 #define pb push_back
20 #define mp make_pair
21 #define lson l,m,rt<<1  
22 #define rson m+1,r,rt<<1|1 
23 typedef long long ll;
24 typedef vector<int> VI;
25 typedef pair<int, int> PII;
26 const ll MOD = 1e9 + 7;
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 2010;
29 // head
30 
31 string s = " QWERTYUIOPASDFGHJKLZXCVBNM ";
32 
33 struct Node {
34     char c, l, r;
35 }node[30];
36 
37 int main() {
38     rep(i, 1, 27) {
39         node[i].c = s[i];
40         node[i].l = s[i - 1];
41         node[i].r = s[i + 1];
42     }
43     node[1].l = 0;
44     node[11].l = 0;
45     node[20].l = 0;
46     node[10].r = 0;
47     node[19].r = 0;
48     node[26].r = 0;
49     int T;
50     cin >> T;
51     while (T--) {
52         char x;
53         string dir;
54         cin >> x >> dir;
55         int pos = s.find(x, 0);
56         char ans;
57         if (dir == "Left") ans = node[pos].l;
58         else ans = node[pos].r;
59         if (ans == 0) cout << "No letter." << endl;
60         else cout << ans << endl;
61     }
62     return 0;
63 }

问题 F: Hmz 的女装

题意:

你有k种flower,你要编一个长度n的花环,每次要求你第l和第r个位置的必须一样,问你有多少种编法

题解:

这道题可以退化成 你有k种flower,你要编一个长度n的花环有多少种方法,然后就是找规律递推了,注意需要用long long

dp[i] = ((k - 1)*dp[i - 2] + (k - 2)*dp[i - 1]) % MOD;

代码:

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <queue>
 5 #include <stack>
 6 #include <cstdio>
 7 #include <string>
 8 #include <vector>
 9 #include <cstdlib>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <functional>
15 using namespace std;
16 #define rep(i,a,n) for (int i=a;i<n;i++)
17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
18 #define all(x) (x).begin(),(x).end()
19 #define pb push_back
20 #define mp make_pair
21 #define lson l,m,rt<<1  
22 #define rson m+1,r,rt<<1|1 
23 typedef long long ll;
24 typedef vector<int> VI;
25 typedef pair<int, int> PII;
26 const ll MOD = 1e9 + 7;
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 1e5 + 7;
29 // head
30 
31 ll dp[MAXN];
32 
33 int main() {
34     ll n, m, k;
35     while (cin >> n >> m >> k) {
36         dp[2] = k - 1;
37         dp[3] = (k - 1)*(k - 2);
38         rep(i, 4, n) dp[i] = ((k - 1)*dp[i - 2] + (k - 2)*dp[i - 1]) % MOD;
39         while (m--) {
40             int l, r;
41             scanf("%d%d", &l, &r);
42             if (l > r) swap(l, r);
43             ll ans = k;
44             ans = (ans*dp[r - l]) % MOD;
45             ans = (ans*dp[n - r + l]) % MOD;
46             printf("%lld\n", ans);
47         }
48     }
49     return 0;
50 }

 

问题 G: 最大子段和

题意:

给你一个字符串,找出奇数长度的最大子段和

题解:

两次循环,分别从第1和第2个循环找一遍就好了

代码:

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <queue>
 5 #include <stack>
 6 #include <cstdio>
 7 #include <string>
 8 #include <vector>
 9 #include <cstdlib>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <functional>
15 using namespace std;
16 #define rep(i,a,n) for (int i=a;i<n;i++)
17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
18 #define all(x) (x).begin(),(x).end()
19 #define pb push_back
20 #define mp make_pair
21 #define lson l,m,rt<<1  
22 #define rson m+1,r,rt<<1|1 
23 typedef long long ll;
24 typedef vector<int> VI;
25 typedef pair<int, int> PII;
26 const ll MOD = 1e9 + 7;
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 1e5 + 7;
29 // head
30 
31 int a[MAXN];
32 
33 int main() {
34     int T;
35     cin >> T;
36     while (T--) {
37         int n;
38         cin >> n;
39         int n1 = 0, n2 = 0;
40         rep(i, 1, n + 1) scanf("%d", a + i);
41         int sum = a[1], b = a[1];
42         for (int i = 2; i < n; i += 2) {
43             b += a[i] + a[i + 1];
44             if (b < a[i + 1]) b = a[i + 1];
45             if (sum < b) sum = b;
46         }
47         int ans = sum;
48         sum = a[2], b = a[2];
49         for (int i = 3; i < n; i += 2) {
50             b += a[i] + a[i + 1];
51             if (b < a[i + 1]) b = a[i + 1];
52             if (sum < b) sum = b;
53         }
54         ans = max(ans, sum);
55         cout << ans << endl;
56     }
57     return 0;
58 }

问题 H: ch追妹

题意:

给你一个图,你在a点,你的女神在b点,你要到你的女神那里,但是你每走一步,你的情敌就会砍断一条路

而且你的情敌还非常聪明,问你能不能到你的女神那里

题解:

这道题其实就是问你a和b之间有没有一条通路,如果没有就肯定到达不了

比如说你刚到b前一个点,差一步就到了,然后砍了

然后你绕远走,又差一步到了,又砍了。。

除非你直接走到了

代码:

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <queue>
 5 #include <stack>
 6 #include <cstdio>
 7 #include <string>
 8 #include <vector>
 9 #include <cstdlib>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <functional>
15 using namespace std;
16 #define rep(i,a,n) for (int i=a;i<n;i++)
17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
18 #define all(x) (x).begin(),(x).end()
19 #define pb push_back
20 #define mp make_pair
21 #define lson l,m,rt<<1  
22 #define rson m+1,r,rt<<1|1 
23 typedef long long ll;
24 typedef vector<int> VI;
25 typedef pair<int, int> PII;
26 const ll MOD = 1e9 + 7;
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 2010;
29 // head
30 
31 int main() {
32     int T;
33     cin >> T;
34     while (T--) {
35         int n, m, a, b;
36         cin >> n >> m >> a >> b;
37         string ans = "chsad";
38         while (m--) {
39             int x, y;
40             cin >> x >> y;
41             if (x == a && y == b || x == b && y == a) ans = "chhappy";
42         }
43         cout << ans << endl;
44     }
45     return 0;
46 }

问题 I: 小天使改名

题意:

给你两个字符串a,b 问你b是不是由交换一对字母得到的

题解:

这道题有坑,就是还要考虑 如果a和b相等,但是a和b没有重复字母的情况,

这种情况也应该输出NO

代码:

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <queue>
 5 #include <stack>
 6 #include <cstdio>
 7 #include <string>
 8 #include <vector>
 9 #include <cstdlib>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <functional>
15 using namespace std;
16 #define rep(i,a,n) for (int i=a;i<n;i++)
17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
18 #define all(x) (x).begin(),(x).end()
19 #define pb push_back
20 #define mp make_pair
21 #define lson l,m,rt<<1  
22 #define rson m+1,r,rt<<1|1 
23 typedef long long ll;
24 typedef vector<int> VI;
25 typedef pair<int, int> PII;
26 const ll MOD = 1e9 + 7;
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 2010;
29 // head
30 
31 int main() {
32     int T;
33     cin >> T;
34     while (T--) {
35         string a, b;
36         cin >> a >> b;
37         VI v;
38         rep(i, 0, a.length()) if (a[i] != b[i]) v.pb(i);
39         if (v.size() == 0) {
40             string s = a;
41             sort(all(s));
42             int fg = 0;    //如果两个字符串一样 但是没有出现相同字符
43             rep(i, 1, s.length()) if (s[i] == s[i - 1]) fg = 1;
44             if (fg) cout << "YES" << endl;
45             else cout << "NO" << endl;
46         }
47         else if (v.size() == 2) {
48             if (a[v[0]] == b[v[1]] && a[v[1]] == b[v[0]]) cout << "YES" << endl;
49             else cout << "NO" << endl;
50         }
51         else cout << "NO" << endl;
52     }
53     return 0;
54 }

问题 J: 爱看电视的LsF

题意:

给你一个坏掉几个数字键的遥控器,问你最少需要按多少次键能从a台到b台

题解:

比赛时模拟了好久之后,队友说这道题可以暴力写。。那就没什么好说的了

注意0的判断

代码:

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <queue>
 5 #include <stack>
 6 #include <cstdio>
 7 #include <string>
 8 #include <vector>
 9 #include <cstdlib>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <functional>
15 using namespace std;
16 #define rep(i,a,n) for (int i=a;i<n;i++)
17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
18 #define all(x) (x).begin(),(x).end()
19 #define pb push_back
20 #define mp make_pair
21 #define lson l,m,rt<<1  
22 #define rson m+1,r,rt<<1|1 
23 typedef long long ll;
24 typedef vector<int> VI;
25 typedef pair<int, int> PII;
26 const ll MOD = 1e9 + 7;
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 2010;
29 // head
30 
31 int good[20];
32 
33 bool check(int x) {
34     if (x == 0) return good[x];
35     while (x) {
36         if (!good[x % 10]) return false;
37         x /= 10;
38     }
39     return true;
40 }
41 
42 int num_len(int x) {
43     int s = x == 0 ? 1 : 0;
44     while (x) x /= 10, s++;
45     return s;
46 }
47 
48 int main() {
49     int n, s, t;
50     while (cin >> n >> s >> t) {
51         rep(i, 0, 10) good[i] = 1;
52         rep(i, 0, n) {
53             int t;
54             cin >> t;
55             good[t] = 0;
56         }
57         int ans = abs(t - s);
58         int step = ans;
59         rep(i, 0, step) {
60             int l = t - i;
61             int r = t + i;
62             if (l >= 0 && check(l)) ans = min(ans, i + num_len(l));
63             if (check(r)) ans = min(ans, i + num_len(r));
64         }
65         cout << ans << endl;
66     }
67     return 0;
68 }

 

河南工业大学2017校赛题解