ACM-ICPC Asia Training League 暑假第一阶段第四场 ABCDH

模拟这个

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 5e6+10;
 5 int t, q, p, m, n;
 6 unsigned int sa, sb, sc;
 7 ll a[N], st[N], top;
 8 ll MAX[N];
 9 unsigned int f() {
10     sa ^= sa<<16;
11     sa ^= sa >> 5;
12     sa ^= sa << 1;
13     unsigned int t = sa;
14     sa = sb;
15     sb = sc;
16     sc ^= t^sa;
17     return sc;
18 }
19 
20 int main() {
21     scanf("%d", &t);
22     for(int ca = 1; ca <= t; ca ++){
23         top = 0;
24         scanf("%d%d%d%d%u%u%u", &n, &p, &q, &m, &sa, &sb, &sc);
25         for(int i = 1; i <= n; i ++) {
26             if(f()%(p+q) < p) {
27                 int tmp = f()%m+1;
28                 st[++top] = tmp;
29                 MAX[top] = max(st[top], MAX[top-1]); 
30                 a[i] = MAX[top];
31             } else{
32                 if(top == 0) a[i] = 0;
33                 else a[i] = MAX[--top];
34             }
35         }
36         ll ans = 0;
37         for(ll i = 1; i <= n; i ++) {
38             ans ^= i*a[i];
39         }
40         printf("Case #%d: %lld\n",ca,ans);
41     }
42     return 0;
43 }

 

B Rolling The Polygon

Bahiyyah has a convex polygon with nn vertices P_0, P_1, \cdots, P_{n-1}P0​,P1​,⋯,Pn−1​ in the counterclockwise order.Two vertices with consecutive indexes are adjacent, and besides, P_0P0​ and P_{n-1}Pn−1​ are adjacent.She also assigns a point QQ inside the polygon which may appear on the border.

Now, Bahiyyah decides to roll the polygon along a straight line and calculate the length of the trajectory (or track) of point QQ.

To help clarify, we suppose P_n = P_0, P_{n+1} = P_1Pn​=P0​,Pn+1​=P1​ and assume the edge between P_0P0​ and P_1P1​ is lying on the line at first.At that point when the edge between P_{i-1}Pi−1​ and P_iPi​ lies on the line, Bahiyyah rolls the polygon forward rotating the polygon along the vertex P_iPi​ until the next edge (which is between P_iPi​ and P_{i+1}Pi+1​) meets the line.She will stop the rolling when the edge between P_nPn​ and P_{n+1}Pn+1​ (which is same as the edge between P_0P0​ and P_1P1​) meets the line again.

Input Format

The input contains several test cases, and the first line is a positive integer TT indicating the number of test cases which is up to 5050.

For each test case, the first line contains an integer n~(3\le n \le 50)n (3≤n≤50) indicating the number of vertices of the given convex polygon.Following nn lines describe vertices of the polygon in the counterclockwise order.The ii-th line of them contains two integers x_{i-1}xi−1​ and y_{i-1}yi−1​, which are the coordinates of point P_{i-1}Pi−1​.The last line contains two integers x_QxQ​ and y_QyQ​, which are the coordinates of point QQ.

We guarantee that all coordinates are in the range of -10^3−103 to 10^3103, and point QQ is located inside the polygon or lies on its border.

Output Format

For each test case, output a line containing Case \#x: y, where xx is the test case number starting from 11, and yyis the length of the trajectory of the point QQ rounded to 33 places.We guarantee that 44-th place after the decimal point in the precise answer would not be 44 or 55.

Hint

The following figure is the the trajectory of the point QQ in the first sample test case.

样例输入

4
4
0 0
2 0
2 2
0 2
1 1
3
0 0
2 1
1 2
1 1
5
0 0
1 0
2 2
1 3
-1 2
0 0
6
0 0
3 0
4 1
2 2
1 2
-1 1
1 0

样例输出

Case #1: 8.886
Case #2: 7.318
Case #3: 12.102
Case #4: 14.537

给定N边形和一个点，转一圈后点走的轨迹路径是多少。
n个弧，半径为QPi
 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 1e5+10;
 5 const double PI = acos(-1);
 6 struct Point{
 7     double x, y;
 8 }p[55];
 9 
10 double dis(Point a, Point b) {
11     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
12 }
13 
14 double ang(Point a, Point b, Point c) {
15     double x1 = dis(a,b);
16     double x2 = dis(c,b);
17     double x3 = dis(a,c);
18     return acos((x1*x1+x2*x2-x3*x3)/(2*x1*x2));
19 }
20 int main() {
21     int t, n;
22     cin >>t;
23     for(int ca = 1; ca <= t; ca ++) {
24         cin >> n;
25         for(int i = 0; i <= n; i ++) {
26             cin >> p[i].x >> p[i].y;
27         }
28         double ans = 0;
29         for(int i = 0; i < n; i ++) {
30             ans += dis(p[n],p[i])*(PI-ang(p[(i-1+n)%n], p[i], p[(i+1+n)%n]));
31         }
32         printf("Case #%d: %.3lf\n",ca,ans);
33     }
34     return 0;
35 }

 

C 签到题

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 55;
 5 char s1[N], s2[N], s3[N];
 6 int main() {
 7     int t;
 8     cin >> t;
 9     for(int i = 1; i <= t; i ++) {
10         int n, m;
11         memset(s1, 0, sizeof(s1));
12         memset(s2, 0, sizeof(s2));
13         memset(s3, 0, sizeof(s3));
14         cin >> n >> m;
15         cin >> s1 >> s2 >> s3;
16         int ans = s1[0]-s2[0];
17         for(int j = 0; s3[j]; j ++) {
18             s3[j] += ans;
19             if(s3[j] < 'A') s3[j] += 26;
20             else if(s3[j] > 'Z') s3[j] -= 26;
21         }
22         printf("Case #%d: %s\n",i,s3);
23     }
24     return 0;
25 }

 

 

D 找规律。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 1e5+10;
 5 
 6 int main() {
 7     int t, n, m;
 8     cin >> t;
 9     for(int ca = 1; ca <= t; ca ++) {
10         cin >> n >> m;
11         double ans1 = 0.5, ans2 = 1.0*(m+1)/(2*m);
12         if(n == 1) ans1 = 1.0;
13         if(m == 1) ans2 = 1.0;
14         printf("Case #%d: %.6lf %.6lf\n",ca,ans1,ans2);
15     }
16     return 0;
17 }

 

H 题

按血量与攻击次数的比值从大到小选

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 1e5+10;
 5 int t, n, x;
 6 int cnt[1010];
 7 struct Nod{
 8     int num, a;
 9 
10 }e[N];
11 
12 bool cmp(Nod a, Nod b) {
13     if(1.0*a.a/a.num !=  1.0*b.a/b.num)return 1.0*a.a/a.num > 1.0*b.a/b.num;
14     else return a.a > b.a;
15 }
16 
17 int main() {
18     for(int i = 1; i <= 1000; i ++) {
19         cnt[i] = cnt[i-1] + i;
20     }
21     cin >> t;
22     for(int ca = 1; ca <= t; ca ++) {
23         cin >> n;
24         ll ans = 0;
25         for(int i = 1; i <= n; i ++) {
26             scanf("%d%d", &x, &e[i].a);
27             ans += e[i].a;
28             e[i].num = lower_bound(cnt+1,cnt+1000,x) - cnt;
29         }
30         sort(e+1,e+1+n,cmp);
31         ll cnt = 0;
32         for(int i = 1; i <= n; i ++) {
33             // printf("%d %d\n",e[i].num,e[i].a);
34             cnt += e[i].num*ans;
35             ans -= e[i].a;
36         }
37         printf("Case #%d: %lld\n",ca,cnt);
38 
39     }
40     return 0;
41 }

 

posted @ 2018-07-19 20:53  starry_sky  阅读(504)  评论(0)  编辑  收藏  举报