# 7-19题解

A - An Easy Physics Problem

没有计算几何关于圆的模板，都是在场写的，赛场上wa了很多法，
因为考虑的不是很周全．刚开始就因为只注意了圆心到两点的距离需要超过半径，
但忽略了一个点直接到另一个点．关于一个点射向另一个点要注意方向，我选取了额外一个点，
制造了一个角．判断角是否相等来判断一个点射向另外一个点．

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <cmath>
  4 using namespace std;
  5 const double eps = 1e-8;
  6 struct Point {
  7     double x, y;
  8     Point(){};
  9     Point(double a, double b) {
 10         x = a, y = b;
 11     }
 12 };
 13
 14 struct Segment {
 15     Point a, b;
 16     Segment(){};
 17     Segment(Point x, Point y){
 18         a = x, b = y;
 19     }
 20 };
 21
 22 double dis(Point a, Point b){
 23     return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
 24 }
 25 //叉积判断点在线段两端．跟eps或者-eps比较会有不同的结果，注意．
 26 double Multi(Point p0, Point p1, Point p2) {
 27     return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
 28 }
 29 //点积
 30 double dMulti(Point p0, Point p1, Point p2) {
 31     return (p1.x - p0.x) * (p2.x - p0.x) + (p1.y - p0.y) * (p2.y - p0.y);
 32 }
 33 //算角度，利用点积算的acos角度，也就是说方向是需要额外判断的
 34 double ang(Point p0, Point p1, Point p2) {
 35     //printf("%lf",dMulti(p0, p1, p2)/dis(p0, p1)/dis(p0, p2));
 36     return acos(dMulti(p0, p1, p2) / dis(p0, p1) / dis(p0, p2));
 37 }
 38 //点在矩形里面．
 39 bool inRect(Point a, Point x, Point y) {
 40     if (max(x.x, y.x) >= a.x && a.x >= min(x.x, y.x)
 41      && max(x.y, y.y) >= a.y && a.y >= min(x.y, y.y)
 42     ) return true;
 43     return false;
 44 }
 45 //处理近似０和eps的关系的
 46 int dblcmp(double m) {
 47     if (fabs(m) <eps) return 0;
 48     return m > 0 ? 1 : -1;
 49 }
 50 double vx,vy;
 51
 52 double distoSegment(Point a, Segment b) {
 53     double ans = Multi(b.a, b.b, a);
 54     return abs(ans / dis(b.a, b.b));
 55 }
 56
 57 Point GetPoint(Point o, int r, Point a) {
 58     if (abs(vy - a.y) > eps) {
 59         double da = (1+vy * vy / vx / vx);
 60         double db = -(2 * o.x - 2 * vy / vx * (a.y - o.y) + (vy * vy / vx / vx * 2 * a.x));
 61         double dc = (o.x * o.x - 2 * vy / vx * a.x * (a.y - o.y)
 62         + (a.y - o.y) * (a.y - o.y) - r * r + vy * vy / vx / vx * a.x * a.x);
 63         double delat = sqrt(db * db - 4 * da * dc);
 64         //printf("%lf %lf %lf %lf",da,db,dc,delat);
 65         double x = (-db + delat) / 2 / da;
 66         double y = vy / vx * (x - a.x) + a.y;
 67         Point huchi = Point(x, y);
 68         double tx = (-db - delat) / 2 / da;
 69         double ty = vy / vx * (tx - a.x) + a.y;
 70         //printf("%lf %lf %lf %lf\n",x,y,tx,ty);
 71         Point chihu = Point(tx, ty);
 72         if (dis(huchi, a) < dis(chihu, a)) {
 73             return huchi;
 74         } else {
 75             return chihu;
 76         }
 77     } else {
 78         double x = o.x + sqrt(r * r - (a.y - o.y) * (a.y - o.y));
 79         double y = a.y;
 80         Point huchi = Point(x,y);
 81         double tx = -o.x + sqrt(r * r - (a.y - o.y) * (a.y - o.y));
 82         double ty = a.y;
 83         Point chihu = Point(y, x);
 84         if (dis(huchi, a) < dis(chihu, a)) {
 85             return huchi;
 86         } else {
 87             return chihu;
 88         }
 89     }
 90 }
 91
 92 int main() {
 93     int T, t = 0; scanf("%d", &T);
 94     while (t++ < T) {
 95         double o_x, o_y;
 96         scanf("%lf%lf", &o_x, &o_y);
 97         Point O = Point(o_x, o_y);
 98         double r; scanf("%lf", &r);
 99         double a_x, a_y, b_x, b_y;
100         scanf("%lf%lf%lf%lf", &a_x, &a_y, &vx, &vy);
101         scanf("%lf%lf", &b_x, &b_y);
102         Point A = Point(a_x, a_y);
103         Point B = Point(b_x, b_y);
104         bool final_a = 0;
105         if (r <= distoSegment(O, Segment(A,B))) {
106             //cout <<"huchi"<<endl;
107             if (abs(vy * (A.x - B.x) - vx * (A.y - B.y)) < eps) {
108                 Point tt = Point(1001, A.y);
109                 if (abs(ang(A, tt, B) - ang(Point(B.x - vx, B.y - vy), Point(1001, B.y - vy), B)) < eps) {
110                     final_a = 1;
111                 }
112             } else {
113                 Point temp = Point(A.x - vx, A.y - vy);
114                 if (r > distoSegment(O, Segment(A, temp))) {
115                     Point jiaodian = GetPoint(O, r, A);
116                     //printf("%lf:%lf",jiaodian.x,jiaodian.y);
117                     double ang1 = ang(jiaodian, O, A);
118                     double ang2 = ang(jiaodian, B, O);
119                     //printf("%lf %lf\n",ang1,ang2);
120                     if (abs(ang1 - ang2) < eps) final_a = 1;
121                 }
122             }
123         } else if (distoSegment(O, Segment(A,B)) < eps) {
124             Point tt = Point(1001, 0);
125             if (abs(ang(O, A, tt) - ang(O, Point(O.x - vx, O.y - vy), tt)) < eps) {
126                 final_a = 1;
127             }
128         } else {
129             //cout << endl<<endl;
130             if (abs(vy * (A.x - B.x) - vx * (A.y - B.y)) < eps) {
131                     Point jiaodian = GetPoint(O, r, A);
132
133                     if (dis(jiaodian, A) > dis(B,A)) {
134
135                     Point tt = Point(1001, A.y);
136                 if (abs(ang(A, tt, B) - ang(Point(B.x - vx, B.y - vy), Point(1001, B.y - vy), B)) < eps) {
137                     final_a = 1;
138                 }
139             }
140             }
141         }
142         if (final_a) printf("Case #%d: Yes\n",t);
143         else printf("Case #%d: No\n",t);
144     }
145     return 0;
146 }

B - Binary Tree
发现最左边的那条链,把除最后一个点都置为负
比如 -1 -2 -4 8 此时和为1
依次把 -1 取正，-2取证，可以得到 3、5、7、9...所有的奇数
把8取右子节点，可以得到2、4、6...所有偶数
根据n的二进制为判断是否要取正，注意奇偶就行

 1 #include <cstdio>
 2 #include <cstring>
 3 using namespace std;
 4 typedef long long LL;
 5
 6 LL n, a[65];
 7 int b[65], k;
 8
 9 LL qpow(LL a, int n) {
10     LL res = 1;
11     while (n) {
12         if (n & 1) res *= a;
13         a *= a;
14         n >>= 1;
15     }
16     return res;
17 }
18
19 int main() {
20     int T; scanf("%d", &T);
21     for (int cas = 1; cas <= T; ++cas) {
22         scanf("%lld%d", &n, &k);
23         memset(b, -1, sizeof(b));
24         for (int i = 1; i <= k; ++i) a[i] = qpow(2, i - 1);
25         if (!(n & 1)) a[k]++, n -= 2;
26         b[k] = 1;
27         int now = 1;
28         while (n >>= 1) {
29             if (n & 1) b[now] = 1;
30             now++;
31         }
32         printf("Case #%d:\n", cas);
33         for (int i = 1; i <= k; ++i)
34             printf("%lld %c\n", a[i], b[i] == 1 ? ‘+‘ : ‘-‘);
35     }
36     return 0;
37 }

F - Friendship of Frog

签到题．手动打表

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4 map<char,int> shit;
 5 char fuck[1005];
 6 int main() {
 7     int n; scanf("%d",&n);
 8     for(int ks = 1; ks <= n; ks++) {
 9         shit.clear();
10         scanf("%s",fuck);
11         int len = int(strlen(fuck));
12         int ans = 1e9;
13         for (int i = 0; i < len; i++) {
14             if(shit.count(fuck[i]))
15                 ans = min(i - shit[fuck[i]], ans);
16             shit[fuck[i]] = i;
17         }
18         if (ans == 1e9) ans = -1;
19         printf("Case #%d: %d\n",ks,ans);
20     }
21 }

K - Kingdom of Black and White

贪心的跑，先考虑一个长度的，这样可以连通两个块．在跑一边关于长度差的，就是a > b长度的2 *(abs(a-b)+1)．跑一遍取最大值

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <vector>
 5 #include <cstring>
 6 #include <cmath>
 7 using namespace std;
 8 const int N = 1e5 + 5;
 9 typedef long long LL;
10
11 char s[N];
12 vector <int> vec;
13
14 int main() {
15     int t; scanf("%d", &t);
16     for (int cas = 1; cas <= t; ++cas) {
17         vec.clear();
18         scanf("%s", s);
19         int n = strlen(s);
20         int cnt = 1;
21         for (int i = 1; i < n; ++i)
22             if (s[i] != s[i - 1]) vec.push_back(cnt), cnt = 1;
23             else cnt++;
24         vec.push_back(cnt);
25
26         // for (int i = 0; i < (int) vec.size(); ++i)
27         //     printf("vec[%d]=%d  ", i + 1, vec[i]);
28         // cout << endl;
29
30
31         LL sum = 0;
32         for (int i = 0; i < (int) vec.size(); ++i) sum += 1LL * vec[i] * vec[i];
33         LL ans = sum, tmp;
34
35         // printf("sum = %lld\n", sum);
36
37         int ed = (int) vec.size() - 1;
38         LL maxn2 = 0;
39         for (int i = 0; i < (int) vec.size(); ++i)
40             if (vec[i] == 1) {
41                 if (i != 0 && i != ed) {
42                     LL a = 1LL * vec[i - 1];
43                     LL b = 1LL * vec[i + 1];
44                     tmp = (a + 1 + b);
45                     tmp = tmp * tmp - (1LL * a * a + b * b + 1);
46                     // printf("a = %lld, b = %lld, tmp = %lld\n", a, b, tmp);
47                     maxn2 = max(maxn2, tmp);
48                 }
49             }
50         LL maxn = 0;
51         for (int i = 1; i < (int) vec.size(); ++i) {
52             maxn = max(maxn, 1LL * 2 * abs(vec[i] - vec[i - 1]) + 2);
53             //printf("maxn = %lld, ", maxn);
54         }
55         ans = max(ans + maxn, ans + maxn2);
56         printf("Case #%d: %lld\n", cas, ans);
57     }
58     return 0;
59 }

L - LCM Walk

关于跳．(x,y)跳到(x,y+lcm(x,y))或者(x+lcm(x,y),y),首先做处理，让gcd(x,y)＝１,然后(x<y)时考虑(x,y)一定是从(x,p)跳到的，

互质所以有(xp+p)=y,就判断一下y%(x+1),这里我把跳０步忽略了，放在最后＋１了．

 1 #include <cstdio>
 2 #include <iostream>
 3 #define ll long long
 4 using namespace std;
 5
 6 ll dfs(int x, int y) {
 7     if (x > y) swap(x, y);
 8     if (x<=1 && y <= 1) return 0;
 9     if (y % (x + 1) != 0) return 0;
10     return 1 + dfs(x, (y / (x + 1)));
11 }
12
13 int gcd(int a, int b) {
14     return b ? gcd(b, a % b) : a;
15 }
16
17 int main() {
18     int t = 0, T; scanf("%d",&T);
19     while (t++ < T) {
20         int x,y;
21         ll ans;
22         scanf("%d%d",&x,&y);
23         int d = gcd(x, y);
24         x /= d; y /= d;
25         if (x > y) swap(x, y);
26         if (y % (x + 1) != 0) ans = 0;
27         else ans = dfs(x, y);
28         printf("Case #%d: %lld\n", t, ans + 1);
29     }
30 }