15.5.5周练
地址http://acm.hust.edu.cn/vjudge/contest/view.action?cid=77192#overview
密码：acmore
赛中出了3道，之后又写了俩。
题目意思是说有n堆石子，Alice只能从中选出连续的几堆来玩Nim博弈,现在问Alice想要获胜有多少种方法（即有多少种选择方式）。
方法是这样的，由于Nim博弈必胜的条件是所有数的抑或值不为0，证明见 点击 ，所以答案就转化为原序列有多少个区间的亦或值为0，用n*(n+1) / 2 减去这个值就可以了。
而求有多少个区间的亦或值为0，实际上就是求对于亦或值的前缀nim[i]，满足nim[i] == nim[j] 的对数，这时只要对nim数组排序就可以算了
详见代码：题解
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define INF 0x3f3f3f3f 16 #define inf (-((LL)1<<40)) 17 #define lson k<<1, L, mid 18 #define rson k<<1|1, mid+1, R 19 #define mem0(a) memset(a,0,sizeof(a)) 20 #define mem1(a) memset(a,-1,sizeof(a)) 21 #define mem(a, b) memset(a, b, sizeof(a)) 22 #define FIN freopen("in.txt", "r", stdin) 23 #define FOUT freopen("out.txt", "w", stdout) 24 #define rep(i, a, b) for(int i = a; i <= b; i ++) 25 26 template<class T> T CMP_MIN(T a, T b) { return a < b; } 27 template<class T> T CMP_MAX(T a, T b) { return a > b; } 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 110000; 36 const int MAXM = 20010; 37 const double eps = 1e-4; 38 //LL MOD = 987654321; 39 40 int n, a[MAXN], x, s, w, T; 41 42 int main() 43 { 44 while(~scanf("%d", &T)) while(T--) { 45 cin >> n >> s >> w; 46 LL ans = (LL)n * (n + 1) / 2; 47 int g = s; 48 rep (i, 1, n) { 49 x = g; 50 if( x == 0 ) { x = g = w; } 51 if( g%2 == 0 ) { g = (g/2); } 52 else { g = (g/2) ^ w; } 53 a[i] = a[i - 1] ^ x; 54 if(a[i] == 0) ans --; 55 } 56 sort(a + 1, a + n + 1); 57 int num = 1; 58 rep (i, 2, n) { 59 if(a[i] == a[i - 1]) { 60 ans -= num; 61 num++; 62 } 63 else num = 1; 64 } 65 cout << ans << endl; 66 } 67 return 0; 68 }
B： ZOJ 3593 One Person Game （扩展欧几里得）
正在搞。。。
乱搞。
1 #include <iostream> 2 #include <stdio.h> 3 #include <algorithm> 4 #include <queue> 5 #include <cmath> 6 #include <cstring> 7 #define mem0(a) memset(a, 0, sizeof(a)) 8 #define mem1(a) memset(a, -1, sizeof(a)) 9 using namespace std; 10 11 char T1[11][10] = {"Geng", "Xin", "Ren", "Gui", "Jia", "Yi", "Bing", "Ding", "Wu", "Ji"}; 12 char T2[11][10] = {"Xin", "Geng", "Ji", "Wu", "Ding", "Bing", "Yi", "Jia", "Gui", "Ren"}; 13 char D1[13][10] = {"shen", "you", "xu", "hai", "zi", "chou", "yin", "mao", "chen", "si", "wu", "wei"}; 14 char D2[13][10] = {"you", "shen", "wei", "wu", "si", "chen", "mao", "yin", "chou", "zi", "hai", "xu"}; 15 16 int N, T; 17 18 int main() 19 { 20 //freopen("in.txt", "r", stdin); 21 while(cin >> T) while(T--) { 22 cin >> N; 23 if(N > 0) printf("%s%s\n", T1[N % 10], D1[N % 12]); 24 else printf("%s%s\n", T2[(-N) % 10], D2[(-N) % 12]); 25 } 26 return 0; 27 }
D:没出
E:ZOJ 3596 Digit Number（DP+BFS）
见 http://www.cnblogs.com/gj-Acit/p/4492762.html
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define INF 0x3f3f3f3f 16 #define inf (-((LL)1<<40)) 17 #define lson k<<1, L, mid 18 #define rson k<<1|1, mid+1, R 19 #define mem0(a) memset(a,0,sizeof(a)) 20 #define mem1(a) memset(a,-1,sizeof(a)) 21 #define mem(a, b) memset(a, b, sizeof(a)) 22 #define FIN freopen("in.txt", "r", stdin) 23 #define FOUT freopen("out.txt", "w", stdout) 24 #define rep(i, a, b) for(int i = a; i <= b; i ++) 25 26 template<class T> T CMP_MIN(T a, T b) { return a < b; } 27 template<class T> T CMP_MAX(T a, T b) { return a > b; } 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 110000; 36 const int MAXM = 20010; 37 const double eps = 1e-4; 38 //LL MOD = 987654321; 39 40 int t, n, m, x; 41 struct Node { 42 bool vis; 43 char num; 44 int pre, cnt; 45 }s[(1<<10) * 1100]; 46 char ans[110000], d[110000]; 47 48 int bfs() 49 { 50 queue<int>q; 51 q.push(0); 52 while(!q.empty()) { 53 int head = q.front();q.pop(); 54 rep (i, 0, 9) { 55 if(head == 0 && i == 0) continue; 56 int mod = (head % 1000 * 10 + i) % n; 57 int tail = ((head / 1000) | (1 << i)) * 1000 + mod; 58 if(s[tail].vis) 59 continue; 60 s[tail].vis = true; 61 s[tail].num = i + '0'; 62 s[tail].pre = head; 63 s[tail].cnt = s[head].cnt + ((head / 1000) & (1 << i) ? 0 : 1); 64 if(s[tail].cnt == m && mod == 0) { 65 return tail; 66 } 67 if(s[tail].cnt <= m) q.push(tail); 68 } 69 } 70 return 0; 71 } 72 73 //calc a / b 74 char* divide(char *a, int len, int b) { 75 mem0(d); 76 int i = 0, cur = 0, l = 0; 77 while(cur < b && i < len) { 78 cur = cur * 10 + a[i++] - '0'; 79 } 80 d[l++] = cur / b + '0'; 81 while(i < len) { 82 cur = cur % b * 10 + a[i++] - '0'; 83 d[l++] = cur / b + '0'; 84 } 85 return d; 86 } 87 88 void print(int ed) { 89 int len = 0; 90 mem0(ans); 91 while(ed) { 92 ans[len++] = s[ed].num; 93 ed = s[ed].pre; 94 } 95 reverse(ans, ans + len); 96 printf("%s=%d*%s\n", ans, n, divide(ans, len, n)); 97 } 98 99 int main() 100 { 101 //FIN; 102 while(cin >> t) while(t--) { 103 cin >> n >> m; 104 mem0(s); 105 if( !(x = bfs()) ) { 106 puts("Impossible"); 107 } 108 else { 109 print(x); 110 } 111 } 112 return 0; 113 }
F：ZOJ 3597 Hit the Target! （线段树扫描线 -- 求矩形所能覆盖的最多点数）
见博文http://www.cnblogs.com/gj-Acit/p/4493091.html
不多说，利用法向量计算平面夹角
1 #include <iostream> 2 #include <stdio.h> 3 #include <algorithm> 4 #include <queue> 5 #include <cmath> 6 #include <cstring> 7 #define mem0(a) memset(a, 0, sizeof(a)) 8 #define mem1(a) memset(a, -1, sizeof(a)) 9 using namespace std; 10 11 const double PI = 4.0 * atan(1.0); 12 13 struct Point { 14 double x, y, z; 15 }a[3]; 16 17 double dot(Point a, Point b) { 18 return a.x * b.x + a.y * b.y + a.z * b.z; 19 } 20 21 Point cross(Point a, Point b) { 22 Point ret; 23 ret.x = a.y * b.z - a.z * b.y; 24 ret.y = a.z * b.x - a.x * b.z; 25 ret.z = a.x * b.y - a.y * b.x; 26 return ret; 27 } 28 29 int T; 30 31 int main() 32 { 33 //freopen("in.txt", "r", stdin); 34 while(cin >> T) while(T--) { 35 double A, B; 36 for(int i = 0; i < 3; i ++) { 37 cin >> A >> B; 38 B = B * PI / 180; A = A * PI / 180; 39 a[i].z = sin(B); 40 a[i].x = cos(B) * cos(A); 41 a[i].y = cos(B) * sin(A); 42 } 43 Point p1 = cross(a[0], a[1]), p2 = cross(a[1], a[2]), p3 = cross(a[2], a[0]); 44 double l1 = sqrt(dot(p1, p1)), l2 = sqrt(dot(p2, p2)), l3 = sqrt(dot(p3, p3)); 45 double s1 = acos(dot(p1, p2) / l1 / l2), s2 = acos(dot(p2, p3) / l2 / l3), s3 = acos(dot(p3, p1) / l3 / l1); 46 double ans = PI * 3 - s1 - s2 - s3; 47 printf("%.2lf\n", ans * 180 / PI); 48 } 49 return 0; 50 }
