2014ACM/ICPC亚洲区鞍山赛区现场赛——题目重现

题目链接

5小时内就搞了5题B、C、D、E，I。
H题想到要打表搞了，可惜时间不够，后面打出表试了几下过了- -
K题过的人也比较多，感觉是个几何旋转+ploya，但是几何实在不行没什么想法

B：这题就是一个大模拟，直接数组去模拟即可，注意细节就能过

C：类似大白上一题红蓝三角形的， 每个数字找一个互质和一个不互质个数，除掉重复就直接除2，然后总的C(n, 3)减去即可，问题在怎么处理一个数字互质和不互质的，其实只要处理出不互质的即可，这步利用容斥原理去搞，每个数字分解成因子，然后计数因子个数，对于每个数而言，为f(1个质因子) - f(2个质因子) + f(3) - f(4) + ...，然后就可以统计出答案了

D：这题首先转化成n个数字去掉k个，惯性最小，那么其实重心的计算方法，就根据物理里面的杠杆原理（反正我是这样想的），由于质量都是1，所以等于所有距离累加起来，然后减去0，在除个数就是等于每个位置距离中心要减少的长度，那么推出公式∑(di−d平均值)2，其实这个就是方差了，问题就转化为n个数字去掉k个，使得方差最大，那么贪心知道越密集越好，所以肯定是连续的n - k个，那么只要从左往右一遍扫过去不断维护即可

E：dp，dp[i][j]表示到第i个位置，在第j列，这样递推过去即可，也算比较水的一题

H：把竖的一列看成一个二进制数，每次就相当与选已有的两个二进制数，去构造一个新的二进制数，然后把这个数丢进已有的数组中，没想到好的办法，写个暴力，让他跑9步（这个跑了1分多钟，10步的话要跑好久啊），然后发现还有6个无解，全赋为10就wa了，然后试了一下，原来有一个解的答案是11，然后就过了- -

I：这题就是个签到题，不能多说

K：应该是几何旋转求出有几种方式，然后ploya定理求解

代码：

B：

#include <cstdio>#include <cstring>#include <map>using namespace std;typedef long long ll;const int N = 5005;int t, n, x;char op[15];int top, a[N], an;map<int, ll> C;void init() {C.clear();top = 0;an = 0;}int find(int u) {for (int i = 0; i < an; i++)if (a[i] == u) return i;}void Add() {scanf("%d", &x);if (C.count(x))printf("same priority.\n");else {C[x] = 0;a[an++] = x;printf("success.\n");}}void Close() {scanf("%d", &x);if (!C.count(x))printf("invalid priority.\n");else {printf("close %d with %I64d.\n", x, C[x]);C.erase(x);int tmp = find(x);an--;for (int i = tmp; i < an; i++)a[i] = a[i + 1];}}void Chat() {scanf("%d", &x);if (an == 0) printf("empty.\n");else {if (top) C[top] += x;else C[a[0]] += x;printf("success.\n");}}void Rotate() {scanf("%d", &x);if (x >= 1 && x <= an) {int tmp = a[x - 1];for (int i = x - 1; i; i--)a[i] = a[i - 1];a[0] = tmp;printf("success.\n");} else printf("out of range.\n");}void Prior() {if (an == 0) printf("empty.\n");else {int Max = 0, Max_v;for (int i = 0; i < an; i++) {if (Max < a[i])Max = a[i], Max_v = i;}int tmp = a[Max_v];for (int i = Max_v; i; i--)a[i] = a[i - 1];a[0] = tmp;printf("success.\n");}}void Choose() {scanf("%d", &x);if (C.count(x)) {int v = find(x);int tmp = a[v];for (int i = v; i; i--)a[i] = a[i - 1];a[0] = tmp;printf("success.\n");} else printf("invalid priority.\n");}void Top() {scanf("%d", &x);if (C.count(x)) {top = x;printf("success.\n");} else printf("invalid priority.\n");}void Untop() {if (top == 0) printf("no such person.\n");else {top = 0;printf("success.\n");}}void gao() {if (top && C[top]) printf("Bye %d: %I64d\n", top, C[top]);for (int i = 0; i < an; i++) {if (a[i] == top) continue;if (C[a[i]] == 0) continue;printf("Bye %d: %I64d\n", a[i], C[a[i]]);}}int main() {scanf("%d", &t);while (t--) {init();scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%s", op);printf("Operation #%d: ", i);if (!strcmp(op, "Add")) Add();if (!strcmp(op, "Close")) Close();if (!strcmp(op, "Chat")) Chat();if (!strcmp(op, "Rotate")) Rotate();if (!strcmp(op, "Prior")) Prior();if (!strcmp(op, "Choose")) Choose();if (!strcmp(op, "Top")) Top();if (!strcmp(op, "Untop")) Untop();}gao();}return 0;}

C：

#include <cstdio>#include <cstring>#include <vector>using namespace std;typedef long long ll;const int N = 100005;int vis[N], prime[N], pn = 0;void getprime() {for (int i = 2; i < N; i++) {if (vis[i]) continue;prime[pn++] = i;for (ll j = (ll)i * i; j < N; j += i)vis[j] = 1;}}int fac[N], fn;void getfac(int n) {fn = 0;if (n == 1) {fac[fn++] = 1;return;}int tmp = n;for (int i = 0; i < pn && prime[i] * prime[i] <= tmp; i++) {if (tmp % prime[i] == 0) {fac[fn++] = prime[i];while (tmp % prime[i] == 0)tmp /= prime[i];}}if (tmp != 1) fac[fn++] = tmp;}vector<int> g[N];int cnt[N];void dfs(int n, int u, int sum, int c) {if (u == fn) {if (c == 0) return;g[n].push_back(sum);cnt[sum] = c;return;}dfs(n, u + 1, sum * fac[u], c + 1);dfs(n, u + 1, sum, c);}void build(int n) {getfac(n);dfs(n, 0, 1, 0);}void init() {getprime();for (int i = 1; i <= 100000; i++)build(i);}int t, n, a[N];int have[N];int main() {init();scanf("%d", &t);while (t--) {scanf("%d", &n);memset(have, 0, sizeof(have));for (int i = 0; i < n; i++) {scanf("%d", &a[i]);for (int j = 0; j < g[a[i]].size(); j++)have[g[a[i]][j]]++;}ll ans = (ll)n * (n - 1) * (n - 2) / 6;ll cao = 0;for (int i = 0; i < n; i++) {ll sb = 0;for (int j = 0; j < g[a[i]].size(); j++) {int facc = g[a[i]][j];if (cnt[facc] % 2) sb += have[facc];else sb -= have[facc];}cao += (sb - 1) * (n - sb);}printf("%I64d\n", ans - cao / 2);}return 0;}

D：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 50005;int t, n, k;double a[N];int main() {scanf("%d", &t);while (t--) {scanf("%d%d", &n, &k);for (int i = 0; i < n; i++) scanf("%lf", &a[i]);if (n == k) {printf("%.10lf\n", 0.0);continue;}sort(a, a + n);k = n - k;double s1 = 0, s2 = 0;for (int i = 0; i < k; i++)s2 += a[i];for (int i = 0; i < k; i++)s1 += a[i] * a[i];double ans = s1 - s2 * s2 / k * 2 + s2  * s2 / k;for (int i = k; i < n; i++) {s1 = s1 - a[i - k] * a[i - k] + a[i] * a[i];s2 = s2 - a[i - k] + a[i];ans = min(ans, s1 - s2 * s2 / k * 2 + s2 * s2 / k);}printf("%.10lf\n", ans);}return 0;}

E：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int INF = 0x3f3f3f3f;const int N = 55;int t, n, m;int g[N][N], dp[N * 2][N], a;int main() {scanf("%d", &t);while (t--) {scanf("%d%d", &n, &m);for (int i = 1; i <= m; i++)for (int j = 1; j <= m; j++)scanf("%d", &g[i][j]);memset(dp, -INF, sizeof(dp));scanf("%d", &a);if (a == -1) for (int i = 1; i <= m; i++) dp[1][i] = 0;else dp[1][a] = 0;int ans = 0;for (int i = 2; i <= n; i++) {scanf("%d", &a);int s, e;if (a == -1) s = 1, e = m;else s = a, e = a;for (int j = s; j <= e; j++) {for (int k = 1; k <= m; k++)dp[i][j] = max(dp[i][j], dp[i - 1][k] + g[k][j]);if (i == n) ans = max(dp[i][j], ans);}}printf("%d\n", ans);}return 0;}

H：

打表程序：

#include <cstdio>#include <cstring>#include <set>#include <vector>using namespace std;const int INF = 0x3f3f3f3f;const int N = 105;int ans[1005];int h[N], hn = 0;int NAND(int a, int b) {return (~(a&b))&((1<<8) - 1);}vector<int> ss;int vis[257];void dfs(int step) {ans[h[hn - 1]] = min(ans[h[hn - 1]], step);if (step == 9) return;int v[257];memset(v, 0, sizeof(v));for (int i = 0; i < hn; i++) {for (int j = i; j < hn ;j++) {int tmp = NAND(h[i], h[j]);if (v[tmp]) continue;v[tmp] = 1;if (vis[tmp]) continue;h[hn++] = tmp;vis[tmp] = 1;dfs(step + 1);vis[tmp] = 0;hn--;}}}int main() {memset(ans, INF, sizeof(ans));vis[240] = vis[204] = vis[170] = vis[255] = vis[0] = 1;h[hn++] = 240; h[hn++] = 204; h[hn++] = 170; h[hn++] = 255; h[hn++] = 0;ans[240] = ans[204] = ans[170] = ans[255] = ans[0] = 1;dfs(1);for (int i = 0; i < 256; i++)printf("sb[%d] = %d;\n", i, ans[i]);return 0;}

表：

#include <cstdio>#include <cstring>const int INF = 0x3f3f3f3f;int t;char str[10];int sb[300];int main() {sb[0] = 1;sb[1] = 8;sb[2] = 7;sb[3] = 5;sb[4] = 7;sb[5] = 5;sb[6] = 8;sb[7] = 5;sb[8] = 6;sb[9] = 9;sb[10] = 4;sb[11] = 6;sb[12] = 4;sb[13] = 6;sb[14] = 5;sb[15] = 2;sb[16] = 7;sb[17] = 5;sb[18] = 8;sb[19] = 5;sb[20] = 8;sb[21] = 5;sb[22] = 11;//就这个是11- -，也就是说如果能优化到跑10步，其实答案就显而易见了sb[23] = 8;sb[24] = 8;sb[25] = 7;sb[26] = 8;sb[27] = 6;sb[28] = 8;sb[29] = 6;sb[30] = 8;sb[31] = 5;sb[32] = 6;sb[33] = 9;sb[34] = 4;sb[35] = 6;sb[36] = 8;sb[37] = 7;sb[38] = 8;sb[39] = 6;sb[40] = 7;sb[41] = 1061109567;sb[42] = 4;sb[43] = 7;sb[44] = 7;sb[45] = 7;sb[46] = 5;sb[47] = 4;sb[48] = 4;sb[49] = 6;sb[50] = 5;sb[51] = 2;sb[52] = 8;sb[53] = 6;sb[54] = 8;sb[55] = 5;sb[56] = 7;sb[57] = 7;sb[58] = 5;sb[59] = 4;sb[60] = 5;sb[61] = 7;sb[62] = 8;sb[63] = 2;sb[64] = 6;sb[65] = 9;sb[66] = 8;sb[67] = 7;sb[68] = 4;sb[69] = 6;sb[70] = 8;sb[71] = 6;sb[72] = 7;sb[73] = 1061109567;sb[74] = 7;sb[75] = 7;sb[76] = 4;sb[77] = 7;sb[78] = 5;sb[79] = 4;sb[80] = 4;sb[81] = 6;sb[82] = 8;sb[83] = 6;sb[84] = 5;sb[85] = 2;sb[86] = 8;sb[87] = 5;sb[88] = 7;sb[89] = 7;sb[90] = 5;sb[91] = 7;sb[92] = 5;sb[93] = 4;sb[94] = 8;sb[95] = 2;sb[96] = 7;sb[97] = 1061109567;sb[98] = 7;sb[99] = 7;sb[100] = 7;sb[101] = 7;sb[102] = 5;sb[103] = 7;sb[104] = 9;sb[105] = 1061109567;sb[106] = 7;sb[107] = 9;sb[108] = 7;sb[109] = 9;sb[110] = 6;sb[111] = 7;sb[112] = 4;sb[113] = 7;sb[114] = 5;sb[115] = 4;sb[116] = 5;sb[117] = 4;sb[118] = 8;sb[119] = 2;sb[120] = 7;sb[121] = 9;sb[122] = 6;sb[123] = 7;sb[124] = 6;sb[125] = 7;sb[126] = 8;sb[127] = 4;sb[128] = 5;sb[129] = 9;sb[130] = 8;sb[131] = 7;sb[132] = 8;sb[133] = 7;sb[134] = 9;sb[135] = 6;sb[136] = 3;sb[137] = 9;sb[138] = 5;sb[139] = 6;sb[140] = 5;sb[141] = 6;sb[142] = 6;sb[143] = 3;sb[144] = 8;sb[145] = 7;sb[146] = 9;sb[147] = 6;sb[148] = 9;sb[149] = 6;sb[150] = 9;sb[151] = 9;sb[152] = 8;sb[153] = 6;sb[154] = 8;sb[155] = 6;sb[156] = 8;sb[157] = 6;sb[158] = 9;sb[159] = 6;sb[160] = 3;sb[161] = 9;sb[162] = 5;sb[163] = 6;sb[164] = 8;sb[165] = 6;sb[166] = 8;sb[167] = 6;sb[168] = 4;sb[169] = 9;sb[170] = 1;sb[171] = 6;sb[172] = 5;sb[173] = 7;sb[174] = 5;sb[175] = 3;sb[176] = 5;sb[177] = 6;sb[178] = 6;sb[179] = 3;sb[180] = 8;sb[181] = 6;sb[182] = 9;sb[183] = 6;sb[184] = 5;sb[185] = 7;sb[186] = 5;sb[187] = 3;sb[188] = 7;sb[189] = 7;sb[190] = 8;sb[191] = 5;sb[192] = 3;sb[193] = 9;sb[194] = 8;sb[195] = 6;sb[196] = 5;sb[197] = 6;sb[198] = 8;sb[199] = 6;sb[200] = 4;sb[201] = 9;sb[202] = 5;sb[203] = 7;sb[204] = 1;sb[205] = 6;sb[206] = 5;sb[207] = 3;sb[208] = 5;sb[209] = 6;sb[210] = 8;sb[211] = 6;sb[212] = 6;sb[213] = 3;sb[214] = 9;sb[215] = 6;sb[216] = 5;sb[217] = 7;sb[218] = 7;sb[219] = 7;sb[220] = 5;sb[221] = 3;sb[222] = 8;sb[223] = 5;sb[224] = 4;sb[225] = 9;sb[226] = 5;sb[227] = 7;sb[228] = 5;sb[229] = 7;sb[230] = 7;sb[231] = 7;sb[232] = 7;sb[233] = 1061109567;sb[234] = 4;sb[235] = 7;sb[236] = 4;sb[237] = 7;sb[238] = 4;sb[239] = 6;sb[240] = 1;sb[241] = 6;sb[242] = 5;sb[243] = 3;sb[244] = 5;sb[245] = 3;sb[246] = 8;sb[247] = 5;sb[248] = 4;sb[249] = 7;sb[250] = 4;sb[251] = 6;sb[252] = 4;sb[253] = 6;sb[254] = 7;sb[255] = 1;scanf("%d", &t);while (t--) {scanf("%s", str);int ans = 0;for (int i = 7; i >= 0; i--) ans = ans * 2 + str[i] - '0';int out = sb[ans];if (out == INF) out = 10;printf("%d\n", out);}return 0;}

I：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;int t, n;const int N = 1005;struct Point {double x, y;int t;void read() {scanf("%d%lf%lf", &t, &x, &y);}} p[N];bool cmp(Point a, Point b) {return a.t < b.t;}double dis(Point a, Point b) {double dx = a.x - b.x;double dy = a.y - b.y;return sqrt(dx * dx + dy * dy);}int main() {scanf("%d", &t);while (t--) {scanf("%d", &n);for (int i = 0; i < n; i++) {p[i].read();}sort(p, p + n, cmp);double ans = 0;for (int i = 1; i < n; i++) {ans = max(ans, dis(p[i], p[i - 1]) / (p[i].t - p[i - 1].t));}printf("%.10lf\n", ans);}return 0;}

H：事后补