标签：count 细节 -- 有一个 cst strcmp wrap 方式 ase
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题过的人也比較多。感觉是个几何旋转+ploya。可是几何实在不行没什么想法
B：这题就是一个大模拟，直接数组去模拟就可以。注意细节就能过
C：类似大白上一题红蓝三角形的。 每一个数字找一个互质和一个不互质个数。除掉反复就直接除2，然后总的C(n, 3)减去就可以，问题在怎么处理一个数字互质和不互质的，事实上仅仅要处理出不互质的就可以。这步利用容斥原理去搞。每一个数字分解成因子，然后计数因子个数。对于每一个数而言，为f(1个质因子) - f(2个质因子) + f(3) - f(4) + ...，然后就能够统计出答案了
D：这题首先转化成n个数字去掉k个，惯性最小。那么事实上重心的计算方法。就依据物理里面的杠杆原理（反正我是这样想的），因为质量都是1。所以等于全部距离累加起来。然后减去0。在除个数就是等于每一个位置距离中心要降低的长度。那么推出公式∑(di?d平均值)2，事实上这个就是方差了，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。那么仅仅要从左往右一遍扫过去不断维护就可以
E：dp，dp[i][j]表示到第i个位置，在第j列，这样递推过去就可以。也算比較水的一题
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;
}
#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;
}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;
}
#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- -，也就是说假设能优化到跑10步，事实上答案就显而易见了
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;
}
#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;
}标签：count 细节 -- 有一个 cst strcmp wrap 方式 ase
原文地址：http://www.cnblogs.com/ljbguanli/p/7253679.html
