CSP模拟-7
集合专练？？？？？逆天！！！！！！！
T1 卷
逆天！！！！！！！！！！！！！！！！！！！！又没看懂题。独立集指集合里的每个点不相连呜呜呜呜呜，我还以为是剩下的点互不相连，直接寄掉。
式子好推，就不推了，咕。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <vector>
#define mod 1000000007
using namespace std;
long long n, w[5211314 >> 3], beg;
long long cnt, head[5211314 >> 3], nex[5211314 >> 3], to[5211314 >> 3];
long double comp = 1, val[5211314 >> 3], dp_log[5211314 >> 3][2];
long long dp[5211314 >> 3][2];
inline void AddPoi(int v1, int v2) {
cnt ++;
nex[cnt] = head[v1];
head[v1] = cnt;
to[cnt] = v2;
return;
}
void DFS(int fa, int now) {
dp_log[now][1] = val[now];
dp[now][1] = w[now];
for (int i = head[now]; i != 0; i = nex[i]) {
if (to[i] != fa) {
DFS(now, to[i]);
if (dp_log[to[i]][0] > dp_log[to[i]][1]) {
dp_log[now][0] += dp_log[to[i]][0];
dp[now][0] = dp[now][0] * dp[to[i]][0] % mod;
}
else {
dp_log[now][0] += dp_log[to[i]][1];
dp[now][0] = dp[now][0] * dp[to[i]][1] % mod;
}
dp_log[now][1] += dp_log[to[i]][0];
dp[now][1] = dp[now][1] * dp[to[i]][0] % mod;
}
}
//dp式子好推，就不多BB了
return;
}
int main() {
cin >> n;
for (int i = 1; i <= n; ++ i) {
dp[i][0] = 1;
}
for (long long i = 1; i <= n; ++ i) {
cin >> w[i];
val[i] = log(w[i]);
//一定要记住啊啊啊啊啊啊啊啊，乘法就转换成log的加
}
for (long long i = 1; i <= n - 1; ++ i) {
long long u, v;
cin >> u >> v;
AddPoi(u, v), AddPoi(v, u);
}
DFS(1, 1);
//记得判断QAQ
if (dp_log[1][0] > dp_log[1][1]) {
cout << dp[1][0] << endl;
}
else {
cout << dp[1][1] << endl;
}
return 0;
}
T2 简单题
确实够简单奥。。。。。
将每个是双倍的点连边，判断链长，处理，咕
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define mod 10000019
using namespace std;
typedef long long ll;
ll base[67], n ,q, basic, m, fac[5211314 << 1], inv[5211314 << 1];
ll sum1, sum2;
inline ll QuickPower(ll a, ll k) {
ll base = a, ans = 1;
while (k != 0) {
if (k & 1 == 1) {
ans = ans * base % mod;
}
base = base * base % mod;
k >>= 1;
}
return ans;
}
inline ll GetInv(ll x) {
if (x == 1 || x == 0) return 1;
return (mod - mod / x) * (GetInv(mod % x)) % mod;
}
inline ll GetC(ll a, ll b) {
if (a < b) return 0;
else return (fac[a] * GetInv(fac[a - b])) % mod * GetInv(fac[b]) % mod;
}
ll Lucas(ll a, ll b) {
if (b == 0) {
return 1;
}
return Lucas(a / mod, b / mod) * GetC(a % mod, b % mod) % mod;
}
int main() {
scanf("%lld%lld", &n, &q);
fac[0] = 1;
for (ll i = 1; i <= mod; ++ i) {
fac[i] = fac[i - 1] * i % mod;
}
base[0] = 1;
for (ll i = 1; i <= 60; ++ i) {
base[i] = base[i - 1] * 2 % (int)4e18;
}
double k = log(n)/log(2);
for (ll i = 0 ; i <= k; ++ i) {
ll l = n / base[i + 1], r = n / base[i], temp;
l += 1;
if (l % 2 == 1) {
temp = (r - l) / 2 + 1;
}
else if (l % 2 == 0) {
if (r % 2 == 1) temp = (r - l) / 2 + 1;
else temp = (r - l) / 2;
}
basic += (i + 1) / 2 * temp;
if ((i + 1) % 2 == 0) {
sum1 += temp;
}
else {
sum2 += temp;
}
}
while (q --) {
scanf("%lld", &m);
if (m < basic) cout << 0 << endl;
else {
cout << QuickPower(2, sum1) % mod * Lucas(sum2, m - basic) % mod << endl;
}
}
return 0;
}
T3 粉丝
在取的数中，每次可做的操作有两种：
- 插入\(\sqrt{n}\)
- 将所有选的数都加一
这样就保证了每种选择方式都能出现，如我们要找 \(\sqrt{n}\) 和 \(\sqrt{n}+3\) 的排列，则可以先插入 \(\sqrt{n}\) 再加两次 \(1\) 然后插一次 \(\sqrt{n}\) 就行
逆天根号分治，总之就是不太会就对了
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define mod 10000019
using namespace std;
typedef long long ll;
ll base[67], n ,q, basic, m, fac[5211314 << 1], inv[5211314 << 1];
ll sum1, sum2;
inline ll QuickPower(ll a, ll k) {
ll base = a, ans = 1;
while (k != 0) {
if (k & 1 == 1) {
ans = ans * base % mod;
}
base = base * base % mod;
k >>= 1;
}
return ans;
}
inline ll GetInv(ll x) {
if (x == 1 || x == 0) return 1;
return (mod - mod / x) * (GetInv(mod % x)) % mod;
}
inline ll GetC(ll a, ll b) {
if (a < b) return 0;
else return (fac[a] * GetInv(fac[a - b])) % mod * GetInv(fac[b]) % mod;
}
ll Lucas(ll a, ll b) {
if (b == 0) {
return 1;
}
return Lucas(a / mod, b / mod) * GetC(a % mod, b % mod) % mod;
}
int main() {
scanf("%lld%lld", &n, &q);
fac[0] = 1;
for (ll i = 1; i <= mod; ++ i) {
fac[i] = fac[i - 1] * i % mod;
}
base[0] = 1;
for (ll i = 1; i <= 60; ++ i) {
base[i] = base[i - 1] * 2 % (int)4e18;
}
double k = log(n)/log(2);
for (ll i = 0 ; i <= k; ++ i) {
ll l = n / base[i + 1], r = n / base[i], temp;
l += 1;
if (l % 2 == 1) {
temp = (r - l) / 2 + 1;
}
else if (l % 2 == 0) {
if (r % 2 == 1) temp = (r - l) / 2 + 1;
else temp = (r - l) / 2;
}
basic += (i + 1) / 2 * temp;
if ((i + 1) % 2 == 0) {
sum1 += temp;
}
else {
sum2 += temp;
}
}
while (q --) {
scanf("%lld", &m);
if (m < basic) cout << 0 << endl;
else {
cout << QuickPower(2, sum1) % mod * Lucas(sum2, m - basic) % mod << endl;
}
}
return 0;
}
总结
T1从第一次到现在就没读过懂题，但暴力总能骗一骗，总是出现能有正解思路但不知道接下来怎么做的情况。
