2019 Multi-University Training Contest 2
Replenishment link: 2019 Multi-University Training Contest The 2
1005 Everything Is Generated In Equal Probability (HDU-6595)
The meaning of problems
A given integer \ (N \) , in \ ([1, N] \ ) randomly generate one \ (n-\) . Then generates length \ (n-\) of the full array \ ([. 1, n-] \) .
The arrangement of a running program, the current program seeking to reverse logarithmic arrangement, then a randomly selected sequence from whole arrangement. The sequences present routine continues recursively until the sequence length \ (0 \) exit, the program returns the total number of reverse pair. Program generates seeking answers expectations.
answer
#include <iostream>
#include <queue>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll mod = 998244353;
ll qpow(ll a, ll n, ll m) {
ll ans = 1;
while (n) {
if (n & 1) {
ans = (ans * a) % m;
}
a = (a * a) % m;
n >>= 1;
}
return ans;
}
int main() {
int n;
ll k = qpow(9, mod - 2, mod);
while (~scanf("%d", &n)) {
ll ans = n * n - 1;
printf("%lld\n", ans * k % mod);
}
return 0;
}1010 Just Skip The Problem (HDU-6600)
\ (Solved \ by \ ZMZ \)
The meaning of problems
Given a number \ (n \) , you can ask many times \ (y_i \) , then you can know \ (n \ xor \ y_i \ ) is equal to \ (y_i \) , the minimum number of inquiries can know \ (n \ ) is the number, find the least number of inquiries asking several programs.
Results of the \ (1e6 + 3 \) modulo.
answer
The answer is \ (the n-! \ MOD \ 1e6 + 3 \) .
When \ (n \ ge 1e6 + 3 \) when the answer to \ (0 \) .
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 5;
const ll mod = 1e6 + 3;
ll fac[maxn];
void init() {
fac[0] = 1;
for (int i = 1; i <= mod; ++i) {
fac[i] = i * fac[i - 1];
fac[i] %= mod;
}
}
int main() {
int n;
init();
while (~scanf("%d", &n)) {
if (n <= mod) {
printf("%lld\n", fac[n]);
}
else {
printf("0\n");
}
}
return 0;
}1011 Keen On Everything But Triangle (HDU-6601)
The meaning of problems
Given \ (n-\) number, \ (Q \) th interrogation.
Each interrogation a given interval \ ([L, R] \ ) maximum perimeter, the inner section can be composed of triangular and asked how much.
answer
Chairman of the enumeration tree
Chairman of the tree seeking Interval \ (k \) large, from the largest, the second largest, the \ (k \) big so continue to enumerate, meet the conditions on the output.
Number not constitute a triangle as a Fibonacci number. Since Fibonacci column \ (40 \) number more than \ (1E9 \) , and therefore will not enumerate the number of times a lot.
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
struct node {
int ls, rs, sum;
} ns[MAXN * 20];
int ct;
int rt[MAXN * 20];
void cpy(int& now, int old) {
now = ++ct;
ns[now] = ns[old];
}
void pushUp(int& now) {
ns[now].sum = ns[ns[now].ls].sum + ns[ns[now].rs].sum;
}
void build(int& now, int l, int r) {
now = ++ct;
ns[now].sum = 0;
if (l == r) return;
int m = (l + r) >> 1;
build(ns[now].ls, l, m);
build(ns[now].rs, m + 1, r);
}
void update(int& now, int old, int l, int r, int x) {
cpy(now, old);
if (l == r) {
ns[now].sum++;
return;
}
int m = (l + r) >> 1;
if (x <= m) update(ns[now].ls, ns[old].ls, l, m, x);
else update(ns[now].rs, ns[old].rs, m + 1, r, x);
pushUp(now);
}
int query(int s, int t, int l, int r, int k) {
if (l == r) return l;
int m = (l + r) >> 1;
int cnt = ns[ns[t].ls].sum - ns[ns[s].ls].sum;
//cout << s << " " << t << " " << cnt << endl;
if (k <= cnt) return query(ns[s].ls, ns[t].ls, l, m, k);
return query(ns[s].rs, ns[t].rs, m + 1, r, k - cnt);
}
void init(int n) {
ct = 0;
build(rt[0], 1, n);
}
int a[MAXN], b[MAXN];
int c[MAXN];
int main() {
int n, m;
while (cin >> n >> m) {
// scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
b[i] = a[i];
}
sort(b + 1, b + n + 1);
int sz = unique(b + 1, b + 1 + n) - b - 1;
init(sz);
for (int i = 1; i <= n; i++) {
a[i] = lower_bound(b + 1, b + 1 + sz, a[i]) - b;
update(rt[i], rt[i - 1], 1, sz, a[i]);
}
while (m--) {
int s, t, k;
scanf("%d%d", &s, &t);
// printf("%d", t - s + 1);
// printf("%d\n", b[query(rt[s - 1], rt[t], 1, sz, k)]);
if(t - s + 1 < 3) printf("-1\n");
else {
int cnt = 0, flag = 0;
for(int i = t - s + 1; i > 0; --i) {
c[cnt] = b[query(rt[s - 1], rt[t], 1, sz, i)];
if(cnt > 1 && c[cnt - 2] < c[cnt - 1] + c[cnt]) {
printf("%lld\n", c[cnt - 2] * 1ll + c[cnt - 1] * 1ll + c[cnt] * 1ll);
flag = 1;
break;
}
++cnt;
}
if(!flag) printf("-1\n");
}
}
}
return 0;
}