D - Factorial Divisibility
对于a[i]>=x的数一定可以整除,考虑a[i]<x的数,因为(x+1)*x! = (x+1)!
统计ai出现的次数, 把他转换为大的阶乘, 如果, 最终1到x - 1, ai的出现次数均为0则说明可以被x!整除
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(fast)
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<cstring>
#include<math.h>
#include<map>
#include<vector>
#include<stack>
#define ms(x,y) memset(x,y,sizeof x);
#define YES cout<<"YES"<<'\n';
#define NO cout<<"NO"<<'\n';
#define endl cout<<'\n';
typedef long long ll;
const int maxn = 5e5 + 100, inf = 1e18;
const int mod = 1e9 + 7;
using namespace std;
ll a[maxn];
void solve() {
int n, x;
cin >> n >> x;
map<int, int>mp;
for (int i = 1; i <= n; i++) {
int y;
cin >> y;
mp[y]++;
}
for (int i = 1; i < x; i++) {
mp[i + 1] += (mp[i] / (i + 1));
mp[i] = mp[i] % (i + 1);
}
for (int i = 1; i < x;i++) {
if (mp[i] != 0) {
NO
return;
}
}
YES
}
signed main()
{
ios::sync_with_stdio(false);
solve();
}
C1 - Make Nonzero Sum (easy version)
思路:奇数一定不可以,偶数一定可以,两两相同放一起,不同分开
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(fast)
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<cstring>
#include<math.h>
#include<map>
#include<vector>
#include<stack>
#define ms(x,y) memset(x,y,sizeof x);
#define YES cout<<"YES"<<'\n';
#define NO cout<<"NO"<<'\n';
#define endl cout<<'\n';
typedef long long ll;
const int maxn=2e5+10,inf = 1e18 ;
const int mod = 1e9 + 7;
using namespace std;
int a[maxn];
struct node {
int l, r;
}x[maxn];
void solve(){
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
if (n & 1) {
cout << -1 << '\n';
return;
}
int cnt = 0;
for (int i = 1; i <= n; i += 2) {
if (a[i] == a[i + 1]) {
x[++cnt].l = i;
x[cnt].r = i + 1;
}
else {
x[++cnt].l = i;
x[cnt].r = i;
x[++cnt].l = i + 1;
x[cnt].r = i + 1;
}
}
cout << cnt << '\n';
for (int i = 1; i <= cnt; i++) {
cout << x[i].l << ' ' << x[i].r << '\n';
}
}
signed main()
{
ios::sync_with_stdio(false);
int t;
cin >> t;
while (t--) {
solve();
}
}
C2. Make Nonzero Sum (hard version)
思路:变成了0,1,-1三种情况,对于去除0后为奇数的一定不可以,偶数的一定可以,
和(easy version)情况相似,相邻的不为0的凑在一起
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(fast)
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<cstring>
#include<math.h>
#include<map>
#include<vector>
#include<stack>
using namespace std;
#define ll long long
#define pcc pair<char, char>
#define pii pair<int, int>
#define inf 0x3f3f3f3f
const int maxn = 200010;
int n, a[maxn];
void solve() {
cin >> n;
int cnt = 0;
for (int i = 1; i <= n; ++i) {
cin >> a[i];
if (a[i]) ++cnt;
}
if (cnt & 1) {
cout << -1 << '\n';
return;
}
vector<pii> res;
int l, r;
for (int i = 1; i <= n;) {
while (i <= n && !a[i]) {
res.push_back({ i, i });
++i;
}
if (i > n) {
break;
}
l = i;
r = i + 1;
if (a[r]) {
if (a[l] == a[r]) {
res.push_back({ l, r });
}
else {
res.push_back({ l, l });
res.push_back({ r, r });
}
}
else {
res.push_back({ l, l });
while (r <= n && !a[r]) {
res.push_back({ r, r });
++r;
}
if (a[l] == a[r]) {
res.back().second = r;
}
else {
res.push_back({ r, r });
}
}
i = r + 1;
}
int len = res.size();
cout << len << '\n';
for (auto p : res) {
cout << p.first << ' ' << p.second << '\n';
}
}
int main() {
ios::sync_with_stdio(false);
int t ;
cin >> t;
while (t--) {
solve();
}
}
G 严肃古板的秩序
思路:?只有12个,暴力,dfs3种情况,防止long long爆掉采用龟速乘
#include<bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(fast)
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<cstring>
#include<math.h>
#include<map>
#include<vector>
#include<stack>
#define ms(x,y) memset(x,y,sizeof x)
typedef long long ll;
const int maxn=2e5+10,INF = 1e18 ;
using namespace std;
int num[20],cnt = 0;
char op[3] = { '+','-','#' }, s1[20];
ll numb = 0;
bool flag = false;
ll qadd(ll a,ll b, ll p){ //龟速乘防止long long 爆掉
ll res = 0;
while (b){
if (b & 1) res = (res + a) % p;
a = (a + a) % p;
b = b >> 1;
}
return res;
}
void dfs(ll u, ll v) {
if (u == cnt - 1) {
if (v == numb) flag = true;
return;
}
if (u >= cnt) return;
if (flag) return;
s1[u] = op[0];
dfs(u + 1, v + num[u + 2]);
if (flag) return;
s1[u] = op[1];
dfs(u + 1, v - num[u + 2]);
if (flag) return;
s1[u] = op[2];
if (v > 0) {
dfs(u + 1, qadd(v, v, num[u + 2]));
}
}
void solve(){
string s;
cin >> s;
for (int i = 0; i < s.size(); i++) {
if (isdigit(s[i])) {
numb =numb*10LL+s[i] - '0';
}
else {
num[++cnt] = numb;
numb = 0LL;
}
}
dfs(0, num[1]);
int cnt1 = 0;
if (flag) {
for (int i = 0; i < s.size(); i++) {
if (s[i] != '?')
cout << s[i];
else {
cout << s1[cnt1++];
}
}
}
else {
cout << -1 << '\n';
}
}
signed main()
{
ios::sync_with_stdio(false);
solve();
}
C 忽远忽近的距离文章来源:https://www.toymoban.com/news/detail-487469.html
题意:构造一个n数组,使得每个2<=|ai-i|<=3;
思路1:4,5,6可构造除7外往后的所有数
n%4==1,4,5搭配;n%4==2,4,6搭配;n%4==3,4,5,6搭配文章来源地址https://www.toymoban.com/news/detail-487469.html
#include<bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(fast)
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<cstring>
#include<math.h>
#include<map>
#include<vector>
#include<stack>
#define ms(x,y) memset(x,y,sizeof x)
#define int long long
typedef long long ll;
const int maxn=2e5+10,INF = 1e18 ;
using namespace std;
int start = 1;
void fun4() {
cout << start + 2<<' ' << start + 3<<' ' << start<<' ' << start + 1<<' ';
start += 4;
}
void fun5() {
cout << start + 3 <<' '<< start + 4 <<' ' << start <<' ' << start + 1<<' ' << start + 2 << ' ';
start+=5;
}
void fun6() {
cout << start + 3 << ' ' << start + 4 << ' ' << start + 5 << ' ' << start << ' ' << start + 1 << ' ' << start + 2 << ' ';
start += 6;
}
void solve(){
int n;
cin >> n;
if (n <= 3 || n == 7) {
cout << -1 << '\n';
return;
}
start = 1;
int x = n / 4;
int y = n % 4;
if (y == 0) {
for (int i = 1; i <= x; i++) {
fun4();
}
}
else if (y == 1) {
for (int i = 1; i <= x - 1; i++) {
fun4();
}
fun5();
}
else if (y == 2) {
for (int i = 1; i <= x - 1; i++) {
fun4();
}
fun5();
}
else if (y == 3) {
for (int i = 1; i <= x - 1; i++) {
fun4();
}
fun5();
fun6();
}
}
signed main()
{
ios::sync_with_stdio(false);
solve();
}
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