信息网站建设情况工作会,如何制作漂亮的微信公众号,做电影网站怎样赚钱吗,网站建设需要什么手续D - Factorial Divisibility 对于a[i]x的数一定可以整除#xff0c;考虑a[i]x的数#xff0c;因为(x1)*x! (x1)! 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 optimiz…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(x1)*x! (x1)! 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)
#includeiostream
#includealgorithm
#includemap
#includeset
#includequeue
#includecstring
#includemath.h
#includemap
#includevector
#includestack
#define ms(x,y) memset(x,y,sizeof x);
#define YES coutYES\n;
#define NO coutNO\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;mapint, intmp;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) {NOreturn;}}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)
#includeiostream
#includealgorithm
#includemap
#includeset
#includequeue
#includecstring
#includemath.h
#includemap
#includevector
#includestack
#define ms(x,y) memset(x,y,sizeof x);
#define YES coutYES\n;
#define NO coutNO\n;
#define endl cout\n;
typedef long long ll;
const int maxn2e510,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01-1三种情况对于去除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)
#includeiostream
#includealgorithm
#includemap
#includeset
#includequeue
#includecstring
#includemath.h
#includemap
#includevector
#includestack
using namespace std;
#define ll long long
#define pcc pairchar, char
#define pii pairint, 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;}vectorpii 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
#includebits/stdc.h
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(fast)
#includeiostream
#includealgorithm
#includemap
#includeset
#includequeue
#includecstring
#includemath.h
#includemap
#includevector
#includestack
#define ms(x,y) memset(x,y,sizeof x)
typedef long long ll;
const int maxn2e510,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*10LLs[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
题意构造一个n数组使得每个2|ai-i|3; 思路14,5,6可构造除7外往后的所有数 n%414,5搭配n%42,4,6搭配n%43,4,5,6搭配
#includebits/stdc.h
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(fast)
#includeiostream
#includealgorithm
#includemap
#includeset
#includequeue
#includecstring
#includemath.h
#includemap#includevector#includestack
#define ms(x,y) memset(x,y,sizeof x)
#define int long long
typedef long long ll;
const int maxn2e510,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 ;start5;
}
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();
}
