【2020牛客NOIP赛前集训营-提高组（第二场）】题解（GCD，包含，前缀，移动）
目录
T1：GCD
title
solution
非常水，看一眼就知道了
首先我们知道每一个数都有唯一的标准整数分解，即拆成若干个质数的幂的乘积
而我们又知道质数彼此互质，\(gcd()=1\)
所以就可以迅速反应到，如果一个数有\(\ge 2\)个质数因子，那么\(gcd\)一定等于\(1\)
否则\(gcd\)就是唯一的质数因子
这样的话，需要寻找的特殊数一定是幂级递增
欧拉筛出\(n\)以内的所有质数，然后进行幂级累乘，时间复杂度就是\(O(nlogn)\)
最后加上没有被计算的数的个数即可，因为这些数每个都只会贡献\(1\)
然后直接干掉这道水题
code
#include <cstdio>
#define ll long long
#define MAXN 10000000
int a, b, cnt;
int prime[MAXN + 5];
bool vis[MAXN + 5];
ll ans[MAXN + 5];
void sieve() {
for( int i = 2;i <= MAXN;i ++ ) {
if( ! vis[i] )
vis[i] = 1, prime[++ cnt] = i;
for( int j = 1;i * prime[j] <= MAXN && j <= cnt;j ++ ) {
vis[i * prime[j]] = 1;
if( i % prime[j] == 0 ) break;
}
}
}
int main() {
sieve();
for( int i = 1;i <= MAXN;i ++ )
ans[i] = 1;
for( int i = 1;i <= cnt;i ++ ) {
int j = 1;
while( 1ll * j * prime[i] <= MAXN ) {
j *= prime[i];
ans[j] = prime[i];
}
}
for( int i = 1;i <= MAXN;i ++ )
ans[i] += ans[i - 1];
scanf( "%d %d", &a, &b );
printf( "%lld\n", ans[b] - ans[a - 1] );
return 0;
}
T2：包含
title
solution
这道题只要了解一点点枚举子集就能\(AC\)
直接暴力一个数枚举子集
自测大数据跑了1.3s以为会T，但是没想到交上去能跑过诶
好了——数据加强了，终于，我被卡掉了5分
正解就是每次随便丢掉任何一个\(1\)，\(O(nlogn)\)
#include <cstdio>
#define MAXN 1000005
int n, m;
bool vis[MAXN];
int main() {
scanf( "%d %d", &n, &m );
for( int i = 1, x;i <= n;i ++ ) {
scanf( "%d", &x );
vis[x] = 1;
}
for( int i = 1e6;i;i -- ) {
if( ! vis[i] ) continue;
for( int j = 0;j < 20;j ++ )
if( i >> j & 1 )
vis[i ^ ( 1 << j )] = 1;
}
for( int i = 1, x;i <= m;i ++ ) {
scanf( "%d", &x );
if( vis[x] ) printf( "yes\n" );
else printf( "no\n" );
}
return 0;
}
code(正解code已补充在上面)
#include <cstdio>
#include <iostream>
using namespace std;
#define MAXN 100005
#define MAXM 1000000
int n, m, cnt, maxx;
int a[MAXN];
bool vis[MAXM + 5];
int main() {
scanf( "%d %d", &n, &m );
for( int i = 1;i <= n;i ++ ) {
scanf( "%d", &a[i] );
maxx = max( maxx, a[i] );
}
for( int i = 1;i <= n;i ++ ) {
int x = a[i];
if( vis[x] ) continue;
vis[x] = 1, cnt ++;
if( cnt == maxx ) continue;
for( int j = x;j;j = ( ( j - 1 ) & x ) )
if( ! vis[j] ) {
vis[j] = 1;
cnt ++;
if( cnt == maxx ) break;
}
}
for( int i = 1, x;i <= m;i ++ ) {
scanf( "%d", &x );
if( vis[x] ) printf( "yes\n" );
else printf( "no\n" );
}
return 0;
}
T3：前缀
title
solution
大模拟！！无需多说慢慢敲，高精直接上
code
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define int long long
#define mod 998244353
#define MAXN 100005
vector < int > pos[30];
char s[MAXN], t[MAXN];
int pi[MAXN][2], num[MAXN];
int n, ans;
int calc( int l, int r, int siz, int len ) {
int cnt = 0, ip;
for( int i = l;i <= r;i ++ )
num[i - l + 1] = t[i] ^ 48;
for( ip = r - l + 1;! num[ip];ip -- );
num[ip] --;
for( int i = ip + 1;i <= r - l + 1;i ++ )
num[i] = 9;
int g = 0, R = 0;
for( int i = 1;i <= r - l + 1;i ++ )
g = R * 10 + num[i], num[++ cnt] = g / siz, R = g % siz;
g = 0;
for( int i = cnt;i;i -- )
g += len * num[i], num[i] = g % 10, g /= 10;
for( int i = 1;i <= cnt;i ++ )
g = ( g * 10 + num[i] ) % mod;
ans = ( ans + g ) % mod;
return R;
}
signed main() {
scanf( "%s", s );
int lens = strlen( s );
for( int i = 0;i < lens;i ++ )
pos[s[i] ^ 96].push_back( i ), pi[i][0] = pos[s[i] ^ 96].size() - 1;
for( int i = 0;i < lens;i ++ )
pos[0].push_back( i ), pi[i][1] = i;
scanf( "%lld", &n );
while( n -- ) {
ans = 0;
scanf( "%s", t );
int lent = strlen( t );
int l = 0, r = 0, now = lens - 1, nxt, flag = 1, f;
for( ;l < lent;l = ++ r ) {
if( t[l] == '*' ) t[l] = 96, f = 1;
else f = 0;
int id = t[l] ^ 96, siz = pos[id].size();
if( ! siz ) { flag = 0; break; }
if( pos[id][siz - 1] <= now ) nxt = pos[id][0];
else nxt = *upper_bound( pos[id].begin(), pos[id].end(), now );
if( now < nxt ) ans = ( ans + nxt - now ) % mod, now = nxt;
else ans = ( ans + lens + nxt - now ) % mod, now = nxt;
if( l + 1 < lent && isdigit( t[l + 1] ) )
for( ;r + 1 < lent && isdigit( t[r + 1] );r ++ );
if( l < r ) {
int R = calc( l + 1, r, siz, lens );
if( R ) {
nxt = pos[id][( pi[now][f] + R ) % siz];
if( now < nxt ) ans = ( ans + nxt - now ) % mod, now = nxt;
else ans = ( ans + lens + nxt - now ) % mod, now = nxt;
}
}
}
if( ! flag ) printf( "-1\n" );
else printf( "%lld\n", ans );
}
return 0;
}
T4：移动
title
solution
用心出题，用脚造数据
有的代码不考虑后退错误贪心都能AC，还有暴力过去的
考虑将时间离散化，变成每个门在哪些时间段会打开，用\(vector+pair\)存储
大概是\(n+m\)个时间段
设\(dp[i]\)表示最早到达 第\(i\)个时间段对应的门 的时间
然后用类似最短路的方法去\(dp\)转移，每次可以向两边转移（前提是这个门和转移到达的门都打开）
code
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define Pair pair < int, int >
#define inf 0x3f3f3f3f
#define MAXN 100005
struct node {
int id, x, t;
node() {}
node( int Id, int X, int T ) {
id = Id, x = X, t = T;
}
};
priority_queue < node, vector < node >, greater < node > > q;
vector < Pair > G[MAXN], tmp;
int n, m;
int id[MAXN << 1], dp[MAXN << 1];
bool cmp( Pair x, Pair y ) {
return x.first < y.first;
}
bool operator > ( node x, node y ) {
return x.t > y.t;
}
void calc( node p, int x ) {
int r = G[p.x][p.id - id[p.x]].second;
int i = lower_bound( G[x].begin(), G[x].end(), make_pair( p.t + 1, 0 ) ) - G[x].begin() - 1;
if( G[x][i].second >= p.t + 1 ) {
if( dp[id[x] + i] > p.t + 1 ) {
dp[id[x] + i] = p.t + 1;
q.push( node( id[x] + i, x, p.t + 1 ) );
}
}
i ++;
while( i < G[x].size() && G[x][i].first <= r + 1 ) {
if( dp[id[x] + i] > G[x][i].first ) {
dp[id[x] + i] = G[x][i].first;
q.push( node( id[x] + i, x, G[x][i].first ) );
}
i ++;
}
}
void solve() {
memset( dp, 0x3f, sizeof( dp ) );
q.push( node( 0, 0, 0 ) );
dp[0] = 0;
while( ! q.empty() ) {
node now = q.top(); q.pop();
if( now.t > dp[now.id] ) continue;
if( now.x > 0 ) calc( now, now.x - 1 );
if( now.x <= n ) calc( now, now.x + 1 );
}
}
int main() {
scanf( "%d %d", &n, &m );
for( int i = 1, a, b, c;i <= m;i ++ ) {
scanf( "%d %d %d", &a, &b, &c );
G[a].push_back( make_pair( b, c ) );
}
G[0].push_back( make_pair( 0, inf ) );
G[n + 1].push_back( make_pair( 0, inf ) );
id[1] = 1;
for( int i = 1;i <= n;i ++ ) {
tmp.clear();
sort( G[i].begin(), G[i].end(), cmp );
int r = -1;
for( int j = 0;j < G[i].size();j ++ ) {
if( G[i][j].first > r + 1 ) tmp.push_back( make_pair( r + 1, G[i][j].first - 1 ) );
r = max( r, G[i][j].second );
}
tmp.push_back( make_pair( r + 1, inf ) );
G[i] = tmp;
id[i + 1] = id[i] + G[i].size();
}
solve();
printf( "%d\n", dp[id[n + 1]] );
return 0;
}
