A. 小斧头
\(O(N^3)\) 20 points
暴力枚举左右端点，然后暴力求区间最值
#include <bits/stdc++.h>
using namespace std;
#define int long long
int read(){...}
int32_t main() {
int n = read() , res = 0;
vector<int> a(n) , b(n);
for( auto & i : a ) i = read();
for( auto & i : b ) i = read();
for( int i = 0 ; i < n ; i ++ ){
for( int j = i ; j < n ; j ++ ){
int maxA = INT_MIN , maxB = INT_MIN;
for( int k = i ; k <= j ; k ++ )
maxA = max( maxA , a[k] ) , maxB = max( maxB , b[k] );
if( maxB >= maxA ) res ++;
}
}
cout << res << "\n";
return 0;
}
\(O(N^2)\) 20 points
依旧是枚举左右端点，但是最值改为求前缀最值。但实际上这个优化微乎其微。
#include <bits/stdc++.h>
using namespace std;
#define int long long
int read(){...}
int32_t main() {
int n = read() , res = 0;
vector<int> a(n) , b(n);
for( auto & i : a ) i = read();
for( auto & i : b ) i = read();
for( int i = 0 ; i < n ; i ++ ){
int maxA = INT_MIN , maxB = INT_MIN;
for( int j = i ; j < n ; j ++ ){
maxA = max( maxA , a[j] ) , maxB = max( maxB , b[j] );
if( maxB >= maxA ) res ++;
}
}
cout << res << "\n";
return 0;
}
\(O(N\log N)\) 70 points
枚举每一个点，求他作为最大值的区间范围。因为范围越大，最大值一定越大，所以满足单调性，可以使用二分来查找。区间最值可以使用ST表实现。
#include <bits/stdc++.h>
using namespace std;
#define int long long
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
int32_t main() {
int n = read();
vector<int> a(n+1) , b(n+1);
for( int i = 1 ; i <= n ; i ++ ) a[i] = read();
for( int i = 1 ; i <= n ; i ++ ) b[i] = read();
vector<int> lg2( n+1 );
lg2[0] = -1;
for( int i = 1 ; i <= n ; i ++ ) lg2[i] = lg2[i>>1] + 1;
int lgN = lg2[n] + 1;
vector<vector<int>> f(n+1,vector<int>(lgN,0));
for( int i = 1 ; i <= n ; i ++ )
f[i][0] = max( a[i] , b[i] );
for( int j = 1 ; j <= lgN ; j ++ )
for( int i = 1 ; i + (1<<j)-1 <= n ; i ++ )
f[i][j] = max( f[i][j-1] , f[i+(1<<j-1)][j-1] );
auto search = [f,lg2,lgN]( int l , int r ){
if( r < l ) return 0ll;
int s = lg2[r-l+1];
return max( f[l][s] , f[r-(1<<s) + 1][s] );
};
int res = 0;
for( int i = 1 , l , r , m , L , R ; i <= n ; i ++ ){
L = R = i;
l = 1 , r = i-1;
while ( l <= r ){
m = ( l + r ) >> 1;
if( search( m , i-1 ) < b[i] ) L = m , r = m - 1;
else l = m + 1;
}
l = i + 1 , r = n;
while( l <= r ){
m = ( l + r ) >> 1;
if( search( i , m ) <= b[i] ) R = m , l = m + 1;
else r = m - 1;
}
if( search(L,R) > b[i] ) continue;
L = i - L + 1 , R = R - i + 1;
res += L * R;
}
cout << res;
return 0;
}
\(O(N)\) 100points
\(f_k\)表示满足条件且\(j=k\)的\((i,j)\)对数量。然后要们找到\(last_k\)，\(last_k\)表示从\(k\)开始向前\(A\)或\(B\)的最大值发生改变的地方。
\[f_k=f_{lask_k} + ( B_k\ge A_k)\times(k-last_k) \]\(last_k\)可以用单调栈求出，所以整体复杂度\(O(N)\)
#include <bits/stdc++.h>
using namespace std;
#define int long long
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
int32_t main() {
int n = read();
vector<int> a(n+1) , b(n+1) , c(n+1);
for( int i = 1 ; i <= n ; i ++ ) a[i] = read();
for( int i = 1 ; i <= n ; i ++ ) b[i] = read();
for( int i = 1 ; i <= n ; i ++ ) c[i] = max( a[i] , b[i] );
stack<int> stk1 , stk2;
vector<int> f(n+1);
int res = 0;
for( int i = 1 , lst ; i <= n ; i ++ ){
while( !stk1.empty() && a[i] > c[stk1.top()] ) stk1.pop();
while( !stk2.empty() && b[i] > c[stk2.top()] ) stk2.pop();
if( b[i] >= a[i] ){
if(stk2.empty()) lst = 0;
else lst = stk2.top();
f[i] = f[lst] + i - lst;
}else{
if( stk1.empty() ) lst = 0;
else lst = stk1.top();
f[i] = f[lst];
}
res += f[i];
stk1.push(i) , stk2.push(i);
}
cout << res;
return 0;
}
B. Floor or xor ?
\(O(N^4)\) 20 points
直接枚举暴力一下就好
#include <bits/stdc++.h>
using namespace std;
#define int long long
int read(){
int x = 0 , f = 1 , ch = getchar();
while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
if( ch == '-' ) f = -1 , ch = getchar();
while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
return x * f;
}
int32_t main(){
int n = read() , T = read() , mod = read();
vector<int> a(n);
for( auto & i : a ) i = read();
int res = 0;
for( int i = 0 ; i < n ; i ++ )
for( int j = 0 ; j < n ; j ++ )
for( int k = 0 ; k < n ; k ++ )
for( int l = 0 ; l < n ; l ++ ){
if( a[i] / a[j] + a[k] / a[l] == T ) res ++;
}
cout << res;
return 0;
}
\(O(N^2)\) 40 points
统计每种\(\frac {A_i}{ A_j}\)，出现的次数，然后\(O(N)\)计算答案。
#include <bits/stdc++.h>
using namespace std;
#define int long long
int read(){
int x = 0 , f = 1 , ch = getchar();
while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
if( ch == '-' ) f = -1 , ch = getchar();
while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
return x * f;
}
int32_t main(){
int n = read() , T = read() , mod = read();
vector<int> a(n);
for( auto & i : a ) i = read();
vector<int> cnt(T+1);
for( int i = 0 , t ; i < n ; i ++ )
for( int j = 0 ; j < n ; j ++ ){
t = a[i] / a[j];
if( t > T ) continue;
cnt[t] ++;
}
int res = 0;
for( int i = 0 , j = T ; i <= T ; i ++ , j -- ){
if( cnt[i] == 0 || cnt[j] == 0 ) continue;
res = ( res + cnt[i]*cnt[j] % mod ) % mod;
}
cout << res;
return 0;
}
\(O(5\times 10^5 \sqrt{5\times 10^5})\) 70 points
考虑40分做法中，复杂度实际上是在统计cnt数组。对于\(\left \lfloor \frac {A_x}{A_y} \right \rfloor\)，如果\(A_x\)确定了，至多只会有\(2\sqrt{ A_x}\)种值。我们可以通过枚举值，计算\(A_y\)的取值范围，以此来统计\(A_y\)的数量以此来计算贡献。\(A_y\)的数量可以用前缀和来计算。
代码不想写了。
\(O(5\times10^5\ln(5\times 10 ^5))\) 100 points
#include <bits/stdc++.h>
using namespace std;
#define int long long
int read(){
int x = 0 , f = 1 , ch = getchar();
while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
if( ch == '-' ) f = -1 , ch = getchar();
while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
return x * f;
}
int32_t main(){
int n = read() , T = read() , mod = read();
vector<int> a(n);
for( auto & i : a ) i = read();
int m = *max_element(a.begin(),a.end());
vector<int> pre(m+1);
for( int i : a ) pre[i] ++;
for( int i = 1 ; i <= m ; i ++ ) pre[i] += pre[i-1];
auto query =[m,pre]( int l , int r ){
if( r < l ) return 0ll;
if( l == 0 ) return pre[r];
return pre[r] - pre[l-1];
};
vector<int> cnt(T+1);
for( int i = 1 , l , r ; i <= m ; i ++ ){
if( query(i,i) == 0 ) continue;
for( int j = 0 ; j <= T && i*j <= m ; j ++ ){
l = j*i , r = min( m , l + i - 1 );
cnt[j] = ( cnt[j] + query( i , i ) * query( l , r ) ) % mod;
}
}
int res = 0;
for( int i = 0 , j = T ; i <= T ; i ++ , j -- ){
if( cnt[i] == 0 || cnt[j] == 0 ) continue;
res = ( res + cnt[i]*cnt[j] % mod ) % mod;
}
cout << res;
return 0;
}
C. 又是一道构造题
When do we know for certain that no solution exists?
If a solution exists, then pretty much any greedy approach works.
首先\(ans[i][j]\)都一定只会最为因子在\(a[i]\)和\(b[j]\)中出现一次，所以有解的条件是\(\Pi a_i = \Pi b_i\)
贪心的策略就是
- 一开始\(ans[i][j]\)全取1
- 如果\(a[i]>1 ,b[j]>1\) ，\(ans[i][j]=\gcd( a[i],b[j])\)
- \(a[i],b[j]\)都除以\(ans[i][j]\)
操作完后判断一下\(a[i],b[j]\)是否全为\(1\)，是就输出答案。
#include <bits/stdc++.h>
using namespace std;
#define int long long
int read(){
int x = 0 , f = 1 , ch = getchar();
while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
if( ch == '-' ) f = -1 , ch = getchar();
while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
return x * f;
}
void solve(){
int n = read() , m = read();
vector<int> a(n) , b(m);
for( auto & i : a ) i = read();
for( auto & i : b ) i = read();
vector<vector<int>> ans(n , vector<int>(m , 1 ) );
for( int i = 0 , d ; i < n ; i ++ )
for( int j = 0 ; j < m ; j ++ ){
d = gcd( a[i] , b[j] );
ans[i][j] = d , a[i] /= d , b[j] /= d;
}
if( *max_element(a.begin(),a.end()) > 1 || *max_element(b.begin(),b.end()) > 1 )
return cout << "-1\n", void();
for( int i = 0 ; i < n ; i ++ )
for( int j = 0 ; j < m ; j ++ )
printf("%lld%c" , ans[i][j] , " \n"[j==m-1] );
return;
}
int32_t main(){
for( int T = read() ; T ; T -- )
solve();
return 0;
}