abc 247 题解
作者:互联网
A - Move Right
给一个四位的二进制,输出右移一位的结果
#include <bits/stdc++.h>
using namespace std;
int n;
string s ;
int main()
{
cin >> s;
cout << '0';
for( int i = 0 ; i <= 2 ; i ++ )
cout << s[i];
}
B - Unique Nicknames
每一个人都有一个姓和一个名,要给每个人取一个昵称
昵称要满足两个条件,首先昵称必修与姓或名相同,其次昵称不能与其他人的姓或名相同
实际上把所有的所有的名和姓塞入到一个 map 中,只要一个人的名或姓是唯一的那么他就是可以起一个昵称
注意的是,会有人的名和姓相同
#include <bits/stdc++.h>
using namespace std;
int n;
map<string , int> st;
string a[105] , b[105];
int main()
{
cin >> n;
for( int i = 1 ; i <= n ; i ++ ) {
cin >> a[i] >> b[i];
st[a[i]] ++ ;
if( a[i] != b[i] ) st[b[i]] ++;
}
for( int i = 1 ; i <= n ; i ++ )
{
if( st[a[i]] == 1 || st[b[i]] == 1) continue;
printf("No\n");
return 0;
}
printf("Yes\n");
return 0;
}
C - 1 2 1 3 1 2 1
题目给定了一个构造字符串的方法,\(s_i=s_{i-1}is_{i-1}\)
因为 n 很小直接构造就好
#include <bits/stdc++.h>
using namespace std;
int n;
string s[20];
int main()
{
cin >> n;
s[1] = "1";
for( int i = 2 ; i <= n ; i ++ )
s[i] = s[i-1] + " " + to_string(i) + " " + s[i-1];
cout << s[n];
return 0;
}
D - Cylinder
首先给一个队列,然后有两种操作
- 插入 c 个 x
- 取出 c 个数并求和
如果单纯的模拟的话会 t,所以队列中存的是一个pair
表示数和数的个数
然后每次我们取出一个pair
如果有剩余的数就插回到队头,所以这里要用到双端队列
#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll n , cnt;
deque< pair<ll,ll> > dq;
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;
}
int main() {
n = read();
for( ll op , m , x , c ; n ; n -- ) {
op = read();
if (op == 1)
x = read(), c = read(), dq.push_back({x, c});
else {
m = read(), cnt = 0;
while (m) {
x = dq.front().first, c = dq.front().second, dq.pop_front();
if (c <= m)
cnt += x * c, m -= c;
else
cnt += x * m , dq.push_front({x, c - m}) , m = 0;
}
printf("%lld\n", cnt);
}
}
return 0;
}
E - Max Min
给定长度为 n 的序列,问有多少个区间的最大值最小值是 x 和 y
首先枚举一下左端点,然后二分出符合条件的最小值和最大值即可
那么如何判区间时候符合条件呢?实际上就是用线段树来维护一下区间最值就好了
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+5;
int n , maxx , miny , a[N] ;
ll cnt ;
struct Node{
int l , r , maxval , minval;
Node *left , *right;
Node( int l , int r , int maxval , int minval , Node * left , Node * right ) : l(l) , r(r) , maxval(maxval) , minval(minval) , left(left) , right(right){};
} *root;
Node * build( int l , int r )
{
if( l == r ) return new Node( l , r , a[l] , a[l] , 0 , 0 );
int mid = ( l + r ) >> 1;
Node * left = build( l , mid ) , * right = build( mid + 1 , r );
return new Node( l , r , max( left -> maxval , right -> maxval) , min( left -> minval , right -> minval ) , left , right );
}
void query( int l , int r , Node * cur , int &maxval , int &minval )
{
if( l == cur -> l && r == cur ->r )
{
maxval = max( maxval , cur -> maxval ) , minval = min( minval , cur -> minval );
return ;
}
int mid = ( cur -> l + cur -> r ) >> 1;
if( l > mid ) query( l , r , cur -> right , maxval , minval );
else if( r <= mid ) query( l , r , cur -> left , maxval , minval );
else query( l , mid , cur -> left , maxval , minval ) , query( mid + 1 , r , cur -> right , maxval , minval );
}
ll erfen1( int x ){
int l = x , r = n , mid , res = -1, maxval , minval;
while( l <= r ){
mid = (l+r)>>1 , maxval = -1 , minval = 0x7fffffff;
query( x , mid , root , maxval , minval );
if( maxval >= maxx && minval <= miny ) res = mid , r = mid - 1;
else l = mid + 1;
}
if( res == -1 ) return -1;
maxval = -1 , minval = 0x7fffffff , query( x , res , root , maxval , minval );
if(maxval == maxx && minval == miny ) return res;
return -1;
}
ll erfen2( int x ){
int l = x , r = n , mid , res = -1, maxval , minval;
while( l <= r ){
mid = (l+r)>>1 , maxval = -1 , minval = 0x7fffffff;
query( x , mid , root , maxval , minval );
if( maxval <= maxx && minval >= miny ) res = mid , l = mid + 1;
else r = mid - 1;
}
if( res == -1 ) return -1;
maxval = -1 , minval = 0x7fffffff , query( x , res , root , maxval , minval );
if(maxval == maxx && minval == miny ) return res;
return -1;
}
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;
}
int main() {
n = read() , maxx = read() , miny = read();
for( int i = 1 ; i <= n ; i ++ ) a[i] = read();
root = build(1,n);
for( ll i = 1 , l , r ; i <= n ; i ++ ){
l = erfen1( i ) , r = erfen2(i);
if( l != -1 && r != -1 ) cnt += r - l + 1;
}
cout << cnt << endl;
return 0;
}
标签:maxval,abc,minval,int,题解,mid,247,ch,right 来源: https://www.cnblogs.com/PHarr/p/16306159.html