BUCOJ周赛5题解

题目链接

A

签到题

#include <bits/stdc++.h>
using namespace std;

int a , b ;
int main()
{
    cin >> a >> b;
    if( a < b ) puts("0");
    else puts("10");
}

B

枚举即可，判断特殊情况恰好跳到x和根本跳不到x

#include<bits/stdc++.h>
using namespace std;


int main()
{
    int n , m ;
    cin >> n >> m;
    int  ans = 0;
    for( int i = 1 , x ; i <= n ; i ++ )
    {
        cin >> x;
        ans += x;
        if( ans < m ) continue;
        else if( ans == m ) cout << i + 1 << endl;
        else cout << i << endl;
        return 0;
    }
    cout << n + 1 << endl;
}

C

多画几次就会发现一定可以分成一半一半

如果(x,y)在两条对角线的焦点上有无数种情况，反之只有(x,y)与对角线交点连线这一种情况

#include <bits/stdc++.h>;
using namespace std;


double n , m , a , b;

int main()
{
    cin >> n >> m >> a >> b;
    printf("%.6lf " , n * m / 2 );
    if( a == n / 2 && b == m / 2 ) cout << 1 << endl;
    else cout << 0 << endl;
}

D

二分 前缀和

如果[l,r]满足，则[l,r+1]也满足一直到[l,n]全部都满足

#include <bits/stdc++.h>
using namespace std;

#define ll long long

const int N = 1e5 + 5; 
int n ;
ll a[N] , cnt , k ;

inline ll read()
{
    ll x = 0;
    char ch = getchar();
    while( ch < '0' || ch > '9' ) ch = getchar();
    while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
    return x;
}

int main()
{
    n = read() , k = read();
    for( int i = 1 ; i <= n ; i ++ ) a[i] = a[ i - 1 ] + read();
    for( int i = 1 , x ; i <= n ; i ++ )
    {
        if( a[n] - a[ i - 1 ] < k ) break;
        cnt += n - ( lower_bound( a + 1 , a + 1 + n , a[ i - 1 ] + k ) - a ) + 1;
    }
    cout << cnt << endl;
    return 0;
}

E

动态规划

设f[i][j]为S到i，T到j有多少个公共子串若s[i]==t[i]，i-1、j-1任意一个子串都可以

f[i][j] =f[i-1][j-1] + f[i-1][j]+f[i][j-1]-f[i-1][j-1]+1

若不相等，则是s[i]和t[j]不能同时选

f[i][j]=f[i-1][j]+f[i][j-1]-f[i-1][j-1]

#include <bits/stdc++.h>
using namespace std;

const int N = 2e3 + 50 , Mod = 1e9 + 7;
int n , m , a[N] , b[N] ;
long long f[N][N];

signed main()
{
    cin >> n >> m;
    for( int i = 1 ; i <= n ; i ++ ) cin >> a[i];
    for( int i = 1 ; i <= m ; i ++ ) cin >> b[i];
    for( int i = 1 ; i <= n ; i ++ )
    {
        for( int j = 1 ; j <= m ; j ++ )
        {
            if( a[i] == b[j] ) f[i][j] = f[ i - 1 ][j] + f[i][ j - 1 ] + 1;
            else f[i][j] = f[ i - 1 ][j] + f[i][ j - 1 ] - f[ i - 1 ][ j - 1 ];
            f[i][j] %= Mod;
        }
    }
    cout << ( f[n][m] + 1 + Mod ) % Mod << endl;
    return 0;
}

F

时间限制是100s，然后手掌面积是一个单峰函数，用三分法枚举即可

#include <bits/stdc++.h>
using namespace std;


const int N = 1e5 + 5 ;
const int dx[] = { 0 , 1 , -1 , 0 , 0 } , dy[] = { 0 , 0 , 0 , 1 , -1 };
int n ;

struct node
{
    double x , y;
    int dir;
}a[N];

inline int read()
{
    int x = 0 , f = 1;
    char 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;
}

inline int getdir()
{
    char c = getchar();
    while( c != 'R' && c != 'L' && c != 'D' && c != 'U' ) c = getchar();

    // R L D U
    // 1 2 3 4
    if( c == 'R' ) return 1;
    if( c == 'L' ) return 2;
    if( c == 'U' ) return 3;
    if( c == 'D' ) return 4;
}

inline double f( double x )
{
    double maxx = -1e10 , minx = 1e10 , maxy = -1e10 , miny = 1e10 , tx , ty;

    for( int i = 1 ; i <= n ; i ++ )
    {
        tx = a[i].x + x * dx[ a[i].dir ] , ty = a[i].y + x * dy[ a[i].dir ];
        maxx = max( maxx , tx ) , maxy = max( maxy ,ty ) , minx = min( minx , tx ) , miny = min( miny , ty );
    }
    return ( maxx - minx ) * ( maxy - miny );
}



int main()
{
    n = read();
    for( int i = 1 ; i <= n ; i ++ ) a[i].x = read() , a[i].y = read() , a[i].dir = getdir();
    double l = 0 , r = 1e12 , ans = 1e18;
    for( int i = 0 ; i < 300 ; i ++ )
    {
        double ml = l + ( r - l ) / 3 , mr = l + 2 * ( r - l ) / 3;
        double ansl = f( ml ) , ansr = f( mr );
        if( ansl > ansr ) l = ml;
        else r = mr ;
        ans = min( { ans , ansl , ansr } );
    }
    printf("%.6lf\n" , ans );
    return 0;
}