SMU 211204 新生训练赛

A 分糖果

做一个简单的取模，然后取min即可

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

int n , l , r ;

int main()
{
    cin >> n >> l >> r;
    r -= l/n*n , l %= n;
    cout << min( n - 1 , r ) << endl;
}

B方格取数

裸暴力的20分为什么没有人写？？？

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

const int dx[] = { 0 , 1 , -1 } , dy[] = { 1 , 0 , 0 };
const int N = 1e3+5;
int n , m , a[N][N] , ans = - 1e9;
bool vis[N][N];

inline 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;
}

inline void dfs( int x , int y , int sum )
{
    if( x == n && y == m )
    {
        ans = max( ans , sum );
        return ;
    }
    for( int i = 0 , fx , fy ; i < 3; i ++ )
    {
        fx = x + dx[i] , fy = y + dy[i];
        if( fx < 1 || fx > n || fy < 1 || fy > m || vis[fx][fy] ) continue;
        vis[fx][fy] = 1 ;
        dfs( fx , fy , sum + a[fx][fy] );
        vis[fx][fy] = 0;
    }
    return ;
}

int main()
{
    n = read() , m = read();
    for( int i = 1 ; i <= n ; i ++ )
        for( int j = 1 ; j <= m ; j ++ ) a[i][j] = read();
    vis[1][1] = 1 ;
    dfs( 1 , 1 , a[1][1] );
    cout << ans << endl;
    return 0;
}

正解就是DP，但是可以写成记忆化搜索，f[i][j][0]表示从上面过来，f[i][j][0]表从从下面过来

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

const LL INF = -1e18;
int n, m;
LL w[1005][1005], f[1005][1005][2];

inline LL dfs(int x, int y, int from)
{
    if (x < 1 || x > n || y < 1 || y > m) return INF;
    if (f[x][y][from] != INF ) return f[x][y][from];
    if (from == 0) f[x][y][from] = max(dfs( x+1 , y , 0 ) , max ( dfs( x , y-1 , 0), dfs( x , y-1 , 1 ) ) ) + w[x][y];
    else f[x][y][from] = max ( dfs( x-1 , y , 1) , max ( dfs( x , y-1 , 0), dfs( x, y-1, 1) ) ) + w[x][y];
    return f[x][y][from];
}


int main() {
	cin >> n >> m;

	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j) cin >> w[i][j] , f[i][j][0] = f[i][j][1] = INF;
    f[1][1][0] = f[1][1][1] = w[1][1];
	cout << dfs(n, m, 1) << endl;
	return 0;
}

C 直播获奖

直接用桶排序，统计分数线即可

#include<bits/stdc++.h>
using namespace std;
int t[605];
int n,w;
int main()
{
	int x;
	cin>>n>>w;
	for(int i=1;i<=n;i++)
	{
		cin>>x;
		t[x]++;
		int sum=0;
		for(int j = 600 ; j >= 0 ; j -- )
		{
			sum+=t[j];
			if(sum >= max( 1 , i * w / 100 ) )
			{
				printf( "%d " , j );
				break;
			}
		}
	}
	return 0;
 }

D 数字游戏

为什么不用bitset

#include<bits/stdc++.h>
using namespace std;
bitset<8> s;
int main()
{
	cin >> s;
	cout << s.count() << endl;
}

E 小凯的疑惑

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

int main()
{
	long long a , b;
	cin >> a >> b;
	cout << a * b - a - b << endl;
}	

F 奶酪

枚举两个点，如果距离小于2r就把两个点用并查集连起来

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


const int N = 1005;
LL t,n,h,r,father[N] = {};
LL sa[N] = {}, sb[N] = {};

struct node
{
	LL x,y,z;
}ball[N] = {};

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

inline LL getfather(int x)
{
	if(x == father[x]) return x;
	return father[x] 		=  getfather(father[x]);
}

inline LL dis(int a,int b)
{
	return 	(ball[a].x-ball[b].x)*(ball[a].x-ball[b].x) + (ball[a].y-ball[b].y)*(ball[a].y-ball[b].y) + (ball[a].z-ball[b].z)*(ball[a].z-ball[b].z);
}

inline void work()
{
	sa[0] = sb[0] = 0;
	n = read(); h = read(); r = read();
	for(register int i = 1;i <= n;i++)
	{
		father[i] = i;
		ball[i].x = read(); ball[i].y = read(); ball[i].z = read();
		if(ball[i].z-r <= 0) sa[++sa[0]] = i;
		if(ball[i].z+r >= h) sb[++sb[0]] = i;
	}
	for(register int i = 1;i < n;i++)
	{
		for(register int j = i+1;j <= n;j++)
		{
			if(dis(i,j) > 4*r*r) continue;
			father[getfather(i)] = getfather(j);
		}
	}
	for(register int i = 1;i <= sa[0];i++)
	{
		for(register int j = 1;j <= sb[0];j++)
		{
			if(getfather(sa[i]) == getfather(sb[j]))
			{
				puts("Yes");
				return ;
			}
		}
	}
	puts("No");
	return ;
}


int main()
{
	t = read();
	while(t--) work();
	return 0;
}

G RESETO

#include <bits/stdc++.h>
using namespace std;
int n , k , w = 0;
bool s[1005];
int main()
{
	cin>>n>>k;
	memset(s,1,sizeof(s));
	for(int i=2;i<=n;i++)
	{
		if(s[i])
		{
			for(int j=i;j<=n;j+=i)
			{
				if(s[j])
				{
					w++ , s[j] = 0;
					if(w==k) cout<<j , exit(0);
				}
			}
		}
	}
}