2022有道小图灵信息学水平测试（二）题解

比赛地址

K

看看有多少个数小于等于 $a_k$ 即可

#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 1010
int n, m, i, j, k, T; 
int a[N]; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	n=read(); k=read(); 
	for(i=1; i<=n; ++i) a[i]=read(); 
	for(i=1; i<=n; ++i) if(a[i]<=a[k]) ++j; 
	printf("%d", j); 
	return 0; 
}

L

显然，0全丢前面，1全丢后面

#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 100010
int n, m, i, j, k, T; 
char a[N]; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	n=read();
	scanf("%s", a+1); 
	// for(i=1; <=n; ++i) a[i]=read(); 
	sort(a+1, a+n+1); 
	for(i=1; i<=n; ++i) printf("%c", a[i]); 
	return 0; 
}

M

用个map先存起来，最后判断在不在里面即可

#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	T=read(); 
	while(T--)
	{
		map<string, int>mp; 
		string s, c; 
		n=read(); 
		for(i=1; i<=n; ++i) 
		{
			cin>>s; mp[s]=1;
			c+=s[0]; 
		}
		printf(mp[c] ? "yes\n" : "no\n"); 
	}
	return 0; 
}

N

DP，设 $dp(i,0/1)$ 代表前 $i$ 枚硬币里最后一枚是反面/正面朝上的方案数

$$\Large dp(i,0)=dp(i-1,1)\\\Large dp(i,1)=dp(i-1, 0)+dp(i-1, 1)$$

#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 40
int n, m, i, j, k, T; 
int dp[N][2]; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	n=read(); 
	dp[1][1]=dp[1][0]=1; 
	for(i=2; i<=n; ++i) 
	{
		dp[i][1]=dp[i-1][1]+dp[i-1][0]; 
		dp[i][0]=dp[i-1][1]; 
	}
	printf("%lld", dp[n][0]+dp[n][1]); 
	return 0; 
}

O

按题意所说排序即可

#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 100010
struct node
{
	int x, y, a, b, c; 
	string s; 
}a[N]; 
int n, m, i, j, k, T; 

bool cmp(node x, node y)
{
	if(x.a==y.a) {
		if(x.b==y.b) {
			if(x.c==y.c) return x.s<y.s; 
			return x.c<y.c; 
		}
		return x.b<y.b; 
	}
	return x.a<y.a; 
}

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	n=read(); 
	for(i=1; i<=n; ++i) 
	{
		m=read(); 
		for(j=1; j<=m; ++j) 
		{
			a[++k].x=i; a[k].y=j; 
			cin>>a[k].s; 
			a[k].a=read(); a[k].b=read(); a[k].c=read(); 
		}
	}
	sort(a+1, a+k+1, cmp); 
	for(i=1; i<=k; ++i) 
		cout<<a[i].s, printf(" %d %d\n", a[i].x, a[i].y); 
	return 0; 
}