本文搬运自本人初中博客园博客，若图片加载不出来，可到原文查看：https://www.cnblogs.com/zhangtingxi/p/16532595.html

比赛地址

P

image

有点恶心

按顺序先填最左，然后斜着，然后向下，然后再斜着，填过不填就是

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
#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 18
int n, m, i, j, k, T; 
int a[N][N]; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	n=read(); 
	for(i=1; i<=n; ++i) a[i][1]=++k; 
	for(i=n, j=1; i>=1; --i, ++j)
		if(!a[i][j]) a[i][j]=++k; 
	for(i=1; i<=n; ++i)
		if(!a[i][n]) a[i][n]=++k; 
	for(i=n, j=n; i>=1; --i, --j)  
		if(!a[i][j]) a[i][j]=++k; 
	for(i=1; i<=n; ++i, printf("\n"))
		for(j=1; j<=n; ++j) 
			if(a[i][j]) printf("%5lld", a[i][j]); 
			else printf("     "); 
	return 0; 
}

Q

image

显然从后往前维护元音数量，遇到辅音就加上当前元音数量

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
#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 1000010
int n, m, i, j, k, T; 
char s[N]; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	scanf("%s", s+1); n=strlen(s+1); 
	for(i=n; i>=1; --i)
	{
		if(s[i]=='a' || s[i]=='e' || s[i]=='i' || s[i]=='o' || s[i]=='u') ++k; 
		else m+=k; 
	}
	printf("%lld\n", m); 
	return 0; 
}

R

image

不能理解为什么答案要用复杂的二分，明明一个map就好了

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
#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 1000010
int n, m, i, j, k, T; 
int a[N]; 
map<int, int>mp; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	n=read(); m=read(); 
	for(i=1; i<=n; ++i) a[i]=read(); 
	for(i=1; i<=m; ++i) mp[read()]=1; 
	for(i=1, k=0; i<=n; ++i) 
		if(mp[a[i]]) k=1, printf("%lld ", a[i]); 
	if(!k) printf("No"); 
	return 0; 
}

S

image

题目都告诉你是记忆化递归了，直接照着打就是了

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
#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 25
int n, m, i, j, k, T; 
int f[N][N]; 

int w(int x, int y)
{
	if(x<=0 || y<=0) return 1; 
	if(x>20 || y>20) return w(20, 20); 
	if(f[x][y]!=-1) return f[x][y]; 
	if(x<=y) return f[x][y]=w(x, y-1)+w(x, y-2); 
	else return f[x][y]=w(x-1, y)+w(x-2, y); 
}

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	memset(f, -1, sizeof(f)); 
	n=read(); m=read(); 
	while(n!=-1 || m!=-1) 
		printf("%lld\n", w(n, m)), n=read(), m=read(); 
	return 0; 
}

T

image

image

**出题人公开赛考高精

知道高精就行了，套个模板直接走人

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
#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 1000
int n, m, i, j, k, T; 
int a[N], b[N]; 
char s[N]; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	// k=read(); printf("%lld", k*k-k); 
	scanf("%s", s+1); n=strlen(s+1); 
	for(i=1, j=n; i<=n; ++i, --j) a[i]=s[j]-'0'; 
	for(i=1; i<=n; ++i) 
		for(j=1; j<=n; ++j) 
			b[i+j-1]+=a[i]*a[j]; 
	m=2*n-1; 
	for(i=1; i<=m; ++i) 
	{
		if(b[i]>=10) b[i+1]+=b[i]/10, b[i]%=10; 
		if(b[i+1]) m=max(m, i+1); 
	}
	while(!b[m] && m) --m; 
	for(i=1; i<=n; ++i) b[i]-=a[i]; 
	for(i=1; i<=m; ++i)
		if(b[i]<0) b[i]+=10, b[i+1]--; 
	while(!b[m] && m) --m; 
	// printf("%lld\n", m); 
	if(!m) printf("0"); 
	else {
		for(i=m; i>=1; --i) printf("%lld", b[i]); 
	}
	return 0; 
}