2022有道小图灵信息学水平测试（一）题解

本来不好意思写这种难度的比赛的，但为了骗流量还是写了

比赛地址

A

循环找最小值

#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; 
double f, mx; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	mx=1e9; 
	for(i=1; i<=8; ++i) 
		scanf("%lf", &f), mx=min(mx, f); 
	printf("%.2lf", mx); 
	return 0; 
}

B

循环一遍，遇到数字就输出

#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); 
	for(i=1; s[i]; ++i)
	 	if(s[i]>='0' && s[i]<='9') printf("%c", s[i]), ++k; 
	if(!k) printf("0"); 
	return 0; 
}

C

每次加上弹高看看高不高过篮高

#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; 
double f; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	n=read(); 
	for(i=1; i<=n; ++i) 
	{
		scanf("%lf", &f); 
		if(f+30.7>=305) ++k; 
	}
	printf("%d", k); 
	return 0; 
}

D

不断循环题目所说过程，然后输出，到1就停止

#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); 
	n=read(); 
	while(n!=1)
	{
		if(n&1) n=n*3+1; 
		else n/=2;  
		printf("%lld ", n); 
	}
	return 0; 
}

E

统计行为1或 \(n\)，列为1或 \(n\) 的和即可

#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); 
	n=read(); m=read(); 
	for(i=1; i<=n; ++i) 
		for(j=1; j<=m; ++j) 
		{
			k=read(); 
			if(i==1 || j==1 || i==n || j==m) T+=k; 
		}
	printf("%lld", T); 
	return 0; 
}

F

两个符合条件的都输出即可

#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; 
double x, y; 

signed main()
{
//	freopen("tiaoshi.in", "r", stdin); 
//	freopen("tiaoshi.out", "w", stdout); 
	n=read(); 
	for(i=1; i<=n; ++i) 
	{
		scanf("%lf%lf", &x, &y); 
		if(x<89.9 || x>150.7) continue; 
		if(y<49.9 || y>90.34) continue; 
		++k; 
	}
	printf("%lld", k); 
	return 0; 
}

G

输出所有数的和

#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); 
	while(n=read()) m+=n; 
	printf("%d", m); 
	return 0; 
}

H

按题意所说加个判断输出即可，每个位置要么空格要么星号

#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); 
	n=read(); 
	for(i=n; i>=1; --i, printf("\n")) 
		for(j=n; j>=1; --j) 
			printf("%c", j<=i ? '*' : ' '); 
	return 0; 
}

I

从大面值到小面值不断除模即可

#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); 
	n=read(); 
	printf("%d ", n/100), n%=100; 
	printf("%d ", n/50), n%=50; 
	printf("%d ", n/20), n%=20; 
	printf("%d ", n/10), n%=10; 
	printf("%d ", n/5), n%=5; 
	printf("%d", n/1); 
	return 0; 
}

J

不算博弈的博弈，显然模4为0就输，否则赢了

#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); 
	printf(read()%4 ? "Yes" : "No"); 
	return 0; 
}