[解题报告]训练日记4：山东大学(SDU)acm新生赛

其他 2021-04-05 15:02:21 阅读次数: 0

前

牛客

题目还是偏简单的 我是零零碎碎用上课时间写的

所以当时可能处理细节不太到位吧 才提交了很多遍

但其实思路都比较简单 就是部分题有坑点吧

中

A

签到题（出题人说的）

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
using namespace std;
#define PI acos(-1)
#define fin freopen("data.txt","r",stdin);
#define LL long long
#define INF 2147483647
#define eps 1e-7
void read(int &x) {
    
    
    char c = getchar(); x = 0;
    while(c < '0' || c > '9') c = getchar();
    while(c <= '9' && c >= '0') x = x*10+c-48, c = getchar();
}

int main() {
    
    
	printf("deer");
	return 0;
}

B

有一个坑点没注意到 2102 这种数据

就是不一定满足在1和26之间就要 因为后面的0没处放

所以0一定要和前面那个一起

//a1 b2 c3 d4 e5 f6 g7 h8
//i9 j10 k11 l12 m13 n14 o15
//p16 q17 r18 s19 t20 u21 v22 w23 x24 y25 z26
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
using namespace std;
#define PI acos(-1)
#define fin freopen("data.txt","r",stdin);
#define LL long long
#define INF 2147483647
#define eps 1e-7
void read(int &x) {
    
    
    char c = getchar(); x = 0;
    while(c < '0' || c > '9') c = getchar();
    while(c <= '9' && c >= '0') x = x*10+c-48, c = getchar();
}

int len , x , top;
string s;

int main() {
    
    
//	fin
	cin>>s;
	len = s.length();	
	top = 0;
	while(1) {
    
    
		if(top!=len-1) {
    
    
			x = (s[top]-'0')*10+(s[top+1]-'0');
			if(x>=1&&x<=26&&s[top+2]!='0')
				top += 2;
			else {
    
    
				x = (s[top]-'0');
				top++;
			}			
		} else {
    
    
			x = (s[top]-'0');
			top++;
		}
		printf("%c",(char)(x+'a'-1));
		if(top>=len)
			break;
	}
	return 0;
}

C

直接统计每个方向的移动次数

然后乘每个方向对xy的操作

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
using namespace std;
#define PI acos(-1)
#define fin freopen("data.txt","r",stdin);
#define LL long long
#define INF 2147483647
#define eps 1e-7
void read(int &x) {
    
    
    char c = getchar(); x = 0;
    while(c < '0' || c > '9') c = getchar();
    while(c <= '9' && c >= '0') x = x*10+c-48, c = getchar();
}

LL num[130] , len , x , y;
string s;
//N 78 E 69 S 83 W 87

int main() {
    
    
//	fin
	cin>>s;
	len = s.length();	
	for(int i=0; i<len; i++)
		num[s[i]]++;
	x -= num[87];
	x += num[69];
	y -= num[83];
	y += num[78];
	printf("%lld %lld",x,y);
	return 0;
}

F

简单排序

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
using namespace std;
#define PI acos(-1)
#define fin freopen("data.txt","r",stdin);
#define LL long long
#define INF 2147483647
#define eps 1e-7
#define L 1005
void read(int &x) {
    
    
    char c = getchar(); x = 0;
    while(c < '0' || c > '9') c = getchar();
    while(c <= '9' && c >= '0') x = x*10+c-48, c = getchar();
}

int n , num[L] , m;

bool cmp(int a , int b) {
    
    
	return a>b;
}

int main() {
    
    
//	fin
	read(n); read(m);
	for(int i=1; i<=n; i++)
		read(num[i]);
	sort(num+1 , num+n+1 , cmp);
	printf("%d",num[m]);
	return 0;
}

H

行和列如果能除尽就去除数的最小值

原理就是稍微平移一下就可以满足条件

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
using namespace std;
typedef long long LL;
#define PI acos(-1)
#define fin freopen("data.txt","r",stdin)
#define INF 2147483647
#define eps 1e-7
#define L 1005
#define Fo(i,a,b) for(LL i=(a),_=(b); i<=_; i++)
#define Ro(i,a,b) for(LL i=(b),_=(a); i>=_; i--)
inline LL read() {
    
    
	LL x=0,f=1;char c=getchar();
    while(!isdigit(c)) {
    
    if(c=='-')f=-f;c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48ll),c=getchar();
    return x*f;
}

int n , m , ans;

int main() {
    
    
//	fin;
	n=read();m=read();
	Fo(i,2,n)
		if(n%i==0) {
    
    
			ans = i;
			break;
		}
	Fo(i,2,m)
		if(m%i==0) {
    
    
			int x=i;
			ans=min(ans,x);
			break;
		}
	printf("%d",ans);
	return 0;
}

I

暴力就可以过啊

宏定义写的有点难看……

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
using namespace std;
typedef long long LL;
#define PI acos(-1)
#define fin freopen("data.txt","r",stdin);
#define INF 2147483647
#define eps 1e-7
#define L 1005
#define Fo(i,a,b) for(LL i=(a),_=b; i<=_; i++)
#define Ro(i,a,b) for(LL i=(b),_=(a); i>=_; i--)
#define _min minn = INF
#define _maxx maxx = -1
#define umin(_,__) _ = min(_ , __)
#define umax(_,__) _ = max(_ , __)
#define pr(_) printf("%d",_)
inline LL read() {
    
    
	LL x=0,f=1;char c=getchar();
    while(!isdigit(c)) {
    
    if(c=='-')f=-f;c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48ll),c=getchar();
    return x*f;
}

int n , minn = INF , ans , maxx = -1 , num[L];

int main() {
    
    
//	fin
	n=read();
	Fo(i,1,n) num[i]=read();
	Fo(i,1,n) {
    
    
		_min;
		Fo(j,1,i)
			umin(minn,num[j]);
		umax(ans,num[i]-minn);
	}
	pr(ans);
	return 0;
}

J

用STL里string的操作模拟就行

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
using namespace std;
typedef long long LL;
#define PI acos(-1)
#define fin freopen("data.txt","r",stdin)
#define INF 2147483647
#define eps 1e-7
#define L 1005
#define Fo(i,a,b) for(LL i=(a),_=(b); i<=_; i++)
#define Ro(i,a,b) for(LL i=(b),_=(a); i>=_; i--)
inline LL read() {
    
    
	LL x=0,f=1;char c=getchar();
    while(!isdigit(c)) {
    
    if(c=='-')f=-f;c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48ll),c=getchar();
    return x*f;
}

queue<LL>z;
queue<LL>a;
LL tim;
string s;

int main() {
    
    
//	fin;
	s="";
	Fo(i,1,52) {
    
    
		int x=read();
		if(i%2!=0)
			z.push(x);
		else
			a.push(x);
	}
	while(1) {
    
    
		LL op , v;
		tim++;
		v=z.front();
		z.pop();
		op=(LL)s.find(v+'0');
		if(op==-1)
			s+=(v+'0');
		else {
    
    
			string del=s.substr(op,s.size()-op);
			s.erase(op,s.size()-op);
			int len=del.size();
			Fo(i,0,len-1) {
    
    
				int x=del[i]-'0';
				z.push(x);
			}
			z.push(v);
		}
		v=a.front();
		a.pop();
		op=(LL)s.find(v+'0');
		if(op==-1)
			s+=(v+'0');
		else {
    
    
			string del=s.substr(op,s.size()-op);
			s.erase(op,s.size()-op);
			int len=del.size();
			Fo(i,0,len-1) {
    
    
				int x=del[i]-'0';
				a.push(x);
			}
			a.push(v);
		}
		if(a.size()==0) {
    
    
			printf("zcl");
			return 0;			
		}
		if(z.size()==0) {
    
    
			printf("ajh");
			return 0;			
		}
		if(tim==1000) {
    
    
			if(z.size()>a.size())
				printf("zcl");
			if(z.size()<a.size())
				printf("ajh");
			if(z.size()==a.size())
				printf("no winner");
		}
	}
	return 0;
}

K

注意！“唯一持有”！

读题啊！

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
using namespace std;
typedef long long LL;
#define PI acos(-1)
#define fin freopen("data.txt","r",stdin)
#define INF 2147483647
#define eps 1e-7
#define L 1005
#define Fo(i,a,b) for(LL i=(a),_=(b); i<=_; i++)
#define Ro(i,b,a) for(LL i=(b),_=(a); i>=_; i--)
inline LL read() {
    
    
	LL x=0,f=1;char c=getchar();
    while(!isdigit(c)) {
    
    if(c=='-')f=-f;c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48ll),c=getchar();
    return x*f;
}

int n , m , k , v[200005] , minn=INF;

int main() {
    
    
//	fin;
	n=read(); m=read(); k=read();
	if(m>=k) {
    
    
		printf("Yes");
		return 0;
	}
	Fo(i,1,n) {
    
    
		v[i]=read();
		minn=min(minn,v[i]);
	}
	if(m<minn) {
    
    
		printf("No");
		return 0;
	}
	Fo(i,1,n)
		if(v[i]<v[i+1]&&v[i]<=m)
			m+=v[i+1]-v[i];
	 if(m>=k)
	 	printf("Yes");
	else
		printf("No");
	return 0;
}

L

其实没必要这么麻烦 完全可以按照坐标去做

但是暴力还是可以出来的

坑点就是 答案不一定在-100到100之间

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
using namespace std;
typedef long long LL;
#define PI acos(-1)
#define fin freopen("data.txt","r",stdin)
#define INF 2147483647
#define eps 1e-7
#define L 1005
#define Fo(i,a,b) for(LL i=(a),_=(b); i<=_; i++)
#define Ro(i,a,b) for(LL i=(b),_=(a); i>=_; i--)
inline LL read() {
    
    
	LL x=0,f=1;char c=getchar();
    while(!isdigit(c)) {
    
    if(c=='-')f=-f;c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48ll),c=getchar();
    return x*f;
}

struct Node {
    
    
	int x , y;
	bool operator != (Node a) {
    
    
		return a.x!=x||a.y!=y;
	}
}; 
Node n1 , n2 , n3;
int len1 , len2 , len3 , flag;

int len(Node a , Node b) {
    
    
	return pow(abs(a.x-b.x),2)+pow(abs(a.y-b.y),2);
}

void sol() {
    
    
	if(len1>len2&&len1>len3)
		flag = 3;
	if(len2>len1&&len2>len3)
		flag = 2;
	if(len3>len1&&len3>len2)
		flag = 1;
}

int main() {
    
    
//	fin;
	n1.x=read(); n1.y=read();
	n2.x=read(); n2.y=read();
	n3.x=read(); n3.y=read();
	len1=len(n1,n2); len2=len(n1,n3); len3=len(n2,n3);
	sol();
	if(flag==1) {
    
    
		Fo(i,-200,200)
			Fo(j,-200,200) {
    
    
				Node x;
				x.x=i; x.y=j;
				if(len(x,n2)==len(x,n3)&&len(x,n2)==len2&&x!=n1) {
    
    
					printf("%d %d",i,j);
					return 0;
				}
			}		
	}
	if(flag==2) {
    
    
		Fo(i,-200,200)
			Fo(j,-200,200) {
    
    
				Node x;
				x.x=i; x.y=j;
				if(len(x,n1)==len(x,n3)&&len(x,n1)==len1&&x!=n2) {
    
    
					printf("%d %d",i,j);
					return 0;
				}
			}		
	}
	if(flag==3) {
    
    
		Fo(i,-200,200)
			Fo(j,-200,200) {
    
    
				Node x;
				x.x=i; x.y=j;
				if(len(x,n2)==len(x,n1)&&len(x,n2)==len3&&x!=n3) {
    
    
					printf("%d %d",i,j);
					return 0;
				}
			}		
	}
	return 0;
}

M

一个细节：答案一定不大于26

比较容易看出来吧

就是如果s和c中都存在的话 这个字母就算一个

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
using namespace std;
typedef long long LL;
#define PI acos(-1)
#define fin freopen("data.txt","r",stdin)
#define INF 2147483647
#define eps 1e-7
#define L 1005
#define Fo(i,a,b) for(LL i=(a),_=(b); i<=_; i++)
#define Ro(i,b,a) for(LL i=(b),_=(a); i>=_; i--)
inline LL read() {
    
    
	LL x=0,f=1;char c=getchar();
    while(!isdigit(c)) {
    
    if(c=='-')f=-f;c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48ll),c=getchar();
    return x*f;
}

int n , m , t[30] , ans , x[30];
string s , c;

int main() {
    
    
//	fin;
	n=read(); m=read();
	cin>>s>>c;
	Fo(i,0,n-1)
		t[s[i]-'a'] = 1;
	Fo(i,0,m-1)
		x[c[i]-'a'] = 1;
	for(int i=0; i<26; i++)
		if(x[i]&&t[i])
			ans++;
	printf("%d",ans);
	return 0;
}

后

题目比较不错 真 新生赛

细节也很好哈哈哈 虽然我有些题写的超级麻烦吧

就这样吧 还好补完了会的题

剩下的待补 溜了溜了

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