4
2
2016
2

April Fool Contest(铪?)

最近一两天做CF的April Fool Contest
感觉非常好玩 实力浪费人生

A. Mysterious numbers - 1
把第二个数字倒一下加上第一个就可以了
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#include<bitset>
#include<vector>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Drp(x,a,b) for (int x=a;x>=b;x--)
#define Cross(x,a) for (int x=head[a];~x;x=next[x])
#define ll long long
#define INF (1<<29)
#define mk(a,b) make_pair(a,b)
#define fr first
#define sc second
using namespace std;
inline ll IN(){
	ll x=0;int ch=getchar(),f=1;
	while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
	if (ch=='-'){f=-1;ch=getchar();}
	while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
inline void OUT(ll x){
	if (x<0) putchar('-'),x=-x;
	if (x>=10) OUT(x/10),putchar(x%10+'0');
		else putchar(x+'0');
}
int main(){
	int a=IN(),b=IN(),cnt;
	char c[29];
	cnt=sprintf(c,"%d",b);
	reverse(c,c+cnt);
	sscanf(c,"%d",&b);
	OUT(a+b);
}
B. Star
题意就是求像图中那样的$n$层的star由几个球构成
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#include<bitset>
#include<vector>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Drp(x,a,b) for (int x=a;x>=b;x--)
#define Cross(x,a) for (int x=head[a];~x;x=next[x])
#define ll long long
#define INF (1<<29)
#define mk(a,b) make_pair(a,b)
#define fr first
#define sc second
using namespace std;
inline ll IN(){
	ll x=0;int ch=getchar(),f=1;
	while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
	if (ch=='-'){f=-1;ch=getchar();}
	while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
inline void OUT(ll x){
	if (x<0) putchar('-'),x=-x;
	if (x>=10) OUT(x/10),putchar(x%10+'0');
		else putchar(x+'0');
}
int main(){
	int n=IN(),Ans=1;
	Rep(i,1,n-1) Ans+=i*12;
	OUT(Ans);
}
C. A Piece of Cake
给跪了……这坨不知道在干啥像烹饪指南一样的东西,
原来是一个叫Chef的编程语言……
意思就是算$\sum_{i=0}^n{i*a_i}$
打个差不多意思的代码就好了。。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#include<bitset>
#include<vector>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Drp(x,a,b) for (int x=a;x>=b;x--)
#define Cross(x,a) for (int x=head[a];~x;x=next[x])
#define ll long long
#define INF (1<<29)
#define mk(a,b) make_pair(a,b)
#define fr first
#define sc second
using namespace std;
inline ll IN(){
	ll x=0;int ch=getchar(),f=1;
	while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
	if (ch=='-'){f=-1;ch=getchar();}
	while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
inline void OUT(ll x){
	if (x<0) putchar('-'),x=-x;
	if (x>=10) OUT(x/10),putchar(x%10+'0');
		else putchar(x+'0');
}
int main(){
	int Ans=0,Cnt=0,n;
	while (~scanf("%d",&n)) Ans+=n*Cnt++;
	OUT(Ans);
}
D. Broken checker
还是给跪了……人力二分
题意就是总共5个点,$1 \sim 5$ 每个数字对应一个答案,
答案的范围是 $1 \sim 3$
我是这样的:
第一次:如果输入>3就让它RE,如果<3就让它TLE,否则输出1
然后直接TLE 2了~非常开森 说明3的答案是1
第二次:输入3就输出1(类似的下面全部省略)如果>3直接RE
如果=2直接MLE,如果是1输出3
然后WA2 说明第二个点输入是1,且1的答案不是3
第三次:输入1输出2,如果输入5输出3,
如果输入4直接RE,如果输入2就MLE
然后RE3 说明1的答案是2,第三个点输入是4
第四次、第五次:枚举了一下4输出1或输出3全都WA3了,说明4的答案是2
第六次:如果输入5输出3,如果输入2就RE
然后RE4,说明第四个点是2
第七次:输入2就输出3,否则就是输入5,也输出3
然后WA5,说明2的答案是3,5的答案不是3
第八次:输入5输出2,WA5,说明5的答案是1
第九次:Accept(热泪盈眶)
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#include<bitset>
#include<vector>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Drp(x,a,b) for (int x=a;x>=b;x--)
#define Cross(x,a) for (int x=head[a];~x;x=next[x])
#define ll long long
#define INF (1<<29)
#define mk(a,b) make_pair(a,b)
#define fr first
#define sc second
using namespace std;
inline ll IN(){
	ll x=0;int ch=getchar(),f=1;
	while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
	if (ch=='-'){f=-1;ch=getchar();}
	while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
inline void OUT(ll x){
	if (x<0) putchar('-'),x=-x;
	if (x>=10) OUT(x/10),putchar(x%10+'0');
		else putchar(x+'0');
}
int n;
int main(){
	n=IN();
	if (n==3) return puts("1"),0;
	if (n==1) return puts("2"),0;
	if (n==4) return puts("2"),0;
	if (n==2) return puts("3"),0;
	if (n==5) return puts("1"),0;
}
E. MYSTERIOUS LANGUAGE
猜语言并且用那种语言编程输出它的名字……
讲道理直接CE一次看编译信息百度就好了
然后不想猜了直接看了题解
题解说是INTERCAL(什么鬼东西)
这题好像萎一点直接用C++输出就好了
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#include<bitset>
#include<vector>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Drp(x,a,b) for (int x=a;x>=b;x--)
#define Cross(x,a) for (int x=head[a];~x;x=next[x])
#define ll long long
#define INF (1<<29)
#define mk(a,b) make_pair(a,b)
#define fr first
#define sc second
using namespace std;
inline ll IN(){
	ll x=0;int ch=getchar(),f=1;
	while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
	if (ch=='-'){f=-1;ch=getchar();}
	while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
inline void OUT(ll x){
	if (x<0) putchar('-'),x=-x;
	if (x>=10) OUT(x/10),putchar(x%10+'0');
		else putchar(x+'0');
}
int main(){
	puts("INTERCAL");
}
F. ucyhf
这题题面加密了一下
比较简单,每个字母向左移10位就可以了
直接放题面咯:
an emirp is a prime number whose reverse is a different prime number. your task is to find the nth emirp.
the input consists of a single integer n (1 ≤ n ≤ 11184) — the one-based index of the emirp to find.
output a single number.
筛一筛然后乱搞一下就好了
最大的是999983所以只处理出小于1000000的就可以了
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#include<bitset>
#include<vector>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Drp(x,a,b) for (int x=a;x>=b;x--)
#define Cross(x,a) for (int x=head[a];~x;x=next[x])
#define ll long long
#define INF (1<<29)
#define mk(a,b) make_pair(a,b)
#define fr first
#define sc second
using namespace std;
inline ll IN(){
	ll x=0;int ch=getchar(),f=1;
	while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
	if (ch=='-'){f=-1;ch=getchar();}
	while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
inline void OUT(ll x){
	if (x<0) putchar('-'),x=-x;
	if (x>=10) OUT(x/10),putchar(x%10+'0');
		else putchar(x+'0');
}
const int N=1000000;
int Mn[N+5],P[100000],Cnt;
char s[100];
int main(){
	Rep(i,2,N){
		if (!Mn[i]) P[Mn[i]=++Cnt]=i;
		Rep(j,1,Cnt){
			if (i*P[j]>N) break;
			Mn[i*P[j]]=j;
			if (j==Mn[i]) break;
		}
	}
	int n=IN(),Nw=0;
	Rep(i,1,Cnt){
		int len=sprintf(s,"%d",P[i]),Tmp;
		reverse(s,s+len);
		sscanf(s,"%d",&Tmp);
		if (Tmp==P[i]) continue;
		int pos=lower_bound(P+1,P+Cnt+1,Tmp)-P;
		if (P[pos]==Tmp){
			Nw++;
			if (Nw==n) return OUT(P[i]),0;
		}
	}
}
G. Mysterious numbers - 2
又是看样例猜题意系列……
$a1和a2$是一个类Fibonacci数列的第0项和第1项
求第$a3$项
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#include<bitset>
#include<vector>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Drp(x,a,b) for (int x=a;x>=b;x--)
#define Cross(x,a) for (int x=head[a];~x;x=next[x])
#define ll long long
#define INF (1<<29)
#define mk(a,b) make_pair(a,b)
#define fr first
#define sc second
using namespace std;
inline ll IN(){
	ll x=0;int ch=getchar(),f=1;
	while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
	if (ch=='-'){f=-1;ch=getchar();}
	while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
inline void OUT(ll x){
	if (x<0) putchar('-'),x=-x;
	if (x>=10) OUT(x/10),putchar(x%10+'0');
		else putchar(x+'0');
}
int a[25];
int main(){
	a[0]=IN(),a[1]=IN();
	int Xs=IN();
	Rep(i,2,Xs) a[i]=a[i-1]+a[i-2];
	OUT(a[Xs]);
}
H. A polyline
题意就是求 $a$ 阶希尔伯特曲线从左下角开始走,走 $b$ 步的空间坐标位置
我们发现,如果把 $a$ 阶希尔伯特曲线分成4个 $a-1$ 阶的,遍历顺序是这样的:
然后如果 $a-1$ 阶再分成四块,遍历顺序是这样的:
然后 $a-2$ 阶和 $a$ 阶是一样的,以此类推。
于是可以分治。
一开始假设阶数都是偶数,做起来比较方便。
如果是奇数的话最后横纵坐标交换一下就好了。
我们定义一个类型为 $pair<int,int>$ 的函数$Solve(a,b)$,
表示在 $a$ 阶曲线上走 $b$ 步的坐标
首先把遍历四块中一块的步数算出来称为$step$,如果$b<step$,
那么说明在左下角,直接返回$Solve(a-1,b)$
否则如果$b<2*step$,先计算出$Solve(a-1,b-step)$,
再根据 $a$ 的奇偶性决定它在左上角还是左下角
否则的话,我们发现遍历的第一第二块和第三第四块是对称的,
那么我们就可以当做它从最终所在坐标开始往回走。
先计算出$Solve(a,4*step-1-b)$,然后根据 $a$ 的奇偶性减一下就好了。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#include<bitset>
#include<vector>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Drp(x,a,b) for (int x=a;x>=b;x--)
#define Cross(x,a) for (int x=head[a];~x;x=next[x])
#define ll long long
#define INF (1<<29)
#define PII pair<int,int>
#define mk(a,b) make_pair(a,b)
#define fr first
#define sc second
using namespace std;
inline ll IN(){
	ll x=0;int ch=getchar(),f=1;
	while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
	if (ch=='-'){f=-1;ch=getchar();}
	while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
inline void OUT(ll x){
	if (x<0) putchar('-'),x=-x;
	if (x>=10) OUT(x/10),putchar(x%10+'0');
		else putchar(x+'0');
}
PII Solve(int Cs,int Nb){
	if (Cs==1){
		if (Nb==0) return mk(0,0);
		if (Nb==1) return mk(1,0);
		if (Nb==2) return mk(1,1);
		if (Nb==3) return mk(0,1);
	}
	int Zs=1<<2*Cs,Sf=Zs/4,Md=1<<Cs-1;
	if (Nb<Sf) return Solve(Cs-1,Nb);
	if (Nb<2*Sf){
		PII Res=Solve(Cs-1,Nb%Sf);
		if (Cs&1) return mk(Res.sc+Md,Res.fr);
			else return mk(Res.sc,Res.fr+Md);
	}
	PII Res=Solve(Cs,Zs-1-Nb);
	if (Cs&1) return mk(Res.fr,(Md<<1)-1-Res.sc);
		else return mk((Md<<1)-1-Res.fr,Res.sc);
}
int main(){
	int a=IN(),b=IN();
	PII Ans=Solve(a,b);
	if (a&1) swap(Ans.fr,Ans.sc);
	OUT(Ans.fr),putchar(' '),OUT(Ans.sc);
}
A. The Great Game
啦啦啦  石头剪刀布  看谁赢得多
8< 是剪刀 ()是石头 []是布
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#include<bitset>
#include<vector>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Drp(x,a,b) for (int x=a;x>=b;x--)
#define Cross(x,a) for (int x=head[a];~x;x=next[x])
#define ll long long
#define INF (1<<29)
#define mk(a,b) make_pair(a,b)
#define fr first
#define sc second
using namespace std;
inline ll IN(){
	ll x=0;int ch=getchar(),f=1;
	while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
	if (ch=='-'){f=-1;ch=getchar();}
	while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
inline void OUT(ll x){
	if (x<0) putchar('-'),x=-x;
	if (x>=10) OUT(x/10),putchar(x%10+'0');
		else putchar(x+'0');
}
char a[25],b[25];
int Ans,la,lb;
int main(){
	scanf("%s",&a);scanf("%s",&b);
	int len=strlen(a);
	for (int i=0;i<len;i+=2){
		if (a[i]=='8') la=0;if (a[i]=='(') la=1;if (a[i]=='[') la=2;
		if (b[i]=='8') lb=0;if (b[i]=='(') lb=1;if (b[i]=='[') lb=2;
		Ans+=((la-lb+1)%3-2)%3+1;
	}
	if (Ans>0) puts("TEAM 1 WINS");
	if (Ans<0) puts("TEAM 2 WINS");
	if (Ans==0) puts("TIE");
}
写不动了不写了不写了
Category: Codeforces | Tags: SB题 二分 字符串 数论 分治 | Read Count: 684
Telangana 10th Class 说:
Aug 18, 2022 03:15:45 AM

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Apr 16, 2023 06:40:36 PM

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