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
Aug 18, 2022 03:15:45 AM
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