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A. Digit Game
給定一串數字, A A A和 B B B輪流選數, A A A只能選奇數位的, B B B只能選偶數位的,最后一個數是奇數則 A A A勝,否則 B B B勝,
- 如果字串長度為偶數,那么就是 B B B最后拿偶數位的數,如果 B B B想贏,就把偶數位的偶數留到最后,換句話說,只要偶數位上有偶數,那么 B B B必勝,
- 如果字串長度為奇數,那么就是 A A A最后拿奇數位的數,如果 A A A想贏,就把奇數位的奇數留到最后,換句話說,只要奇數位上有奇數,那么 A A A必勝,
for _ in range(int(input())):
n=int(input())
s=input()
f=0
if n%2==0:#剩偶數位置的數
for i in range(1,n,2):
if int(s[i])%2==0:
print(2)
f=1
break
if f==0:
print(1)
else:
for i in range(0,n,2):
if int(s[i])%2!=0:
print(1)
f=1
break
if f==0:
print(2)
B. Stairs
就是個找規律的題,
觀察發現建
30
30
30個樓梯時,資料已經到
1
e
18
1e18
1e18了,
那就先打表,把這些樓梯需要的塊的數量都打出來,然后記個前綴和,二分查詢即可,
//C++17
#include<bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define ll long long
#define ull unsigned long long
#define rint register int
#define ld long double
#define db double
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define rep1(i,a,n) for (rint i=a;i<n;i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define per1(i,a,n) for (rint i=a;i>n;i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define sd(x) scanf("%d",&(x))
#define slld(x) scanf("%lld",&(x))
#define sdd(x,y) scanf("%d%d",&(x),&(y))
#define sc(s) scanf("%s",(s))
#define pd(x) printf("%d\n",(x))
#define plld(x) printf("%lld\n",(x))
#define pdk(x) printf("%d ",(x))
const int inf=0x3f3f3f3f;
namespace IO{
char ibuf[1<<21],*ip=ibuf,*ip_=ibuf;
char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21);
inline char gc(){
if(ip!=ip_)return *ip++;
ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin);
return ip==ip_?EOF:*ip++;
}
inline void pc(char c){
if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf;
*op++=c;
}
inline ll read(){
register ll x=0,ch=gc(),w=1;
for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1;
for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48;
return w*x;
}
template<class I>
inline void write(I x){
if(x<0)pc('-'),x=-x;
if(x>9)write(x/10);pc(x%10+'0');
}
class flusher_{
public:
~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);}
}IO_flusher;
}
using namespace IO;
const int N=1e5+10;
ull a[N],b[N],x;
int t;
ll q (ll a, ll b){
ll ret=1;
while(b){
if(b&1)
ret=ret*a;
a=a*a;
b=b>>1;
}
return ret;
}
int main()
{
t=read();
a[1]=1;
rep(i, 2, 30) a[i]=a[i-1]*2+q(2,2*(i-1));
rep(i, 2, 30) b[i]=b[i-1]+a[i];
while(t--){
x=read();
ll c=lower_bound(b+1, b+31 ,x)-b-1;
write(c);
pc('\n');
}
return 0;
}
C. Killjoy
先預處理,每個數都減感染,
- 0 0 0:如果值全為 0 0 0,那么 0 0 0次,
-
1
1
1:
- 如果值的總和為 0 0 0,那么 1 1 1次,
- 如果有 1 1 1個或多個為 0 0 0的數,那么 1 1 1次,為什么呢?因為這幾個數為 0 0 0的直接被感染了,然后它們可以隨意變化數值,變成其他分數,讓它們也感染,
- 2 2 2:其余情況,
//C++17
#include<bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define ll long long
#define ull unsigned long long
#define rint register int
#define ld long double
#define db double
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define rep1(i,a,n) for (rint i=a;i<n;i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define per1(i,a,n) for (rint i=a;i>n;i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define sd(x) scanf("%d",&(x))
#define slld(x) scanf("%lld",&(x))
#define sdd(x,y) scanf("%d%d",&(x),&(y))
#define sc(s) scanf("%s",(s))
#define pd(x) printf("%d\n",(x))
#define plld(x) printf("%lld\n",(x))
#define pdk(x) printf("%d ",(x))
const int inf=0x3f3f3f3f;
namespace IO{
char ibuf[1<<21],*ip=ibuf,*ip_=ibuf;
char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21);
inline char gc(){
if(ip!=ip_)return *ip++;
ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin);
return ip==ip_?EOF:*ip++;
}
inline void pc(char c){
if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf;
*op++=c;
}
inline ll read(){
register ll x=0,ch=gc(),w=1;
for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1;
for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48;
return w*x;
}
template<class I>
inline void write(I x){
if(x<0)pc('-'),x=-x;
if(x>9)write(x/10);pc(x%10+'0');
}
class flusher_{
public:
~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);}
}IO_flusher;
}
using namespace IO;
const int N=1e5+10;
ll t,a[N],b,n,x,ans1,ans2,sum,flag;
int main()
{
t=read();
while(t--){
n=read();x=read();
ans1=0,sum=0,flag=0;
mset(a, 0);
rep(i, 1, n){
a[i]=read();
a[i]-=x;
if(!a[i]){
ans1++;
flag=1;
}
sum+=a[i];
}
if(ans1==n) write(0);
else if (sum==0 || flag) write(1);
else write(2);
pc('\n');
}
return 0;
}
D1. Sage’s Birthday (easy version) && D2. Sage’s Birthday (hard version)
一開始 e a s y easy easy我按 p y py py做的,原本想繼續用 p y py py水過去得了,結果做到 h a r d hard hard的時候,發現 R E RE RE了,吃驚的我直接轉 C C C艸,就過了,,
e a s y easy easy
n=int(input())
lst=list(map(int,input().split()))
lst.sort()
tmp=[i for i in range(1,n+1)]
cnt=0
for i in range(1,n,2):
tmp[i]=lst[cnt]
cnt+=1
if n%2==0:
print(cnt-1)
else:
print(cnt)
for i in range(0,n,2):
tmp[i]=lst[cnt]
cnt+=1
print(*tmp)
h a r d hard hard
//C++17
#include<bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define ll long long
#define ull unsigned long long
#define rint register int
#define ld long double
#define db double
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define rep1(i,a,n) for (rint i=a;i<n;i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define per1(i,a,n) for (rint i=a;i>n;i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define sd(x) scanf("%d",&(x))
#define slld(x) scanf("%lld",&(x))
#define sdd(x,y) scanf("%d%d",&(x),&(y))
#define sc(s) scanf("%s",(s))
#define pd(x) printf("%d\n",(x))
#define plld(x) printf("%lld\n",(x))
#define pdk(x) printf("%d ",(x))
const int inf=0x3f3f3f3f;
namespace IO{
char ibuf[1<<21],*ip=ibuf,*ip_=ibuf;
char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21);
inline char gc(){
if(ip!=ip_)return *ip++;
ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin);
return ip==ip_?EOF:*ip++;
}
inline void pc(char c){
if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf;
*op++=c;
}
inline ll read(){
register ll x=0,ch=gc(),w=1;
for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1;
for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48;
return w*x;
}
template<class I>
inline void write(I x){
if(x<0)pc('-'),x=-x;
if(x>9)write(x/10);pc(x%10+'0');
}
class flusher_{
public:
~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);}
}IO_flusher;
}
using namespace IO;
const int N=1e5+10;
ll a[N],b[N],n,cnt,ans;
int main()
{
n=read();
rep(i, 1, n){
a[i]=read();
}
sort(a+1,a+1+n);
for(int i=2;i<=n;i+=2){
b[i]=a[++cnt];
}
for(int i=1;i<=n;i+=2){
b[i]=a[++cnt];
}
for(int i=2;i<=n;i+=2){
if(b[i-1]>b[i]&&b[i]<b[i+1])
ans++;
}
printf("%lld\n",ans);
rep(i, 1, n) printf("%lld ",b[i]);
return 0;
}
C
F
CF
CF上綠了,,知道很垃圾了
完結
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