话说好久没打CF div 2了
自从上次lxl毒瘤round惨掉rating之后，三个月没打CF
这场质量一般吧，水题偏多
A. Olympiad
题意：找出除了0有多少个不同的数？（应该是这样的题面吧，雾）
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ll long long
#define mem(x,num) memset(x,num,sizeof x)
#define reg(x) for(int i=last[x];i;i=e[i].next)
using namespace std;
inline ll read()
{
ll f=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,a[106],ans=0;
int main()
{
n=read();
rep(i,1,n)a[i]=read();
sort(a+1,a+1+n);
rep(i,1,n)if(a[i]!=a[i-1])ans++;
cout<<ans<<endl;
return 0;
}
B. Vile Grasshoppers
题意：给出两个数x,y，找出y范围内最大的一个数满足不能被2->x整除
话说我莫名其妙理解错题意了，一直找最小，不停WA4 ＴＡＴ
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ll long long
#define mem(x,num) memset(x,num,sizeof x)
#define reg(x) for(int i=last[x];i;i=e[i].next)
#define lson (x<<1)
#define rson (x<<1|1)
using namespace std;
inline ll read(){
ll f=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int x,y;
int main()
{
x=read(),y=read();
dep(i,y,x+1){
bool flag=1;
for(int j=2;j<=min((int)trunc(sqrt(i))+1,x);j++)
if(i%j==0){flag=0;break;}
if(flag){cout<<i;return 0;}
}
printf("-1");
return 0;
}
C. Save Energy!
好励志的题目名字？雾
公式题，不过多解释，反正题目说的神烦，什么开关啊鬼东西
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ll long long
#define mem(x,num) memset(x,num,sizeof x)
#define reg(x) for(int i=last[x];i;i=e[i].next)
#define lson (x<<1)
#define rson (x<<1|1)
using namespace std;
inline ll read(){
ll f=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
ll t,k,d,ans;
int main()
{
cin>>k>>d>>t;
ll x=k/d*d+(k%d!=0)*d;t*=2ll;
ll y=t/(2ll*k+(x-k));ans+=y*x;t-=y*(2ll*k+(x-k));
if(t>=2ll*k){ans+=k+(t-2ll*k);printf("%lld.0\n",ans);}
else{
ans+=t/2;
if(t%2)printf("%lld.5\n",ans);else printf("%lld.0\n",ans);
}
return 0;
}
D. Sleepy Game
题目标题更加有趣（逃
题意：给定一张N个点M条边的有向图和源点S，从S开始是否存在一条长度为奇数到达没有出边的点的路径
直接dfs判断就好了啊
vis[i][0/1]表示到第i个点经过边数为奇/偶数
然后要特判存在环的情况
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ll long long
#define mem(x,num) memset(x,num,sizeof x)
#define reg(x) for(int i=last[x];i;i=e[i].next)
#define lson (x<<1)
#define rson (x<<1|1)
#define mid ((l+r)>>1)
#define pi acos(-1)
#define mp make_pair
#define pb push_back
#define pa pair<int,int>
#define fi first
#define se second
#define lowbit(x) ((x)&(-x))
#define mod (ll)1e9+7
using namespace std;
inline ll read(){
ll f=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline ll getgcd(ll x,ll y){return !x?y:getgcd(y%x,x);}
inline ll getlcm(ll x,ll y){return x/getgcd(x,y)*y;}
inline ll qpow(int x,int y){
ll re=1;
while(y){
if(y&1)re=(re*x)%mod;
y>>=1,x=(x*x)%mod;
}
return re;
}
//**********head by yjjr**********
const int maxn=1e5+6;
vector<int>a[maxn],b;
int vis[maxn][2],pre[maxn][2],vs[maxn],n,m,s;
void dfs(int fr,int x,int d){
if(vis[x][d]==1)return;
vis[x][d]=1;pre[x][d]=fr;
for(int i=0;i<a[x].size();i++)dfs(x,a[x][i],d^1);
}
inline bool loop(int x){
vs[x]=1;
bool flag=0;
for(int i=0;i<a[x].size();i++){
if(vs[a[x][i]]==1)flag=1;else if(vs[a[x][i]]==0)flag=loop(a[x][i]);
if(flag)return 1;
}
vs[x]=2;
return 0;
}
int main()
{
n=read(),m=read();
rep(i,1,n){
int num=read(),x;
rep(j,1,num)a[i].pb(x=read());
}
s=read();
dfs(0,s,0);
rep(i,1,n)
if(vis[i][1]&&a[i].size()==0){
int tmp=i,t=1;
b.pb(tmp);
while(pre[tmp][t]!=0){
b.pb(pre[tmp][t]);
tmp=pre[tmp][t];
t^=1;
}
puts("Win");
dep(j,b.size()-1,0)printf("%d ",b[j]);
return 0;
}
if(loop(s))puts("Draw");else puts("Lose");
return 0;
}
E. Lock Puzzle
题意：给定字符串S和目标串T，每次变换过程为取S后缀进行翻转，之后将翻转后的放到原串前面，问操作的过程
构造题
假设字符串S为a–b–，a为T的后缀，ba也为T的后缀，设b的下标为x
那么先将n翻转，变为–b–a(reverse)
然后将x-1翻转，变为a–b–
最后将1翻转，变为ba
所以一定可以在3*n次操作内解决
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ll long long
#define mem(x,num) memset(x,num,sizeof x)
#define reg(x) for(int i=last[x];i;i=e[i].next)
#define lson (x<<1)
#define rson (x<<1|1)
#define mid ((l+r)>>1)
#define pi acos(-1)
#define mp make_pair
#define pb push_back
#define pa pair<int,int>
#define fi first
#define se second
#define lowbit(x) ((x)&(-x))
using namespace std;
inline ll read(){
ll f=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline ll getgcd(ll x,ll y){return !x?y:getgcd(y%x,x);}
inline ll getlcm(ll x,ll y){return x/getgcd(x,y)*y;}
inline ll qpow(int x,int y,int p){
ll re=1;
while(y){
if(y&1)re=(re*x)%p;
y>>=1,x=(x*x)%p;
}
return re;
}
//**********head by yjjr**********
const int maxn=2e3+6;
char s[maxn],t[maxn];int n;
vector<int>a;
void shift(int x){
if(!x)return;
reverse(s,s+n);
reverse(s+x,s+n);
a.pb(x);
}
int main()
{
n=read();scanf("%s%s",s,t);
rep(i,0,n-1){
int j=i;
while(j<n&&s[j]!=t[n-i-1])j++;
if(j==n){puts("-1");return 0;}
shift(n);shift(j);shift(1);
}
printf("%d\n",a.size());
for(int i=0;i<a.size();i++)printf("%d ",a[i]);
return 0;
}
总结
水题round
但是公式题一定要注意式子正确性，而且注意实数的类型转化
仔细考虑英文的题意，注意每一个细节
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