Codeforces Round #426 (Div. 2) A B C

A
水题
检测是否为顺时针逆时针或同时符合

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define slld(n) scanf("%lld",&n)#define slldd(n,m) scanf("%lld%lld",&n,&m)#define sllddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define slf(n) scanf("%lf",&n)#define slff(n,m) scanf("%lf%lf",&n,&m)#define slfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define llans() printf("%lld",ans)#define llanss() printf("%lld ",ans)#define llansn() printf("%lld\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<v.size();++i)#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))typedef long long ll;typedef unsigned long long ull;const int mod = 1000000000+7;const double eps=1e-7;const int inf=0x3f3f3f3f;const ll infl = 1e16;const int maxn=  300+10;const int maxm = 40000+10;int in(int &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}char ch[4]={'^','>','v','<'};int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    char s[5],s2[5];    ss(s),ss(s2);    int t;    sd(t);    t%=4;    int idx =0 ;    while(ch[idx]!=s[0])idx++;    int a=0,b=0;    if(ch[ (idx+t)%4 ]==s2[0] )a=1;    int check  = (idx-t+4)%4;    if(ch[ (idx-t+4)%4]==s2[0])b=1;    if(a&&!b)puts("cw");    else if(!a&&b)puts("ccw");    else puts("undefined");    return 0;}

B
从后往前处理一下这个字母是不是最后一个
然后再按顺序更新一下此时打开的门数

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define slld(n) scanf("%lld",&n)#define slldd(n,m) scanf("%lld%lld",&n,&m)#define sllddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define slf(n) scanf("%lf",&n)#define slff(n,m) scanf("%lf%lf",&n,&m)#define slfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define llans() printf("%lld",ans)#define llanss() printf("%lld ",ans)#define llansn() printf("%lld\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<v.size();++i)#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))typedef long long ll;typedef unsigned long long ull;const int mod = 1000000000+7;const double eps=1e-7;const int inf=0x3f3f3f3f;const ll infl = 1e16;const int maxn=  1000000+10;const int maxm = 40000+10;int in(int &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}char s[maxn];int rightt[maxn];int pos[30];bool vis[30];int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    int n,k;    sdd(n,k);    ss(s+1);    for(int i=n;i>=1;--i)    {        int idx = s[i]-'A';        rightt[i] = pos[idx];        pos[idx] = i;    }    int now = 0;    int ans = 0;    for(int i=1;i<=n;++i)    {        int idx= s[i]-'A';        if(rightt[i]&&!vis[idx])        {            vis[idx]=1;            now++;            if(now>ans)ans=now;        }        else if(!rightt[i])        {            if(!vis[idx])            {                ans = max(ans,now+1);            }            else            {                now--;            }        }    }    if(ans>k)puts("YES");    else puts("NO");    return 0;}

C
每次操作对a b 总共乘3次
显然结果的a*b可以开出整数立方根

而对于每次操作的数 a b都至少被乘一次
所以一定能整除这个立方根

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define slld(n) scanf("%lld",&n)#define slldd(n,m) scanf("%lld%lld",&n,&m)#define sllddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define slf(n) scanf("%lf",&n)#define slff(n,m) scanf("%lf%lf",&n,&m)#define slfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define llans() printf("%lld",ans)#define llanss() printf("%lld ",ans)#define llansn() printf("%lld\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<v.size();++i)#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))typedef long long ll;typedef unsigned long long ull;const int mod = 1000000000+7;const double eps=1e-7;const int inf=0x3f3f3f3f;const ll infl = 1e16;const int maxn=  1000000+10;const int maxm = 40000+10;int in(ll &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    int n;    sd(n);    for(int i=1;i<=n;++i)    {        ll a,b;//        slldd(a,b);        in(a),in(b);        ll pro = a*b;        ll l=1, r= 1000000;        bool ok=0;        ll ans = 0;        while(l<=r)        {            ll mid = (l+r)>>1;            if(mid*mid*mid==pro)            {                ans = mid;                ok=1;                break;            }            else if(mid*mid*mid>pro)r=mid-1;            else l=mid+1;        }        if(ans&&b%ans)ok=0;        if(ans&&a%ans)ok=0;        puts(ok?"Yes":"No");    }    return 0;}