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

时间：2023-03-08 21:31:51

A. The Useless Toy

题目链接：http://codeforces.com/contest/834/problem/A

思路： 水题

实现代码：

#include<bits/stdc++.h>

#include<iostream>

#include<cstdio>

#include<string>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<map>

#include<queue>

#include<stack>

#include<set>

#include<list>

using namespace std;

#define ll long long

#define sd(x) scanf("%d",&x)

#define sdd(x,y) scanf("%d%d",&x,&y)

#define sddd(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define sf(x) scanf("%s",x)

#define ff(i,x,y) for(int i = x;i <= y;i ++)

#define fj(i,x,y) for(int i = x;i >= y;i --)

#define mem(s,x) memset(s,x,sizeof(s));

#define pr(x) printf("%d",x);

const int Mod = 1e9+;

const int inf = 1e9;

const int Max = 1e5+;

void exgcd(ll a,ll b,ll& d,ll& x,ll& y){if(!b){d=a;x=;y=;}else{exgcd(b,a%b,d,y,x);y-=x*(a/b);}}

ll inv(ll a,ll n){ll d, x, y;exgcd(a,n,d,x,y);return (x+n)%n;}

int gcd(int a,int b)  {  return (b>)?gcd(b,a%b):a;  }

int lcm(int a, int b)  {  return a*b/gcd(a, b);   }

ll mod(ll x,ll y){

    return x-(x/y)*y;

}

int main()

{

    ios::sync_with_stdio();

    cin.tie();

    char a[];

    ll k,b[];

    cin>>a[]>>a[];

    cin>>k;

    for(int i=;i<=;i++){

        if(a[i]=='v') b[i] = ;

        if(a[i]=='<') b[i] = ;

        if(a[i]=='^') b[i] = ;

        if(a[i]=='>') b[i] = ;

    }

    ll ans1 = mod(k+b[],);

    ll ans2 = mod(b[]-k,);

    while(ans2<)

    ans2+=;

    if(b[]==ans1&&b[]==ans2)

        cout<<"undefined"<<endl;

    else if(b[]==ans2)

        cout<<"ccw"<<endl;

    else if(b[]==ans1)

        cout<<"cw"<<endl;

    return ;

}

B. The Festive Evening

题目链接：http://codeforces.com/contest/834/problem/B

思路; 水题

实现代码：

#include<bits/stdc++.h>

#include<iostream>

#include<cstdio>

#include<string>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<map>

#include<queue>

#include<stack>

#include<set>

#include<list>

using namespace std;

#define ll long long

#define sd(x) scanf("%d",&x)

#define sdd(x,y) scanf("%d%d",&x,&y)

#define sddd(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define sf(x) scanf("%s",x)

#define ff(i,x,y) for(int i = x;i <= y;i ++)

#define fj(i,x,y) for(int i = x;i >= y;i --)

#define mem(s,x) memset(s,x,sizeof(s));

#define pr(x) printf("%d",x);

const int Mod = 1e9+;

const int inf = 1e9;

const int Max = 1e5+;

void exgcd(ll a,ll b,ll& d,ll& x,ll& y){if(!b){d=a;x=;y=;}else{exgcd(b,a%b,d,y,x);y-=x*(a/b);}}

ll inv(ll a,ll n){ll d, x, y;exgcd(a,n,d,x,y);return (x+n)%n;}

int gcd(int a,int b)  {  return (b>)?gcd(b,a%b):a;  }

int lcm(int a, int b)  {  return a*b/gcd(a, b);   }

ll mod(ll x,ll y){    return x-(x/y)*y;}

char s[];

map<char,int>mp;

set<char>st;

int main()

{

    int n,m,cnt=,i,vis[];

    ios::sync_with_stdio();

    cin.tie();

       cin>>n>>m;

        mp.size();

       for(i=;i<n;i++){

        cin>>s[i];

        mp[s[i]]=i;

       }

    for(i=;i<n-;i++){

        st.insert(s[i]);

        if(st.size()>m){

            cout<<"YES"<<endl;

            return ;

        }

        if(mp[s[i]]==i)

            st.erase(s[i]);

    }

    cout<<"NO"<<endl;

    return ;

}

C. The Meaningless Game

题目链接：http://codeforces.com/contest/834/problem/C

思路：二分，a,b分别*k或*k^2,那么a*b必为k*(k^2)*(k1^2)*k1*.....也就是 (k*k1*..)^3,直接二分搜a*b是否为某个数的三次方，输入输出数据比较大，用scanf和print.要不超时

实现代码：

#include<bits/stdc++.h>

#include<iostream>

#include<cstdio>

#include<string>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<map>

#include<queue>

#include<stack>

#include<set>

#include<list>

using namespace std;

#define ll long long

#define sd(x) scanf("%d",&x)

#define sdd(x,y) scanf("%d%d",&x,&y)

#define sddd(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define sf(x) scanf("%s",x)

#define ff(i,x,y) for(int i = x;i <= y;i ++)

#define fj(i,x,y) for(int i = x;i >= y;i --)

#define mem(s,x) memset(s,x,sizeof(s));

#define pr(x) printf("%d",x);

const int Mod = 1e9+;

const int inf = 1e9;

const int Max = 1e5+;

void exgcd(ll a,ll b,ll& d,ll& x,ll& y){if(!b){d=a;x=;y=;}else{exgcd(b,a%b,d,y,x);y-=x*(a/b);}}

ll inv(ll a,ll n){ll d, x, y;exgcd(a,n,d,x,y);return (x+n)%n;}

int gcd(int a,int b)  {  return (b>)?gcd(b,a%b):a;  }

int lcm(int a, int b)  {  return a*b/gcd(a, b);   }

int main()

{

    ll n,i,mid,l,r,a,b;

    cin>>n;

    for(i=;i<n;i++){

        scanf("%I64d %I64d",&a,&b);

        l=;r=;

        int flag = ;

        while(l<=r){

            mid = (l+r)/;

            //cout<<Pow(mid,3)<<" "<<a*b<<endl;

            if(mid*mid*mid==a*b){

                flag = ;break;}

            if(mid*mid*mid>a*b)

                r = mid-;

            if(mid*mid*mid<a*b)

                l = mid+;

        }

        if(flag&&a*a%b==&&b*b%a==)

            printf("Yes\n");

        else

        printf("No\n");

    }

    return ;

}

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