Codeforces Round #685 (Div. 2)

最新推荐文章于 2020-11-22 21:18:52 发布

原创最新推荐文章于 2020-11-22 21:18:52 发布 · 153 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#templates #powershell #soa #wxpython #macbook

本文提供了 Codeforces Round #685 (Div.2) 的 A、B、C、D 四题的解题思路及代码实现，涵盖数字处理、字符串匹配和游戏策略等方面。

Codeforces Round #685 (Div. 2)

A - Subtract or Divide

可以除掉很大的因数,答案只会是0，1，2，3

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define ls u<<1
#define rs u<<1|1
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
ll n, x;

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}


int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    n = read();
    while (n--) {
        x = read();
        if (x == 1) puts("0");
        else if (x == 2) puts("1");
        else if (x % 2 == 0 || x == 3) puts("2");
        else puts("3");
    }

    // #ifndef ONLINE_JUDGE
    //     system("pause");
    // #endif
}

View Code

B - Non-Substring Subsequence

暴力判断两边l、r两边的情况

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define ls u<<1
#define rs u<<1|1
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int _, n, q;
char s[N];

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}


int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    _ = read();
    while (_--) {
        n = read();
        q = read();
        scanf("%s", s + 1);
        while (q--) {
            int l, r;
            bool flag = 0;
            l = read(); r = read();
            for (int i = 1; i < l; ++i) {
                if (s[i] == s[l]) {
                    flag = 1;
                    break;
                }
            }
            for (int i = r + 1; i <= n; ++i) {
                if (s[i] == s[r]) {
                    flag = 1;
                    break;
                }
            }
            if (flag) puts("YES");
            else puts("NO");
            
        }
    }

    // #ifndef ONLINE_JUDGE
    //     system("pause");
    // #endif
}

View Code

C - String Equality

开两个桶，一直把状态往后移，判断下位置是否合法即可

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define ls u<<1
#define rs u<<1|1
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 2e6 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int _, n, k;
char s[N], t[N];
int a[30], b[30];

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}


int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    _ = read();
    while (_--) {
        mm(a, 0);
        mm(b, 0);
        n = read(); k = read();
        scanf("%s", s + 1);
        scanf("%s", t + 1);

        rep(i, 1, n) a[s[i] - 'a' + 1]++;
        rep(i, 1, n) b[t[i] - 'a' + 1]++;

        bool flag = 0;
        rep(i, 1, 26) {
            if (a[i] < b[i]) {
                flag = 1;
                break;
            }
            a[i] -= b[i];
            b[i] = 0;
            if (a[i] % k) {
                flag = 1;
                break;
            }
            a[i + 1] += a[i];
        }
        if (a[26] != b[26]) {
            flag = 1;
        }
        if (flag) puts("NO");
        else puts("YES");
    }

    // #ifndef ONLINE_JUDGE
    //     system("pause");
    // #endif
}

View Code

D - Circle Game

可以发现对于后手来说最优解一定是对照先手走对角线，只需判断 d - floor(d /(sqrt(2)*k))和 k 的大小即可

会有精度问题所以不能直接比较，可以通过便利来求出最终的区间来规避精度问题

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define ls u<<1
#define rs u<<1|1
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
ll _, d, k;
int sg[N];

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}

int main()
{
    _ = read();
    while (_--)
    {
        d = read(); k = read();
        ll t = 0;
        while (t*k*t*k + t*k*t*k <= d*d)
            ++t;
        --t;
        ++t; --t;
        ll u = floor(sqrt(d*d - t*t*k*k) / k) - t;
        if (u & 1) puts("Ashish");
        else puts("Utkarsh");

        // mm(sg, 0);
        // ll r = d / k;
        // for (int i = r; i >= 0; --i) {
        //     ll h = d * d - i * i * k * k;
        //     double hh = sqrt(h);
        //     h = floor(hh);
        //     ll tp = h / k;
        //     if (i == r) {
        //         cout << h << '\n';
        //         cout << hh << '\n';
        //         cout << tp << '\n';
        //         puts("");
        //     }
        //     if (i == r) {
        //         if (tp & 1) sg[i] = 1;
        //         else sg[i] = 2;
        //     } else {
        //         if (sg[i + 1] == 2) sg[i] = 1;
        //         else {
        //             sg[i] = 2;
        //         }
        //     }
        // }
        // // cout << sg[2] << '\n';
        // if (sg[0] == 2) puts("Utkarsh");
        // else puts("Ashish");
    }
    return 0;
}

View Code