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Codeforces Round #598 (Div. 3)

时间：2019-11-05 21:37:06      阅读：113      评论：0      收藏：0      [点我收藏+]

标签：break   amp   div   double   stream   set   freopen   map   sig   

Codeforces Round #598 (Div. 3)

A. Payment Without Change

  • 思路：水题 没开long long wa2（考场降智

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

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

ll q, a, b, n, s, x, y;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> q;
    while (q -- ){
        cin >> a >> b >> n >> s;
        if (a * n + b < s){
            cout << "NO\n";
            continue;
        }
        x = s / n, y = s % n;
        if (y <= b)
            cout << "YES\n";
        else
            cout << "NO\n";
    }
    return 0;
}

B. Minimize the Permutation

  • 思路：模拟

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

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

const int N = 110;

int q, n;
int a[N], pos[N];
bool vis[N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> q;
    while (q -- ){
        cin >> n;
        memset(vis, false, sizeof(vis));
        for (int i = 1; i <= n; i ++ ){
            cin >> a[i];
            pos[a[i]] = i;
        }
        for (int i = 1; i <= n; i ++ ){
            for (int j = pos[i]; j >= 2; j -- ){
                if (vis[j])
                    break;
                if (a[j] > a[j - 1])
                    break;
                swap(a[j], a[j - 1]);
                swap(pos[a[j]], pos[a[j - 1]]);
                vis[j] = true;
            }
        }
        for (int i = 1; i < n; i ++ )
            cout << a[i] << " ";
        cout << a[n] << "\n";
    }
    return 0;
}

C. Platforms Jumping

  • 思路：考场降智 当时没想出来 赛后马上想出来了 疯狂改了条件就过了

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

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

const int N = 1010;

int n, m, d, cnt;
int c[N], ans[N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m >> d;
    for (int i = 1; i <= m; i ++ ){
        cin >> c[i];
        cnt += c[i];
    }
    int left = n - cnt;
    int div = left / (m + 1);
    int mod = left % (m + 1);
    if (mod)
        div ++ ;
    if (div >= d){
        cout << "NO\n";
        return 0;
    }
    if (n - cnt <= m + 1){
        int tmp = n - cnt, tot = 1;
        for (int i = 1; i <= n; i ++ ){
            if (tmp){
                i ++ ;
                tmp -- ;
            }
            for (int j = i; j < i + c[tot]; j ++ )
                    ans[j] = tot;
            i += c[tot] - 1;
            tot ++ ;
        }
    }
    else{
        int x = (n - cnt) / (m + 1), y = (n - cnt) % (m + 1), tot = 1;
        for (int i = 1; i <= n; i ++ ){
            if (y){
                i ++ ;
                y -- ;
            }
            i += x;
            for (int j = i; j < i + c[tot]; j ++ )
                    ans[j] = tot;
            i += c[tot] - 1;
            tot ++ ;
        }
    }
    cout << "YES\n";
    for (int i = 1; i < n; i ++ )
        cout << ans[i] << " ";
    cout << ans[n] << "\n";
    return 0;
}

D. Binary String Minimizing

  • 思路：暴力 每次优先把0交换到最前面

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

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

const int N = 1e6 + 10;

ll q, n, k, cnt1, tot;
ll cnt[N], pos[N];
string s;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> q;
    while (q -- ){
        cin >> n >> k >> s;
        cnt1 = 0, tot = 0;
        for (int i = 0; i < n; i ++ ){
            if (s[i] == '1')
                cnt1 ++ ;
            else{
                cnt[tot] = cnt1;
                pos[tot] = i;
                tot ++ ;
            }
        }
        for (int i = 0; i < tot; i ++ ){
            if (cnt[i] <= k){
                swap(s[pos[i]], s[pos[i] - cnt[i]]);
                k -= cnt[i];
            }
            else{
                swap(s[pos[i]], s[pos[i] - k]);
                break;
            }
        }
        cout << s << "\n";
    }
    return 0;
}

F. Equalizing Two Strings

  • 思路：如果两个串有不同的元素直接输出NO 如果有出现次数大于等于2的元素 直接输出YES 然后判断两个串的逆序数之和是否为偶数即可

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

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

int merge(char *str, char *ass, int beg, int mid, int end){
    int cnt = 0;
    int i = beg, j = mid + 1, k = beg;
    while (i <= mid && j <= end){
        if (str[i] > str[j]){
            cnt += mid - i + 1;
            ass[k ++ ] = str[j ++ ];
        }
        else
            ass[k ++ ] = str[i ++ ];
    }
    while (i <= mid)
        ass[k ++ ] = str[i ++ ];
    while (j <= end)
        ass[k ++ ] = str[j ++ ];
    for (int l = beg; l < k; l ++ )
        str[l] = ass[l];
    return cnt;
}

int calc(char *str, char *ass, int beg, int end){
    int cnt = 0;
    if (end > beg){
        int mid = (end + beg) / 2;
        cnt += calc(str, ass, beg, mid);
        cnt += calc(str, ass, mid + 1, end);
        cnt += merge(str, ass, beg, mid, end);
    }
    return cnt;
}

const int N = 2e5 + 10;

int q, n;
char s[N], t[N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> q;
    while (q -- ){
        cin >> n >> s >> t;
        int cnt1[30] = {0}, cnt2[30] = {0};
        for (int i = 0; i < n; i ++ ){
            cnt1[s[i] - 'a'] ++ ;
            cnt2[t[i] - 'a'] ++ ;
        }
        bool flag1 = true, flag2 = false;
        for (int i = 0; i < 26; i ++ ){
            if (cnt1[i] != cnt2[i]){
                flag1 = false;
                break;
            }
        }
        if (!flag1){
            cout << "NO\n";
            continue;
        }
        for (int i = 0; i < 26; i ++ ){
            if (cnt1[i] >= 2 || cnt2[i] >= 2){
                flag2 = true;
                break;
            }
        }
        if (flag2){
            cout << "YES\n";
            continue;
        }
        char tmp1[N], tmp2[N];
        int c1 = calc(s, tmp1, 0, n - 1), c2 = calc(t, tmp2, 0, n - 1);
        if ((c1 + c2) % 2 == 0)
            cout << "YES\n";
        else
            cout << "NO\n";
    }
    return 0;
}

Codeforces Round #598 (Div. 3)

标签：break   amp   div   double   stream   set   freopen   map   sig   

原文地址：https://www.cnblogs.com/Misuchii/p/11801549.html
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