【Codeforces】Codeforces Round #551 (Div. 2)

Codeforces Round #551 (Div. 2)

算是放弃颓废决定好好打比赛好好刷题的开始吧

A. Serval and Bus

处理每个巴士最早到站且大于t的时间

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
    }
    while(c >= '0' && c <= '9') {
    	res = res * 10 +c - '0';
    	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,t;
int c[105];
void Solve() {
    read(N);read(t);
    int s,d;
    for(int i = 1 ; i <= N ; ++i) {
        read(s);read(d);
        if(s >= t) c[i] = s;
        else c[i] = s + ((t - s - 1) / d + 1) * d;
    }
    int ans = 1;
    for(int i = 2 ; i <= N ; ++i) {
        if(c[ans] - t > c[i] - t) ans = i;
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

B. Serval and Toy Bricks

保证有解直接在俯视图每个1的位置填上行列上限的最小值即可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
    }
    while(c >= '0' && c <= '9') {
    	res = res * 10 +c - '0';
    	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,M,H;
int f[105],l[105];
void Solve() {
    read(N);read(M);read(H);
    for(int i = 1 ; i <= M ; ++i) {
        read(f[i]);
    }
    for(int i = 1 ; i <= N ; ++i) {
        read(l[i]);
    }
    int a = 0;
    for(int i = 1 ; i <= N ; ++i) {
        for(int j = 1 ; j <= M ; ++j) {
            read(a);
            if(!a) out(0);
            else out(min(l[i],f[j]));
            space;
        }
        enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

C. Serval and Parenthesis Sequence

如果每个位置都不合法，只需要左右两端是匹配的括号，然后对于剩下的部分做括号匹配即可

就是从前往后扫，如果遇到左括号扔进一个栈，遇到问号扔到一个栈，遇到右括号优先找一个左括号匹配，然后问好匹配左括号，然后问号两两匹配，不用在乎先后顺序，因为对于一个完整的匹配()，如果我在这个括号里插入一个(，在最右再多一个)，还是一个完整匹配

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
    }
    while(c >= '0' && c <= '9') {
    	res = res * 10 +c - '0';
    	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    	out(x / 10);
    }
    putchar('0' + x % 10);
}
char s[300005];
int N;
int sta[2][300005],top[2];
void Solve() {
    read(N);
    scanf("%s",s + 1);
    if(s[1] == ')' || s[N] == '(') {puts(":(");return;}
    if(N & 1) {puts(":(");return;}
    s[1] = '(';s[N] = ')';
    for(int i = 2 ; i < N ; ++i) {
        if(s[i] == '(') sta[0][++top[0]] = i;
        else if(s[i] == '?') sta[1][++top[1]] = i;
        else {
            if(top[0]) --top[0];
            else if(top[1]){
                int u = sta[1][top[1]--];
                s[u] = '(';
            }
            else {puts(":(");return;}
        }
    }
    while(top[0]) {
        int u = sta[1][top[1]--],v = sta[0][top[0]--];
        if(u < v) {puts(":(");return;}
        s[u] = ')';
    }
    if(top[1] & 1) {puts(":(");return;}
    for(int i = 1 ; i <= top[1] ; ++i) {
        int u = sta[1][i];
        if(i & 1) s[u] = '(';
        else s[u] = ')';
    }
    for(int i = 1 ; i <= N ; ++i) {
        putchar(s[i]);
    }
    enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D. Serval and Rooted Tree

我们可以求出每个点最大能取到第几大的

设这个值是\(dp[u]\)

如果一个节点是min，答案就是这个节点所有儿子的dp[u] + 1的总和-1

如果一个节点是max，答案就是这个节点所有儿子中最小的dp[u]

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 300005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
    }
    while(c >= '0' && c <= '9') {
    	res = res * 10 +c - '0';
    	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,op[MAXN],f[MAXN],dp[MAXN],K;
vector<int> son[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(op[i]);
    for(int i = 2 ; i <= N ; ++i) read(f[i]);
    for(int i = N ; i >= 1 ; --i) {
        if(son[i].size() == 0) {dp[i] = 0;++K;}
        else {
            int sum = 0,mm = N;
            for(auto s : son[i]) {
                mm = min(dp[s],mm);
                sum += dp[s] + 1;
            }
            if(!op[i]) dp[i] = sum - 1;
            else dp[i] = mm;
        }
        son[f[i]].pb(i);
    }
    out(K - dp[1]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E. Serval and Snake

我们直接问每一行或一列（认为是一行的下边界或一列的右边界）的线切开了多少大蛇，如果某一行切开的蛇和上一行切开的蛇相差是奇数，那么这行必定有个头或者有个尾

如果行列都有这样的，那么判断这两行两列交出的四个点哪个点连向四周的边是奇数

如果只有行或列有，那么可以通过二分，就是最小列的使得这一行的切割点是奇数

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 300005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
    }
    while(c >= '0' && c <= '9') {
    	res = res * 10 +c - '0';
    	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
int r[1005],c[1005];
vector<int> ac,ar;
vector<pii > ans;
int Query(int x1,int y1,int x2,int y2) {
    printf("? %d %d %d %d\n",x1,y1,x2,y2);
    fflush(stdout);
    int x;read(x);return x;
}
void Solve() {
    read(N);
    for(int i = 1 ; i < N ; ++i) {
        r[i] = Query(1,1,i,N);
        c[i] = Query(1,1,N,i);
    }
    for(int i = 1 ; i <= N ; ++i) {
        if((r[i] ^ r[i - 1]) & 1) ar.pb(i);
        if((c[i] ^ c[i - 1]) & 1) ac.pb(i);
    }
    if(ar.size() >= 2 && ac.size() >= 2) {
        for(int i = 0 ; i < 2 ; ++i) {
            for(int j = 0 ; j < 2 ; ++j) {
                if(Query(ar[i],ac[j],ar[i],ac[j]) == 1) ans.pb(mp(ar[i],ac[j]));
            }
        }
    }
    else if(ar.size() >= 2) {
        int l = 1,r = N;
        while(l < r) {
            int mid = (l + r) >> 1;
            if(Query(ar[0],1,ar[0],mid) & 1) r = mid;
            else l = mid + 1;
        }
        ans.pb(mp(ar[0],l));ans.pb(mp(ar[1],l));
    }
    else if(ac.size() >= 2) {
        int l = 1,r = N;
        while(l < r) {
            int mid = (l + r) >> 1;
            if(Query(1,ac[0],mid,ac[0]) & 1) r = mid;
            else l = mid + 1;
        }
        ans.pb(mp(l,ac[0]));ans.pb(mp(l,ac[1]));
    }
    putchar('!');space;out(ans[0].fi);space;out(ans[0].se);space;
    out(ans[1].fi);space;out(ans[1].se);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F. Serval and Bonus Problem

老了，这么普通的一道计数都切不动

就是转化成插入一个点，使得这个点至少在k个之间

可以用dp完成，相当于给2*n + 1个点分配身份

就是\(f[i][j][x]\)x为0或1表示有没有插入特殊点，到了第\(i\)个点

然后如果\(j >= k , x = 0\)，那么可以转移\(f[i][j][x] \rightarrow f[i + 1][j][x]\)

或者加上一个新区间\(f[i][j][x] \rightarrow f[i + 1][j + 1][x]\)

或者结束一个区间\(f[i][j][x] \rightarrow f[i + 1][j - 1][x] \times j\)

方案总数是\(\frac{(2n + 1)!}{n!2^{n}}\)因为两个点是等价的就直接除上\(2^n\)，然后n个区间的标号是没有顺序的，所以再除上一个\(n!\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 300005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
    }
    while(c >= '0' && c <= '9') {
    	res = res * 10 +c - '0';
    	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 998244353;
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
        if(c & 1) res = mul(res,t);
        t = mul(t,t);
        c >>= 1;
    }
    return res;
}
int k,n,l;
int dp[4005][2005][2],fac[4005];
void Solve() {
    read(n);read(k);read(l);
    dp[0][0][0] = 1;
    for(int i = 0 ; i < 2 * n + 1; ++i) {
        for(int j = 0 ; j <= n ; ++j) {
            for(int x = 0 ; x <= 1 ; ++x) {
                if(!dp[i][j][x]) continue;
                if(j >= k && !x) update(dp[i + 1][j][1],dp[i][j][x]);
                if(j >= 1) update(dp[i + 1][j - 1][x],mul(dp[i][j][x],j));
                update(dp[i + 1][j + 1][x],dp[i][j][x]);
            }
        }
    }
    fac[0] = 1;
    for(int i = 1 ; i <= 2 * n + 1 ; ++i) fac[i] = mul(fac[i - 1],i);
    int ans = l;
    ans = mul(ans,dp[2 * n + 1][0][1]);
    ans = mul(ans,fpow(fac[2 * n + 1],MOD - 2));
    ans = mul(ans,fac[n]);
    ans = mul(ans,fpow(2,n));
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}