【AtCoder】ARC068

ARC 068

C - X: Yet Another Die Game

显然最多的就是一次6一次5

最后剩下的可能需要多用一次6或者6和5都用上

#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('
')
#define eps 1e-10
#define MAXN 3005
//#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);
}
int64 s,n;
void Solve() {
    read(s);
    n = (s / 11) * 2;
    s %= 11;
    if(s > 0 && s <= 6) n += 1;
    if(s > 6) n += 2;
    out(n);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - Card Eater

就是奇数的卡片最后肯定能全消掉，只剩一个

偶数的卡片最后会剩两个，看看两两配对，最后会不会剩一个，剩一个证明肯定需要少一种数，否则就是原来序列中不同的数的个数

#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('
')
#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);
}
map<int,int> zz;
int N;
int a[MAXN];
void Solve() {
    read(N);
    int cnt = 0,p = 0;
    for(int i = 1 ; i <= N ; ++i) {
	read(a[i]);
	zz[a[i]]++;
    }
    for(auto t : zz) {
	++cnt;
	if(t.se % 2 == 0) p ^= 1;
    }
    out(cnt - p);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - Snuke Line

对于一个d来说，我们把大于等于d的区间全部删掉

然后给(l - 1)标记成+1，(r)标记成-1，这些区间里能被d访问到的个数是

d - 1的前缀和2d - 1的前缀和，3d - 1的前缀和....

直接树状数组维护就好了

#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('
')
#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,M;
int tr[MAXN],ans[MAXN];
pii p[MAXN];
int lowbit(int x) {return x & (-x);}
void insert(int x,int v) {
    ++x;
    while(x <= M + 1) {
	tr[x] += v;
	x += lowbit(x);
    }
}
int query(int x) {
    int v = 0;++x;
    while(x > 0) {
	v += tr[x];
	x -= lowbit(x);
    }
    return v;
}
void Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= N ; ++i) {
	read(p[i].fi);read(p[i].se);
	--p[i].fi;
	insert(p[i].fi,1);insert(p[i].se,-1);
    }
    sort(p + 1,p + N + 1,[](pii a,pii b){return a.se - a.fi < b.se - b.fi;});
    int id = N;
    int cnt = 0;
    for(int i = M ; i >= 1 ; --i) {
	while(id >= 1 && p[id].se - p[id].fi >= i) {
	    insert(p[id].fi,-1);insert(p[id].se,1);
	    ++cnt;--id;
	}
	ans[i] = cnt;
	int t = i;
	while(t <= M) {
	    ans[i] += query(t - 1);
	    t += i;
	}
    }
    for(int i = 1 ; i <= M ; ++i) {out(ans[i]);enter;}
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - Solitaire

大意是有一个双端队列，从1到N往里面扔数，再从队首或者队尾取数，要求第K个必须是1，求方案数

这个就看什么样的是合法的，我们发现如果当前没选到1，已经选了i个数，最小的是j，我们要么就选一个比j还小的，要么选一个当前没选过的最大的

这个可以用前缀和优化去dp

当选到第k个之后，就是剩下的序列要么选最大的，要么选最小的，看看乘上2的多少次方就行了

#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('
')
#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 MOD = 1000000007;
int K,N;
int dp[2005][2005],sum[2005];
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 Solve() {
    
    read(N);read(K);
    dp[0][N + 1] = 1;
    for(int i = 1 ; i <= K ; ++i) {
	sum[N + 2] = 0;
	for(int j = N + 1 ; j >= 1 ; --j)  sum[j] = inc(sum[j + 1],dp[i - 1][j]);
	for(int j = 1 ; j <= N ; ++j) {
	    if(i == K && j != 1) continue;
	    if(i < K && j == 1) continue;
	    dp[i][j] = sum[j + 1];
	    if((N - j + 1) > (i - 1)) dp[i][j] = inc(dp[i][j],dp[i - 1][j]);
	}
    }
    int t = 1;
    for(int i = 1 ; i < N - K ; ++i) {
	t = mul(t,2);
    }
    out(mul(dp[K][1],t));enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}