2019ccpc女生赛 9/11

闲暇之余并没有好好补题而和小南瓜一起开了一场女生赛x
然而自己菜死了，没看清I的题目，还分析错了E的复杂度，导致最后只有9题

A

/* ***********************************************
Author        :BPM136
Created Time  :2019/8/2 13:31:09
File Name     :A.cpp
************************************************ */

#include 
#include 
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define filein(x) freopen(#x".in","r",stdin)
#define fileout(x) freopen(#x".out","w",stdout)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define mkdfile() freopen("in.txt","w",stdout);
#define setlargestack(x) int _SIZE=x<<20;char *_PPP=(char*)malloc(_SIZE)+_SIZE;__asm__("movl %0, %%esp\n" :: "r"(_PPP));
#define read2(a,b) read(a),read(b)
#define read3(a,b,c) read(a),read(b),read(c)
#define readg(_x1,_y1,_x2,_y2) read(_x1),read(_y1),read(_x2),read(_y2)
#define USE_CIN_COUT ios::sync_with_stdio(0)

using namespace std;

int random(int l, int r) {
	static std::random_device rd;
	struct timeb timeSeed;
	ftime(&timeSeed);
	size_t seed = timeSeed.time * 1000 + timeSeed.millitm;  // milli time
	static std::mt19937 gen(seed);
	std::uniform_int_distribution<> u(l, r);
	return u(gen);
}

typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int  ui;
typedef unsigned long long ull;
typedef pair<int, int> pii;

namespace fastIO{  
    #define BUF_SIZE 100000  
    #define OUT_SIZE 100000  
    //fread->read  
    bool IOerror=0;  
    inline char nc(){  
		static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;  
		if (p1==pend){  
			p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);  
			if (pend==p1){IOerror=1;return -1;}  
			//{printf("IO error!\n");system("pause");for (;;);exit(0);}  
		}  
		return *p1++;  
	}  
	inline bool blank(char ch){return ch==32||ch==10||ch==13||ch==9;}  
	inline bool enter(char ch){return ch==10||ch==13;}
	inline void read(int &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (sign)x=-x;  
	}  
	inline void read(ll &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (sign)x=-x;  
	}  
	inline void read(double &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (ch==46){  
			double tmp=1; ch=nc();  
			for (;ch>=48&&ch<=57;ch=nc())tmp/=10.0,x+=tmp*(ch-48);  
		}  
		if (sign)x=-x;  
	}  
	inline void read(char *s){  
		char ch=nc();  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;  
		*s=0;  
	}  
	inline void readln(char *s) {
		char ch=nc();
		for (;blank(ch);ch=nc());
		if(IOerror)return;
		for(;!enter(ch)&&!IOerror;ch=nc())*s++=ch;
		*s=0;
	}
	inline void read(char &c){  
		for (c=nc();blank(c);c=nc());  
		if (IOerror){c=-1;return;}  
	} 
#undef OUT_SIZE  
#undef BUF_SIZE  
}
using fastIO::read;

int main() {
	int n;
	scanf("%d",&n);
	double sum = 0;
	for(int i = 0;i < n;i++) {
		int x;
		scanf("%d",&x);
		if(sum >= 100 && sum < 150) x *= 0.8;
		else if(sum >= 150 && sum < 400) x *= 0.5;
		sum += x;
	}
	printf("%.2f\n",sum);
	return 0;
}

B

其实可以根号n的复杂度，但是实际上直接算sum然后枚举约数也是根号n复杂度（可以思考下为什么

/* ***********************************************
Author        :BPM136
Created Time  :2019/8/2 13:53:50
File Name     :B.cpp
************************************************ */

#include 
#include 
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define USE_CIN_COUT ios::sync_with_stdio(0)
#define filein(x) freopen(#x".in","r",stdin)
#define fileout(x) freopen(#x".out","w",stdout)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define mkd(x) freopen(#x".in","w",stdout);

using namespace std;

int random(int l, int r) {
	static std::random_device rd;
	struct timeb timeSeed;
	ftime(&timeSeed);
	size_t seed = timeSeed.time * 1000 + timeSeed.millitm;  // milli time
	static std::mt19937 gen(seed);
	std::uniform_int_distribution<> u(l, r);
	return u(gen);
}

typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef pair<int, int> pii;

int main() {
	ll n;
	cin >> n;
	ll sum = (n + 1) * n / 2;
	for (ll i = 2; i * i <= sum; ++i) 
		if (sum % i == 0) {
			cout << sum / i << '\n';
			return 0;
		}
    return 0;
}

C

考虑最开始就分配好每个都是1，然后不足的部分慢慢+1挪过去，因为开口朝上的二次函数的delta y是递增的，所以是对的

/* ***********************************************
Author        :BPM136
Created Time  :2019/8/2 16:28:30
File Name     :C.cpp
************************************************ */

#include 
#include 
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define USE_CIN_COUT ios::sync_with_stdio(0)
#define filein(x) freopen(#x".in","r",stdin)
#define fileout(x) freopen(#x".out","w",stdout)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define mkd(x) freopen(#x".in","w",stdout);

using namespace std;

int random(int l, int r) {
	static std::random_device rd;
	struct timeb timeSeed;
	ftime(&timeSeed);
	size_t seed = timeSeed.time * 1000 + timeSeed.millitm;  // milli time
	static std::mt19937 gen(seed);
	std::uniform_int_distribution<> u(l, r);
	return u(gen);
}

typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;

int const N = 100005;

struct node {
	int a, b, c;
	
	ll calc(int x) {
		return a * x * x + b * x + c;
	}
} F[N];

int pr[N];

int main() {
	USE_CIN_COUT;
	int n, m;
	cin >> n >> m;
	ll ans = 0;
	auto q = priority_queue<pli, vector<pli>, greater<pli>>();
	for (int i = 1; i <= n; ++i) {
		cin >> F[i].a >> F[i].b >> F[i].c;
		ans += F[i].calc(1);
		q.push(make_pair(F[i].calc(2) - F[i].calc(1), i));
		pr[i] = 1;
	}
	m -= n;
	while (m--) {
		auto now = q.top();
		q.pop();
		ans += now.first;
		auto i = now.second;
		pr[i]++;
		q.push(make_pair(F[i].calc(pr[i] + 1) - F[i].calc(pr[i]), i));
	}
	cout << ans << '\n';
    return 0;
}

D

傻叉树剖

/* ***********************************************
Author        :BPM136
Created Time  :2019/8/2 15:34:50
File Name     :D.cpp
************************************************ */

#include 
#include 
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define USE_CIN_COUT ios::sync_with_stdio(0)
#define filein(x) freopen(#x".in","r",stdin)
#define fileout(x) freopen(#x".out","w",stdout)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define mkd(x) freopen(#x".in","w",stdout);

using namespace std;

int random(int l, int r) {
	static std::random_device rd;
	struct timeb timeSeed;
	ftime(&timeSeed);
	size_t seed = timeSeed.time * 1000 + timeSeed.millitm;  // milli time
	static std::mt19937 gen(seed);
	std::uniform_int_distribution<> u(l, r);
	return u(gen);
}

typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef pair<int, int> pii;

int const N = 100005;

vector<int> G[N];
int a[N], n;

ll sum[N << 2];
bool flag[N << 2];
int id[N];
int ql, qr;

void build(int k, int l, int r) {
	if (l == r) {
		sum[k] = a[id[l]];
		flag[k] = sum[k] == 1;
		return;
	}
	int mid = (l + r) >> 1;
	build(k << 1, l, mid);
	build(k << 1 | 1, mid + 1, r);
	sum[k] = sum[k << 1] + sum[k << 1 | 1];
	flag[k] = flag[k << 1] & flag[k << 1 | 1];
}

void update(int k, int l, int r) {
	if (l == r) {
		sum[k] = sqrt(sum[k]);
		if (sum[k] == 1)
			flag[k] = 1;
		return;
	}
	if (ql <= l && r <= qr && flag[k]) 
		return;
	int mid = (l + r) >> 1;
	if (ql <= mid)
		update(k << 1, l, mid);
	if (qr > mid)
		update(k << 1 | 1, mid + 1, r);
	sum[k] = sum[k << 1] + sum[k << 1 | 1];
	flag[k] = flag[k << 1] & flag[k << 1 | 1];
}

ll query(int k, int l, int r) {
	if (ql <= l && r <= qr) 
		return sum[k];
	int mid = (l + r) >> 1;
	ll ret = 0;
	if (ql <= mid)
		ret += query(k << 1, l, mid);
	if (qr > mid)
		ret += query(k << 1 | 1, mid + 1, r);
	return ret;
}

int fa[N], dep[N], top[N], dfn[N], sz[N], son[N], cnt;

void dfs1(int x) {
	sz[x] = 1;
	int mx = 0;
	son[x] = x;
	for (auto v : G[x]) {
		if (v == fa[x])
			continue;
		fa[v] = x;
		dep[v] = dep[x] + 1;
		dfs1(v);
		sz[x] += sz[v];
		if (sz[v] > mx) {
			mx = sz[v];
			son[x] = v;
		}
	}
}

void dfs2(int x, int tp) {
	top[x] = tp;
	dfn[x] = ++cnt;
	id[cnt] = x;
	if (son[x] != x)
		dfs2(son[x], tp);
	for (auto v : G[x]) {
		if (v == fa[x] || v == son[x])
			continue;
		dfs2(v, v);
	}
}

void _update(int x, int y) {
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) 
			swap(x, y);
		ql = dfn[top[x]];
		qr = dfn[x];
		update(1, 1, n);
		x = fa[top[x]];
	}
	if (dep[x] < dep[y])
		swap(x, y);
	ql = dfn[y];
	qr = dfn[x];
	update(1, 1, n);
}

ll _query(int x, int y) {
	ll ret = 0;
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) 
			swap(x, y);
		ql = dfn[top[x]];
		qr = dfn[x];
		ret += query(1, 1, n);
		x = fa[top[x]];
	}
	if (dep[x] < dep[y])
		swap(x, y);
	ql = dfn[y];
	qr = dfn[x];
	ret += query(1, 1, n);
	return ret;
}

int main() {
	USE_CIN_COUT;
	int m;
	cin >> n >> m;
	for (int i = 1; i <= n; ++i) 
		cin >> a[i];
	for (int i = 1; i < n; ++i) {
		int x, y;
		cin >> x >> y;
		G[x].push_back(y);
		G[y].push_back(x);
	}
	dep[1] = 1;
	dfs1(1);
	dfs2(1, 1);
	build(1, 1, n);
	while (m--) {
		int op, x, y;
		cin >> op >> x >> y;
		if (op == 0) 
			_update(x, y);
		else
			cout << _query(x, y) << '\n';
	}
    return 0;
}

E

其实就是每个点维护一个map，里面是自己子树的颜色，然后每次枚举一个点的所有儿子，和他的兄弟们去算贡献即可，我竟然以为是$O(n^2)$，谈姐姐提醒了才知道，我是傻叉

F

$f_{i,j,k}$表示到第i个位置，一共有j段，末尾的颜色为k的时候的最小次数，然后分成两种情况转移即可x

/* ***********************************************
Author        :BPM136
Created Time  :2019/8/2 14:02:11
File Name     :F.cpp
************************************************ */

#include 
#include 
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define USE_CIN_COUT ios::sync_with_stdio(0)
#define filein(x) freopen(#x".in","r",stdin)
#define fileout(x) freopen(#x".out","w",stdout)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define mkd(x) freopen(#x".in","w",stdout);

using namespace std;

int random(int l, int r) {
	static std::random_device rd;
	struct timeb timeSeed;
	ftime(&timeSeed);
	size_t seed = timeSeed.time * 1000 + timeSeed.millitm;  // milli time
	static std::mt19937 gen(seed);
	std::uniform_int_distribution<> u(l, r);
	return u(gen);
}

typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const int maxn = 100010;
char s[maxn];

struct Seg {
	int len;
	int a;
}ss[maxn];

int dp[maxn][30][15];

void update(int i,int j,int k,int val) {
	if(dp[i][j][k] == -1) dp[i][j][k] = val;
	else dp[i][j][k] = min(dp[i][j][k],val);
}

int upper(int a,int b) {
	if(a%b == 0) return a/b;
	return a/b+1;
}

int main() {
	int n,l,K;
	scanf("%d%d%d",&n,&l,&K);
	scanf("%s",s);
	int cnt = 0;
	ss[cnt].len = 1;
	ss[cnt].a = s[0]-'a';
	for(int i = 1;i < n;i++) {
			cnt++;
			ss[cnt].len = 1;
			ss[cnt].a = s[i]-'a';
	}
	memset(dp,-1,sizeof(dp));
	for(int i = 0;i < 26;i++) dp[0][i][1] = upper(ss[0].len,l);
	dp[0][ss[0].a][1] = 0;
	int p = -1,pre = 0,cur = ss[0].len;
	for(int i = 1;i <= cnt;i++) {
		cur += ss[1].len;
		while(cur-pre > l) pre += ss[++p].len;
		if(p == -1) {
			for(int j = 0;j < 26;j++) dp[i][j][1] = 1;
		}
		else if(p == i) {
			int num = upper(ss[i].len,l);
			for(int j = 0;j < 26;j++) {
				for(int k = 1;k <= K;k++) {
					if(dp[p-1][j][k] == -1) continue;
					update(i,j,k,dp[p-1][j][k]+num);
				}
			}
		}
		else {
			for(int j = 0;j < 26;j++) {
				for(int k = 1;k <= K;k++) {
					if(dp[p][j][k] == -1) continue;
					update(i,j,k,dp[p][j][k]+1);
				}
			}
		}

		for(int j = 0;j < 26;j++) {
			for(int k = 1;k <= K;k++) {
				if(dp[i-1][j][k] == -1) continue;
				if(j == ss[i].a) update(i,j,k,dp[i-1][j][k]);
				else update(i,ss[i].a,k+1,dp[i-1][j][k]);
			}
		}
	}

	int ans = 1e9;
	for(int i = 0;i < 26;i++) {
		for(int j = 1;j <= K;j++) 
			if(dp[cnt][i][j] != -1) ans = min(ans,dp[cnt][i][j]);
	}
	printf("%d\n",ans);
    return 0;
}

G

/* ***********************************************
Author        :BPM136
Created Time  :2019/8/2 13:36:51
File Name     :G.cpp
************************************************ */

#include 
#include 
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define filein(x) freopen(#x".in","r",stdin)
#define fileout(x) freopen(#x".out","w",stdout)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define mkdfile() freopen("in.txt","w",stdout);
#define setlargestack(x) int _SIZE=x<<20;char *_PPP=(char*)malloc(_SIZE)+_SIZE;__asm__("movl %0, %%esp\n" :: "r"(_PPP));
#define read2(a,b) read(a),read(b)
#define read3(a,b,c) read(a),read(b),read(c)
#define readg(_x1,_y1,_x2,_y2) read(_x1),read(_y1),read(_x2),read(_y2)
#define USE_CIN_COUT ios::sync_with_stdio(0)

using namespace std;

int random(int l, int r) {
	static std::random_device rd;
	struct timeb timeSeed;
	ftime(&timeSeed);
	size_t seed = timeSeed.time * 1000 + timeSeed.millitm;  // milli time
	static std::mt19937 gen(seed);
	std::uniform_int_distribution<> u(l, r);
	return u(gen);
}

typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int  ui;
typedef unsigned long long ull;
typedef pair<int, int> pii;

namespace fastIO{  
    #define BUF_SIZE 100000  
    #define OUT_SIZE 100000  
    //fread->read  
    bool IOerror=0;  
    inline char nc(){  
		static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;  
		if (p1==pend){  
			p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);  
			if (pend==p1){IOerror=1;return -1;}  
			//{printf("IO error!\n");system("pause");for (;;);exit(0);}  
		}  
		return *p1++;  
	}  
	inline bool blank(char ch){return ch==32||ch==10||ch==13||ch==9;}  
	inline bool enter(char ch){return ch==10||ch==13;}
	inline void read(int &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (sign)x=-x;  
	}  
	inline void read(ll &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (sign)x=-x;  
	}  
	inline void read(double &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (ch==46){  
			double tmp=1; ch=nc();  
			for (;ch>=48&&ch<=57;ch=nc())tmp/=10.0,x+=tmp*(ch-48);  
		}  
		if (sign)x=-x;  
	}  
	inline void read(char *s){  
		char ch=nc();  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;  
		*s=0;  
	}  
	inline void readln(char *s) {
		char ch=nc();
		for (;blank(ch);ch=nc());
		if(IOerror)return;
		for(;!enter(ch)&&!IOerror;ch=nc())*s++=ch;
		*s=0;
	}
	inline void read(char &c){  
		for (c=nc();blank(c);c=nc());  
		if (IOerror){c=-1;return;}  
	} 
#undef OUT_SIZE  
#undef BUF_SIZE  
}
using fastIO::read;

int main() {
	int n;
	while(~scanf("%d",&n)) {
		const double PI = acos(-1.0);
		double sita = 2*PI/n;
		double x = sqrt(2-2*cos(sita));
		double area = n*sin(sita)/2;
		area += x*(1-cos(sita/2))/2;
		printf("%.6f\n",area);
	}
	return 0;
}

H

注意去重，以及这题只有12个小时（草

/* ***********************************************
Author        :BPM136
Created Time  :2019/8/2 14:17:15
File Name     :H.cpp
************************************************ */

#include 
#include 
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define USE_CIN_COUT ios::sync_with_stdio(0)
#define filein(x) freopen(#x".in","r",stdin)
#define fileout(x) freopen(#x".out","w",stdout)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define mkd(x) freopen(#x".in","w",stdout);

using namespace std;

int random(int l, int r) {
	static std::random_device rd;
	struct timeb timeSeed;
	ftime(&timeSeed);
	size_t seed = timeSeed.time * 1000 + timeSeed.millitm;  // milli time
	static std::mt19937 gen(seed);
	std::uniform_int_distribution<> u(l, r);
	return u(gen);
}

typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef pair<int, int> pii;

struct tim {
	int h, m, s;
	void in() {
		cin >> h >> m >> s;
		if (h >= 12)
			h -= 12;
		assert(0 <= h && h < 12);
	}
	void out() {
		cout << h << ' ' << m << ' ' << s << '\n';
	}
	bool operator != (tim const& oth) const {
		return h != oth.h || m != oth.m || s != oth.s;
	}
} S;

int dis(tim const& x, tim const& y) {
	int s1 = x.h * 60 * 60 + x.m * 60 + x.s;
	int s2 = y.h * 60 * 60 + y.m * 60 + y.s;
	int ds = s2 - s1;
	if (ds < 0)
		ds += 60 * 60 * 12;
	assert(ds >= 0);
	return ds;
}

bool cmp(tim const& x, tim const& y) {
	int dx = dis(S, x), dy = dis(S, y);
	return dx < dy;
}

int main() {
	USE_CIN_COUT;
	int n;
	cin >> n;
	S.in();
	auto b = vector<tim>(n);
	for (auto& v : b)
		v.in();
	auto a = vector<tim>();
	for (int i = 0; i < n; ++i) 
		if (S != b[i]) 
			a.push_back(b[i]);

	sort(a.begin(), a.end(), cmp);
	int ans = min(dis(S, a[SZ(a) - 1]), dis(a[0], S));
	for (int i = 0; i < SZ(a); ++i) {
		int nxt = i + 1;
		if (nxt == SZ(a))
			nxt = 0;
		if (nxt == i)
			continue;
		ans = min(ans, dis(S, a[i]) * 2 + dis(a[nxt], S));
		ans = min(ans, dis(a[nxt], S) * 2 + dis(S, a[i]));
	}
	cout << fixed << setprecision(2) << ans * 6.00 << '\n';
    return 0;
}

I

最开始以为$a_i$是不固定的x
如果固定的话只需要7维表示每个的情况即可，然后$O(k)$算贡献

J

考虑横着某个位置切一刀即可x

/* ***********************************************
Author        :BPM136
Created Time  :2019/8/2 13:48:17
File Name     :J.cpp
************************************************ */

#include 
#include 
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define USE_CIN_COUT ios::sync_with_stdio(0)
#define filein(x) freopen(#x".in","r",stdin)
#define fileout(x) freopen(#x".out","w",stdout)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define mkd(x) freopen(#x".in","w",stdout);

using namespace std;

int random(int l, int r) {
	static std::random_device rd;
	struct timeb timeSeed;
	ftime(&timeSeed);
	size_t seed = timeSeed.time * 1000 + timeSeed.millitm;  // milli time
	static std::mt19937 gen(seed);
	std::uniform_int_distribution<> u(l, r);
	return u(gen);
}

typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef pair<int, int> pii;

int main() {
	USE_CIN_COUT;
	ll n;
	while (cin >> n) {
		ll ans = 7 + n * (6 + 6 + n - 1) / 2;
		cout << ans << '\n';
	}
    return 0;
}

K

草为啥数据是多组数据但是题面没有说明啊(#`O′)
显然只有n和m都是4的倍数的时候才可以
然后复制下样例就完事了x

/* ***********************************************
Author        :BPM136
Created Time  :2019/8/2 14:35:08
File Name     :K.cpp
************************************************ */

#include 
#include 
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define USE_CIN_COUT ios::sync_with_stdio(0)
#define filein(x) freopen(#x".in","r",stdin)
#define fileout(x) freopen(#x".out","w",stdout)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define mkd(x) freopen(#x".in","w",stdout);

using namespace std;

int random(int l, int r) {
	static std::random_device rd;
	struct timeb timeSeed;
	ftime(&timeSeed);
	size_t seed = timeSeed.time * 1000 + timeSeed.millitm;  // milli time
	static std::mt19937 gen(seed);
	std::uniform_int_distribution<> u(l, r);
	return u(gen);
}

typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef pair<int, int> pii;

string s[4];
char ans[15][15];

void color_fill(int x, int y) {
	for (int i = 0; i < 4; ++i)
		for (int j = 0; j < 4; ++j)
			ans[x + i][y + j] = s[i][j];
}

int main() {
	s[0] = "1113";
	s[1] = "2133";
	s[2] = "2243";
	s[3] = "2444";
	int n, m;
	while (cin >> n >> m) {
		if (n % 4 || m % 4) {
			cout << "no response" << '\n';
			continue;
		}
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= m; ++j) 
				if (i % 4 == 1 && j % 4 == 1) 
					color_fill(i, j);
		for (int i = 1; i <= n; ++i) {
			for (int j = 1; j <= m; ++j) 
				cout << ans[i][j];
			cout << '\n';
		}
	}
    return 0;
}