赛中一人写了4题,有很大水分。
F作为签到题也是一个结论,虽然并没有猜出来但是也水过了。
J题是伽马函数,(虽然看着像二项式展开),计算了前几项OEIS也水过了,估计现场赛会推给队友吧。
I题是HDOJ3551一般图最大匹配的简化版?我用最大流水过了,出题人说最大流做法是错误的。
H是拉格朗日插值?我用费用流+map水过了……
//#pragma comment(linker, "/STACK:102400000,102400000")
#include
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
//#define int ll
const int maxn=2e5+10,inf=0x3f3f3f3f,mod=1000000007;
const int MAXN = 100010;
struct SuffixArray
{//倍增算法 O(nlogn)
int n;//包括末尾0在内的字符数
int Rank[MAXN], height[MAXN], sa[MAXN], r[MAXN];
//rnk从0开始
//sa从1开始,因为最后一个字符(最小的)排在第0位
//height从1开始,因为表示的是sa[i - 1]和sa[i]
int wa[MAXN], wb[MAXN], wv[MAXN], ws_[MAXN];
//Suffix函数的参数m代表字符串中字符的取值范围,是基数排序的一个参数,如果原序列都是字母可以直接取128,如果原序列本身都是整数的话,则m可以取比最大的整数大1的值
//待排序的字符串放在r数组中,从r[0]到r[n-1],长度为n
//为了方便比较大小,可以在字符串后面添加一个字符,这个字符没有在前面的字符中出现过,而且比前面的字符都要小
//同上,为了函数操作的方便,约定除r[n-1]外所有的r[i]都大于0,r[n-1]=0
//函数结束后,结果放在sa数组中,从sa[0]到sa[n-1],sa[0]=n-1
void Suffix(int m)
{
int i, j, k, *x = wa, *y = wb, *t;
//对长度为1的字符串排序
//一般来说,在字符串的题目中,r的最大值不会很大,所以这里使用了基数排序
//如果r的最大值很大,那么把这段代码改成快速排序
for(i = 0; i < m; ++i) ws_[i] = 0;
for(i = 0; i < n; ++i) ws_[x[i] = r[i]]++;//统计字符的个数
for(i = 1; i < m; ++i) ws_[i] += ws_[i - 1];//统计不大于字符i的字符个数
for(i = n - 1; i >= 0; --i) sa[--ws_[x[i]]] = i;//计算字符排名
//基数排序
//x数组保存的值相当于是rank值
for(j = 1, k = 1; k < n; j<<=1, m = k)
{
//j是当前字符串的长度,数组y保存的是对第二关键字排序的结果
//第二关键字排序
for(k = 0, i = n - j; i < n; ++i) y[k++] = i;//第二关键字为0的排在前面
for(i = 0; i < n; ++i) if(sa[i] >= j) y[k++] = sa[i] - j;//长度为j的子串sa[i]应该是长度为2 * j的子串sa[i] - j的后缀(第二关键字),对所有的长度为2 * j的子串根据第二关键字来排序
for(i = 0; i < n; ++i) wv[i] = x[y[i]];//提取第一关键字
//按第一关键字排序 (原理同对长度为1的字符串排序)
for(i = 0; i < m; ++i) ws_[i] = 0;
for(i = 0; i < n; ++i) ws_[wv[i]]++;
for(i = 1; i < m; ++i) ws_[i] += ws_[i - 1];
for(i = n - 1; i >= 0; --i) sa[--ws_[wv[i]]] = y[i];//按第一关键字,计算出了长度为2 * j的子串排名情况
//此时数组x是长度为j的子串的排名情况,数组y仍是根据第二关键字排序后的结果
//计算长度为2 * j的子串的排名情况,保存到数组x
t = x;
x = y;
y = t;
for(x[sa[0]] = 0, i = k = 1; i < n; ++i)
x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + j] == y[sa[i] + j]) ? k - 1 : k++;
//若长度为2 * j的子串sa[i]与sa[i - 1]完全相同,则他们有相同的排名
}
}
void build_height()
{
int i, j, k = 0;
for(i = 0; i < n; i++)
Rank[sa[i]] = i;
for(i = 0; i < n; i++)
{
if(k) k--;
int j = sa[Rank[i]-1];
while(r[i+k] == r[j+k]) k++;
height[Rank[i]] = k;
}
}
void debug()
{
printf(" 名次 下标 height 后缀串\n");
for(int i=0;i<n;i++)
{
printf("%5d%5d%5d ",i,sa[i],height[i]);
for(int j=sa[i];j<n;j++)
{
printf("%-3d ",r[j]);
}
printf("\n");
}
}
} sa;
signed main()
{
std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#ifdef DEBUG
freopen("input.in", "r", stdin);
// freopen("output.out", "w", stdout);
#endif
int n;
string s;
while(cin>>n>>s)
{
int a,b;
a=b=n;
sa.n=n+1;
sa.r[n]=n+1;
for(int i=n-1;i>=0;i--)
{
if(s[i]=='a')
{
if(a<n)
sa.r[i]=a-i;
else
sa.r[i]=n;
a=i;
}
else{
if(b<n)
sa.r[i]=b-i;
else
sa.r[i]=n;
b=i;
}
}
// for(int i=0;i<=n;i++)
// cout<
// cout<
sa.Suffix(n+2);
for(int i=n;i>=0;i--)
if(sa.sa[i]!=n)
cout<<sa.sa[i]+1<<' ';
cout<<endl;
}
return 0;
}
//#pragma comment(linker, "/STACK:102400000,102400000")
#include
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
//#define int ll
const int maxn=2e5+10,inf=0x3f3f3f3f,mod=1000000007;
void show(int flag,bool tra)
{
if((flag==1&&!tra)||(flag==2&&tra))
cout<<"<"<<endl;
else if((flag==2&&!tra)||(flag==1&&tra))
cout<<">"<<endl;
else
cout<<"="<<endl;
}
signed main()
{
std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#ifdef DEBUG
freopen("input.in", "r", stdin);
// freopen("output.out", "w", stdout);
#endif
string a,b;
while(cin>>a)
{
cin>>b;
int ok=0;//1小,2大,3等
bool tra=0;
int n=a.length(),m=b.length();
int lim=min(n,m);
if(n>m)
{
tra=1;
swap(n,m);
swap(a,b);
}
b+=b;
m<<=1;
for(int i=0;i<m;i++)
{
if(a[i%n]<b[i])
{
ok=1;
break;
}
else if(a[i%n]>b[i])
{
ok=2;
break;
}
}
show(ok,tra);
}
return 0;
}
//#pragma comment(linker, "/STACK:102400000,102400000")
#include
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
//#define int ll
const int maxn=2e5+10,inf=0x3f3f3f3f,mod=1000000007;
ll gcd(ll a,ll b)
{
if(a<b)
swap(a,b);
ll r;
while((r=a%b)){
a=b;
b=r;
}
return b;
}
ll lcm(ll a,ll b)
{
return (a*b)/gcd(a,b);
}
map<int,int>mp;
struct MCMF
{
struct Edge
{
int from, to, cap, flow, cost;
Edge(int u, int v, int c, int f, int w)
: from(u), to(v), cap(c), flow(f), cost(w) {}
};
int n, m;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn]; //是否在队列中
int d[maxn]; //bellmanford
int p[maxn]; //上一条弧
int a[maxn]; //可改进量
void init(int n)
{
this->n = n;
for (int i = 0; i <= n; i++)
G[i].clear();
edges.clear();
}
void AddEdge(int from, int to, int cap, int cost)
{
// edges.emplace_back(Edge(from, to, cap, 0, cost));
// edges.emplace_back(Edge(to, from, 0, 0, -cost));
edges.push_back(Edge(from, to, cap, 0, cost));
edges.push_back(Edge(to, from, 0, 0, -cost));
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BellmanFord(int s, int t, int& flow, ll& cost)
{
for (int i = 0; i < n; i++)
d[i] = inf;
memset(inq, 0, sizeof(inq));
d[s] = 0;
inq[s] = 1;
p[s] = 0;
a[s] = inf;
queue<int> q;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
inq[u] = 0;
for (int i = 0; i < G[u].size(); i++)
{
Edge& e = edges[G[u][i]];
if (e.cap > e.flow && d[e.to] > d[u] + e.cost)
{
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if (!inq[e.to])
{
q.push(e.to);
inq[e.to] = 1;
}
}
}
}
// cout<<"DEBUG:BellmanFord\n";
// cout<<"flow"<
if (d[t] == inf)
return false; // 当没有可增广的路时退出
flow += a[t];
cost += (ll)d[t] * (ll)a[t];
mp[d[t]*a[t]]+=a[t];
for (int u = t; u != s; u = edges[p[u]].from)
{
edges[p[u]].flow += a[t];
edges[p[u] ^ 1].flow -= a[t];
}
return true;
}
int MincostMaxflow(int s, int t, ll& cost)
{
int flow = 0;
cost = 0;
while(BellmanFord(s, t, flow, cost));
return flow;
}
} mm;
int cost[maxn];
signed main()
{
// std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#ifdef DEBUG
freopen("input.in", "r", stdin);
// freopen("output.out", "w", stdout);
#endif
ll n,m,u,v,w,c,q;
while(~scanf("%lld%lld",&n,&m))
{
mm.init(n+10);
mp.clear();
for(int i=1;i<=m;i++)
{
scanf("%lld%lld%lld",&u,&v,&cost[i]);
// cin>>u>>v>>cost[i];
mm.AddEdge(u,v,1,cost[i]);
}
ll tot=0,mf=mm.MincostMaxflow(1,n,tot);
scanf("%lld",&q);
// cin>>q;
while(q--)
{
scanf("%lld%lld",&u,&v);
// cin>>u>>v;
if(mf*u<v)
puts("NaN");
else{//容量乘以u,跑v的流量
ll now=0,x=0,y=v;
for(auto &i:mp)
{
// printf("%d,%d\n",i.first,i.second);
if(v>i.second*u)
{
v-=i.second*u;
x+=i.first*u;
}
else{
x+=i.first*v/i.second;
break;
}
}
ll g=gcd(x,y);
printf("%lld/%lld\n",x/g,y/g);
}
}
}
return 0;
}
//#pragma comment(linker, "/STACK:102400000,102400000")
#include
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
//#define int ll
const int maxn=2e5+10,inf=0x3f3f3f3f,mod=1000000007;
struct edge
{
int u,v;
edge(int u=0,int v=0):
u(u),v(v){}
};
struct Dinic
{
struct Edge
{
int from, to, cap, flow;
Edge(int u, int v, int c, int f):
from(u), to(v), cap(c), flow(f) {}
};
int n, m, s, t; //结点数,边数(包括反向弧),源点编号和汇点编号
vector<Edge> edges; //边表。edge[e]和edge[e^1]互为反向弧
vector<int> G[maxn]; //邻接表,G[i][j]表示节点i的第j条边在e数组中的序号
bool vis[maxn]; //BFS使用
int d[maxn]; //从起点到i的距离
int cur[maxn]; //当前弧下标
void init(int n)
{
this->n = n;
for (int i = 0; i < n; i++) G[i].clear();
edges.clear();
}
void AddEdge(int from, int to, int cap)
{
// edges.emplace_back(Edge(from, to, cap, 0));//魔改蔡队模板
// edges.emplace_back(Edge(to, from, 0, 0)); //反向弧
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BFS()
{
memset(vis, 0, sizeof(vis));
memset(d, 0, sizeof(d));
queue<int> q;
q.push(s);
d[s] = 0;
vis[s] = 1;
while (!q.empty())
{
int x = q.front();
q.pop();
for (int i = 0; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow)
{
vis[e.to] = 1;
d[e.to] = d[x] + 1;
q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x, int a)//x为当前点,a为当前边上流量
{
if (x == t || a == 0)//到达目标/流量为0
return a;
int flow = 0, f;
for (int& i = cur[x]; i < G[x].size(); i++)
{ //从上次考虑的弧
Edge& e = edges[G[x][i]];
if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)
{
e.flow += f;
edges[G[x][i] ^ 1].flow -= f;
flow += f;
a -= f;
if (a == 0)
break;
}
}
if(!flow)//不知道可不可以加
d[x] = -1;//炸点优化必不可少,证明不必要的点下一次就不用遍历
return flow;
}
int Maxflow(int s, int t)
{
this->s = s, this->t = t;
int flow = 0;
while (BFS())
{
memset(cur, 0, sizeof(cur));
flow += DFS(s,inf);
}
return flow;
}
} di;
int d[maxn],deg[maxn];
signed main()
{
std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#ifdef DEBUG
freopen("input.in", "r", stdin);
// freopen("output.out", "w", stdout);
#endif
int n,m;
while(cin>>n>>m)
{
int s=2*n+1,t=2*n+2,tot=0;
for(int i=1;i<=n;i++)
cin>>d[i],tot+=d[i];
vector<edge>g(m);
di.init(2*n+2);
for(auto &e:g)
{
cin>>e.u>>e.v;
di.AddEdge(e.u,e.v+n,1);
di.AddEdge(e.v,e.u+n,1);
}
for(int i=1;i<=n;i++)
{
di.AddEdge(s,i,d[i]);
di.AddEdge(n+i,t,d[i]);
}
cout<<(di.Maxflow(s,t)==tot?"Yes":"No")<<endl;
}
return 0;
}
//#pragma comment(linker, "/STACK:102400000,102400000")
#include
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
//#define int ll
const int maxn=2e6+10,inf=0x3f3f3f3f,mod=998244353;
ll fac[maxn],a[maxn];
ll quick(ll a,ll b){//快速幂
ll ans=1;
while(b){
if(b&1)ans=ans*a%mod;
a=a*a%mod;
b/=2;
}
return ans%mod;
}
ll ccc(ll n,ll m){//求组合数
return (fac[n] * quick(fac[m], mod - 2) % mod * quick(fac[n - m], mod - 2) % mod)%mod;
}
void initccc()
{
fac[0] = 1;
for (int i = 1; i <maxn ;i++){
fac[i] = fac[i - 1] * i % mod;
}
}
signed main()
{
std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#ifdef DEBUG
freopen("input.in", "r", stdin);
// freopen("output.out", "w", stdout);
#endif
ll n;
initccc();
while(cin>>n)
{
cout<<fac[n]*fac[n]%mod*quick(fac[2*n+1],mod-2)%mod<<endl;
// cout<<(quick(fac[2*n+1]*quick(fac[n]*fac[n]%mod,mod-2), mod - 2))%mod<
}
return 0;
}