A、给你一堆数字，你可以从这些数字取出任意多个组合成一个新的数字，问你这个新的数字在k进制下的最后一位一共有多少种情况
首先考虑k进制下的最后一位是这个数字对k取余，然后有一个扩展贝祖定理：
a1x1+...+anxn=d*gcd(a1...an)
其中d是正整数
那么把gcd求出来然后算20000次倍数丢进set即可
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define scs(a) scanf("%s",a) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;} bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read(){ char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x){ if(x<0)putchar('-'),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int gcd(int a, int b){ if(a==0) return b; if(b==0) return a; if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1; else if(!(b&1)) return gcd(a,b>>1); else if(!(a&1)) return gcd(a>>1,b); else return gcd(abs(a-b),min(a,b)); } int lcm(int x,int y){return x*y/gcd(x,y);} int n,k,a[maxn]; set<int> s; signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>n>>k; cin>>a[1]; int g=a[1]; re(i,2,n) cin>>a[i],g=gcd(g,a[i]); for(int i=1;i<=200000;++i) s.insert((g*i)%k); cout<<s.size()<<endl; for(auto i=s.begin();i!=s.end();++i) cout<<*i<<' '; return 0; }
B、给你一棵树，每个点有点权a，同时给定从根节点到该节点所有点权a的和s
现在把某些s擦除了，需要你重建树使得a的和最小，不能重建输出-1，点权必须大于等于0
输出-1的条件很好说，如果父节点的s小于当前节点s，输出-1
简单的贪心做法是能填0则填0，然后递归地计算儿子
但是这样的做法有个问题，如果你在当前层填了0，那么可能在儿子上会产生额外消耗，看一个样例：
5
1 2 2 3
1 -1 2 3 -1
这里如果把三号节点填上0，那么四号和五号节点的a总和是3，但是如果三号节点填上1，四号和五号节点的总和是1
这意味着我们在填数字的时候应该尽量令父节点的a值更大，那么才有可能使得儿子的a值更小
需要知道儿子数量不会比1小
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define scs(a) scanf("%s",a) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;} bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read(){ char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x){ if(x<0)putchar('-'),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int gcd(int a, int b){ if(a==0) return b; if(b==0) return a; if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1; else if(!(b&1)) return gcd(a,b>>1); else if(!(a&1)) return gcd(a>>1,b); else return gcd(abs(a-b),min(a,b)); } int lcm(int x,int y){return x*y/gcd(x,y);} int n; int p[maxn],s[maxn],a[maxn]; vei g[maxn]; bool can=1; void dfs(int x,int f){ if(!can) return; if(s[x]==-1){ int mn=inf; fo(i,0,g[x].size()){ if(g[x][i]!=f) mn=min(mn,s[g[x][i]]); } // cout<<mn<<endl; if(mn>s[f]&&mn!=inf) s[x]=mn,a[x]=mn-s[f]; else s[x]=s[f],a[x]=0; } else{ if(s[f]>s[x]){ can=0; return; } else{ a[x]=s[x]-s[f]; } } fo(i,0,g[x].size()){ if(g[x][i]!=f) dfs(g[x][i],x); } } signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); p[1]=1; cin>>n; re(i,2,n) cin>>p[i],g[p[i]].pub(i); re(i,1,n) cin>>s[i]; dfs(1,0); int ans=0; re(i,1,n) ans+=a[i]; if(can) cout<<ans; else cout<<-1; return 0; }
C、签到，给你一个字符串，你需要删去一个连续子串使得剩下的串的字符集大小为1，保证原串至少有两种不同的字符
计算一下前缀最长字符集为1的子串和后缀最长字符集为1的子串，累加一下答案即可
注意特判一下前缀和后缀的字符一模一样的情况，乘法取个余即可
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define scs(a) scanf("%s",a) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=998244353; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;} bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read(){ char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x){ if(x<0)putchar('-'),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int gcd(int a, int b){ if(a==0) return b; if(b==0) return a; if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1; else if(!(b&1)) return gcd(a,b>>1); else if(!(a&1)) return gcd(a>>1,b); else return gcd(abs(a-b),min(a,b)); } int lcm(int x,int y){return x*y/gcd(x,y);} int n; string s; signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>n; cin>>s; s='#'+s; int c1=1,ans=0; char c=s[1]; re(i,2,n){ if(s[i]==c) c1++; else break; } int c2=1; c=s[n]; rre(i,n-1,1){ if(s[i]==c) c2++; else break; } ans=c1+c2; if(s[1]==s[n]){ // cout<<"nmsl"<<endl; ans+=(c1*c2)%MOD; ans%=MOD; } ans+=1; cout<<ans; return 0; }
D、给你平面上的一堆点，你需要找到一个圆使得这个圆与y=0至多只有一个交点，并且这个圆包含了所有的点
输出最小的半径，如果不存在这样的圆输出-1
考虑-1的情况，如果y=0两侧都有点则一定不可能，剩下的所有情况都可以画出一个巨型的圆使得它满足条件
现在可以二分圆的半径了
圆心位于y=R这条直线上，R是二分的半径，求出以所有点为圆心，R为半径的圆与y=R的交线，如果这些交线至少有一个公共交点，那么返回true
这个可能有点绕，考虑一个圆，半径为r包含了某个点的情况，那么圆心一定在以这个点为圆心，r为半径的圆里，画图康康就知道了
剩下的是计算交线的位置，维护左右区间即可，如下图：
注意特判一下l小于0的情况
以及double的二分最好限定次数，每次令l=m或者r=m即可，不需加加减减
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define scs(a) scanf("%s",a) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} //bool chk(int now){} //int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;} bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read(){ char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x){ if(x<0)putchar('-'),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int gcd(int a, int b){ if(a==0) return b; if(b==0) return a; if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1; else if(!(b&1)) return gcd(a,b>>1); else if(!(a&1)) return gcd(a>>1,b); else return gcd(abs(a-b),min(a,b)); } int lcm(int x,int y){return x*y/gcd(x,y);} int n; struct point{ db x,y; void read(){ cin>>x>>y; } }p[maxn]; bool chk(db R){ db l=-inf,r=inf; re(i,1,n){ db x=p[i].x,y=p[i].y; if(2*y*R-y*y<0.0) return 0; db len=sqrt(2*y*R-y*y); if(x-len>r||x+len<l) return 0; l=max(l,x-len),r=min(r,x+len); } // cout<<R<<" nmsl"<<endl; return 1; } db half(db l,db r){ int cnt=1000; while(cnt--){ db m=(l+r)/2.0; if(chk(m)) r=m; else l=m; } return l; } signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cout<<fixed<<setprecision(10); cin>>n; bool neg=0,pos=0; re(i,1,n){ p[i].read(); if(p[i].y>0.0) pos=1; else neg=1; } if(pos&&neg){ cout<<-1; return 0; } re(i,1,n){ p[i].y=fabs(p[i].y); } db ans=half(0,1e17); // cout<<ans<<" nmsl"<<endl; cout<<ans; return 0; }
E、一张有向图，给定起点，问你最少需要加多少条边使得从s可以抵达所有的城市
SALA第一场的原题，只要先缩点然后考虑入度为0的且不为起点的点有几个就完事了
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=5005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read(){ char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x){ if(x<0)putchar('-'),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int n,m,ss,x,y; vei g[maxn],scc[maxn],gg[maxn]; int dfn[maxn],low[maxn]; bool ins[maxn]; stack<int> s; int timer=0,cnt=0,c[maxn]; int deg[maxn],ans=0; void build(){ re(i,1,n){ fo(j,0,g[i].size()){ if(c[i]!=c[g[i][j]]) deg[c[g[i][j]]]++; } } re(i,1,cnt) if(deg[i]==0&&i!=c[ss]) ans++; } void tarjan(int x){ dfn[x]=low[x]=++timer; s.push(x),ins[x]=1; fo(i,0,g[x].size()){ int y=g[x][i]; if(!dfn[y]){ tarjan(y); low[x]=min(low[x],low[y]); } else if(ins[y]) low[x]=min(low[x],dfn[y]); } if(dfn[x]==low[x]){ ++cnt;int y; do{ y=s.top(); s.pop(); ins[y]=0; c[y]=cnt; scc[cnt].pub(y); }while(x!=y); } } signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>n>>m>>ss; re(i,1,m) cin>>x>>y,g[x].pub(y); re(i,1,n) if(!dfn[i]) tarjan(i); // re(i,1,n) cout<<c[i]<<' ';cout<<endl; build(); cout<<ans; return 0; }