SGU 114
题意：求一个点到其他点的距离总和最小，距离的定义是x轴距离乘以那个点的人数p
收获：带权中位数，按坐标排序，然后扫一遍，最后权值超过或等于总权值的一半时的那个点就是答案，证明暂无
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 1e5+5; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; struct P{ ll x,p; bool operator <(const P c)const{ return x<c.x; } }a[maxn]; int main(){ int n; scanf("%d",&n); ll sum = 0; rep(i,0,n) scanf("%lld%lld",&a[i].x,&a[i].p),sum+=a[i].p; sort(a,a+n); ll cnt = 0; rep(i,0,n) { cnt += a[i].p; if(2*cnt>=sum) return printf("%lld ",a[i].x),0; } return 0; }
SGU 175
题意:定义一个phi函数，phi（1,2，.....，n） = phi(n,n-1,...,n/2+1) + phi(n/2,...,1);
问你初始在第p个位置的元素最后在哪
收获：二分去模拟，处理好对应的下标
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 1e5+5; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; int ans; void phi(int l,int n,int p){ // dd(l)dd(n)de(p) if(n==1){ ans = l; return ; } int mid = n/2; if(mid>=p) phi(n-mid+l,mid,mid-p+1); else phi(l,n-mid,n-p+1); } int main(){ int n,p; scanf("%d%d",&n,&p); phi(1,n,p); printf("%d ",ans); return 0; }
SGU 231
题意：问你有多少给<a,b>对，a<=b,且a，b为素数，且a+b<=n，且a+b也是素数，你会发现a一定是2（如果是其他的话，那么a+b就一定是偶数，所以a+b肯定不是素数），因为2是唯一一个偶数素数，且2是最小素数
收获：2是唯一一个偶数素数
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 1e5+5; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; const int N = 1e6+5; bool isPrime[N]; int prim[80000]; vector<int> ans; int num = 0; int n; void prime(){ memset(isPrime,true,sizeof(isPrime)); isPrime[0] = isPrime[1] = false; for(int i=2 ; i<=N ; i++){ if( isPrime[i] ) prim[num++] = i; for(int j=0 ; j<num ; j++){ if( i*prim[j]>N ) break; isPrime[ i*prim[j] ] = false; if( i%prim[j] == 0 ) break; } } // de(num) } bool ok(int x,int y){ if(x+y>n) return false; return isPrime[x+y]; } int main(){ prime(); scanf("%d",&n); rep(i,0,num){ if(ok(2,prim[i])) ans.pb(prim[i]); } printf("%d ",sz(ans)); rep(i,0,sz(ans)) printf("2 %d ",ans[i]); return 0; }
SGU 134
题意：求树的重心，重心的定义就是去除这点，然后分成的各个联通块的最大点数最小
收获：计算最大子树节点数
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 16005; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; int n,u,v; int minsubtree = inf; int d[maxn]; vector<int> G[maxn]; vector<int> ans; void dfs(int u,int p){ int mn = 0; d[u] = 1; rep(i,0,sz(G[u])){ int v = G[u][i]; if(v==p) continue; dfs(v,u); d[u] += d[v]; mn = max(mn,d[v]); } mn = max(mn,n-d[u]); // dd(u)de(mn) minsubtree=min(mn,minsubtree); } void get_ans(int u,int p){ int mn = 0; d[u] = 1; rep(i,0,sz(G[u])){ int v = G[u][i]; if(v==p) continue; get_ans(v,u); d[u] += d[v]; mn = max(mn,d[v]); } mn = max(mn,n-d[u]); // dd(u)de(mn) if(mn==minsubtree) ans.pb(u); } int main(){ scanf("%d",&n); rep(i,1,n) scanf("%d%d",&u,&v),G[u].pb(v),G[v].pb(u); // if(n==1) return printf("0 0 1 "),0; dfs(1,-1);mt(d,0);get_ans(1,-1); sort(all(ans)); printf("%d %d ",minsubtree,sz(ans)); rep(i,0,sz(ans)) printf("%d%c",ans[i]," "[i+1==sz(ans)]); return 0; } //sgu对内存的要求真高
SGU 180
题意：求逆序对
收获：无
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 65539; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; int a[maxn]; int tmp[maxn<<1]; ll msort(int a[],int l,int r){ if(l==r) return 0; int i,j,k,m=(l+r)>>1; ll t = msort(a,l,m) + msort(a,m+1,r); for(i=l,j=m+1,k=l;k<=r;++k){ if(i<=m&&a[i]<=a[j]||j>r) tmp[k]=a[i++]; else tmp[k]=a[j++],t+=m-i+1; } memcpy(a+l,tmp+l,(r-l+1)*sizeof(a[0])); return t; } int main(){ int n; scanf("%d",&n); rep(i,1,n+1) scanf("%d",a+i); printf("%lld ",msort(a,1,n)); return 0; }