今日SGU 5.9

SGU 297

题意：就是求余数

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+;
const int maxn = 1e5+;
const double eps = 1e-;
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==?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=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=,f=;char ch=getchar();
    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while (ch>=''&&ch<=''){x=x*+ch-'';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 main(){
    int n,m;
    scanf("%d%d",&n,&m);
    int sum=,x;
    rep(i,,m) scanf("%d",&x),sum+=x;
    printf("%d\n",sum%n);
    return ;
}

SGU 152

题意：求每个数占这些数总和的百分比*100，不是整数的可以向上或者向下取整，然后要求最后百分比和为100

收获：先全部非整数的向下取整，然后不够的就从这些非整的+1最后凑出100

#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+;
const int maxn = 1e4+;
const double eps = 1e-;
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==?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=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=,f=;char ch=getchar();
    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while (ch>=''&&ch<=''){x=x*+ch-'';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];
bool fg[maxn];
int main(){
    int n,sum=,cnt=;
    scanf("%d",&n);
    int h = ;
    rep(i,,n) scanf("%d",&a[i]),sum+=a[i];
    mt(fg,false);
    rep(i,,n){
        if(100.0*a[i]/sum!=*a[i]/sum) fg[i] = true;
        h-=*a[i]/sum;
    }
    rep(i,,n){
//        de(fg[i])
        if(fg[i]){
            if(h) printf("%d%c",*a[i]/sum+," \n"[i+==n]),h--;
            else printf("%d%c",*a[i]/sum," \n"[i+==n]);
        }else {
            printf("%d%c",*a[i]/sum," \n"[i+==n]);
        }
    }
    return ;
}

SGU 124

题意：求一个点和一个简单多边形的关系，在边界，在外面，在内部

收获：dfs组成简单多边形，然后多边形和点关系的模板

#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+;
const int N = 1e4+;
const double eps = 1e-;
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==?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=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=,f=;char ch=getchar();
    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
int tot = ,cnt = ;

struct P {
    double x,y;
    P() {}
    P(double x, double y) {
        this->x = x;
        this->y = y;
    }
    P operator + (const P &c) const {
        return P(x + c.x, y + c.y);
    }
    P operator - (const P &c) const {
        return P(x - c.x, y - c.y);
    }
    P operator * (const db &c) const {
        return P(x * c, y * c);
    }
    P operator / (const db &c) const {
        return P(x / c, y / c);
    }
};

bool vis[N];
P p[N];

db x(P a){
    return a.x;
}

db y(P a){
    return a.y;
}

void print(P p) {
    printf("%lf %lf\n",p.x,p.y);
}

int sign(double x) {
    return (x>eps)-(x<-eps);
}
db dot(P a, P b) {
    return x(a) * x(b) + y(a) * y(b);
}
double cross(P a, P b) {
    return x(a) * y(b) - x(b) * y(a);
}

//判断线段是否规范相交（交点不在任一个端点上）
bool isSS0(P a1, P a2, P b1, P b2) {
    double c1 = cross(a2 - a1, b1 - a1), c2 = cross(a2 - a1, b2 - a1),
       c3 = cross(b2 - b1, a1 - b1), c4 = cross(b2 - b1, a2 - b1);
    return sign(c1) * sign(c2) <  && sign(c3) * sign(c4) < ;
}
//判断线段是否不规范相交
bool isSS1(P a1, P a2, P b1, P b2) {
    double c1 = cross(a2 - a1, b1 - a1), c2 = cross(a2 - a1, b2 - a1),
       c3 = cross(b2 - b1, a1 - b1), c4 = cross(b2 - b1, a2 - b1);
    return sign(max(x(a1), x(a2)) - min(x(b1), x(b2))) >=  &&
           sign(max(x(b1), x(b2)) - min(x(a1), x(a2))) >=  &&
           sign(max(y(a1), y(a2)) - min(y(b1), y(b2))) >=  &&
           sign(max(y(b1), y(b2)) - min(y(a1), y(a2))) >=  &&
           sign(c1) * sign(c2) <=  && sign(c3) * sign(c4) <= ;
}
//判断点是否在线段上（不包括端点）
bool onS0(P p, P a, P b) {
    return sign(cross(p - a, b - a)) ==  && sign(dot(p - a, p - b)) < ;
}
//判断点是否在线段上（包括端点）
bool onS1(P p, P a, P b) {
    return sign(cross(p - a, b - a)) ==  && sign(dot(p - a, p - b)) <= ;
}
//判断点和多边形关系 边上-1 外0 内1
int Pinploy(P o, P *p, int n) {
    int res = ;
    rep(i, , n) {
        P u = p[i], v = p[(i + ) % n];
        if(onS1(o, u, v)) return -;
        int k = sign(cross(v - u, o - u));
        int d1 = sign(y(u) - y(o));
        int d2 = sign(y(v) - y(o));
        if(k >  && d1 <=  && d2 > ) ++res;
        if(k <  && d2 <=  && d1 > ) --res;
    }
    return res != ;
}
map<pii,int> m;
map<int,pii> mm;
vector<int> G[N];
int id(pii a){
    if(m.count(a)) return m[a];
    else m[a]=(tot++);
    mm[tot-] = a;
    return m[a];
}
void add(int u,int v){
    G[u].pb(v),G[v].pb(u);
}
void dfs(int u){
    vis[u]=true;
    p[cnt++]=P(mm[u].fi,mm[u].se);
    rep(i,,sz(G[u])){
        int v=G[u][i];
        if(vis[v]) continue;
        dfs(v);
    }
}
int main(){
    int n,xx,yy;
    scanf("%d",&n);
    rep(i,,n){
        int a,b,c,d;
        scanf("%d%d%d%d",&a,&b,&c,&d);
        int t = id(mp(a,b));
        int tt = id(mp(c,d));
        add(t,tt);
    }
    dfs();
//    de(cnt)
    scanf("%d%d",&xx,&yy);
    P pp = P(xx,yy);
    int ans = Pinploy(pp,p,cnt);
    if(ans==) puts("INSIDE");
    if(ans==) puts("OUTSIDE");
    if(ans==-) puts("BORDER");
    return ;
}