2014 Multi-University Training Contest 8小记
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1002 计算几何
最大的速度才可能拥有无限的面积。
最大的速度的点 求凸包, 凸包上的点( 注意不是端点 ) 才拥有无限的面积
注意 : 凸包上如果有重点则不满足。
另外最大的速度为0也不行的。
int cmp(double x){ if(fabs(x) < 1e-8) return 0 ; if(x > 0) return 1 ; return -1 ;}struct point{ int x , y ; point(){} point(int _x , int _y):x(_x) , y(_y){} friend bool operator == (const point &a , const point &b){ return cmp(a.x - b.x) == 0 && cmp(a.y - b.y) == 0 ; } friend double operator ^ (const point &a , const point &b){ return a.x * b.y - a.y * b.x ; } point operator - (point o){ return point(x - o.x , y - o.y) ; }};struct Poly{ vector<point> p ; Poly(){} Poly(int s = 0){ p.resize(s) ; }} ;bool cmpless(const point &a , const point &b){ return cmp(a.x - b.x) < 0 || cmp(a.x - b.x == 0) && cmp(a.y - b.y) < 0 ;}Poly convex_hull(vector<point> a){ Poly src(2 * a.size() + 5) ; sort(a.begin() , a.end() , cmpless) ; a.erase(unique(a.begin() , a.end()) , a.end()) ; int m = 0 ; for(int i = 0 ; i < a.size() ; i++){ while(m > 1 && cmp( (src.p[m-1] - src.p[m-2]) ^ (a[i] - src.p[m-2]) ) <= 0) m-- ; src.p[m++] = a[i] ; } int k = m ; for(int i = a.size() - 2 ; i >= 0 ; i--){ while(m > k && cmp( (src.p[m-1] - src.p[m-2]) ^ (a[i] - src.p[m-2])) <= 0) m-- ; src.p[m++] = a[i] ; } src.p.resize(m) ; if(a.size() > 1) src.p.resize(m-1) ; return src ;}struct node{ point a ; int v ; int id ; int recover ;}man[508] ;int answer[508] ;int main(){ int n , T = 1 , mxv , i , j ; while(cin>>n && n){ mxv = -1 ; for(i = 0 ; i < n ; i++){ scanf("%d%d%d" , &man[i].a.x , &man[i].a.y , &man[i].v) ; man[i].id = i ; mxv = max(mxv , man[i].v) ; } printf("Case #%d: " , T++) ; if(mxv == 0){ for(i = 0 ; i < n ; i++) putchar('0') ; puts("") ; continue ; } for(i = 0 ; i < n ; i++) man[i].recover = 0 ; for(i = 0 ; i < n ; i++){ if(man[i].v != mxv) continue ; for(j = i+1 ; j < n ; j++){ if(man[i].v == man[j].v && man[i].a == man[j].a){ man[i].recover = man[j].recover = 1 ; } } } vector<node> Man ; vector<point> lis ; for(i = 0 ; i < n ; i++){ if(man[i].v == mxv){ Man.push_back(man[i]) ; lis.push_back(man[i].a) ; } } Poly po = convex_hull(lis) ; memset(answer , 0 , sizeof(answer)) ; for(i = 0 ; i < Man.size() ; i++){ node now = Man[i] ; if(now.recover == 1) continue ; for(j = 0 ; j < po.p.size() ; j++){ if(cmp( (now.a-po.p[j]) ^ (now.a - po.p[(j+1)%po.p.size()] ) ) == 0){ answer[now.id] = 1 ; break ; } } } for(i = 0 ; i < n ; i++) printf("%d" , answer[i]) ; puts("") ; } return 0 ;}
1006 贪心
typedef long long LL ;LL h , a , b , k ;int ok(){ if(h - a < 1) return 1 ; if(h - k * a + (k-1) * b < 1 ) return 1 ; if(- k * a + (k+1) * b < 0) return 1 ; return 0 ;}int main(){ int T = 1 ; while(cin>>h>>a>>b>>k){ if(h == 0 && a==0 && b==0 && k==0 ) break ; printf("Case #%d: " , T++) ; if(ok()) puts("YES") ; else puts("NO") ; } return 0 ;}
1008 找规律
typedef long long LL ;LL a , k ;int main(){ LL i , A , n , t , pt , T = 1 ; while(cin>>a>>k){ if(a == 0 && k == 0) break ; A = a ; pt = -1 ; for(i = 1 ; i <= k ; i++){ if(A % i == 0) t = A / i ; else { A = ((A / i) + 1 ) * i ; t = A / i ; } if(t == pt) break ; pt = t ; } printf("Case #%d: " , T++) ; cout<< t * k << endl ; } return 0 ;}
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