Codeforces Round #437 Div. 1
作者:互联网
A:显然构造一组只包含1和2面值的数据即可。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } signed main() { #ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif int n=read(); cout<<n*2-1<<' '<<2<<endl; cout<<1<<' '<<2; return 0; //NOTICE LONG LONG!!!!! }
B:显然两种pizza的分配应尽可能贴近第二种更优和第一种更优的人数关系。于是在这个边界附近±1暴力枚举一下,然后贪心非常显然。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 100010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,m; ll t=0,ans=0; struct data { int x,y,z; bool operator <(const data&a) const { return y-x>a.y-a.x; } }a[N]; bool cmp(const data&x,const data&y) { return abs(x.y-x.x)>abs(y.y-y.x); } void check(ll k) { if (k<0||k>t) return; ll x=1ll*k*m,y=(t-k)*m,s=0; for (int i=1;i<=n;i++) if (a[i].y>a[i].x) { if (a[i].z<=x) s+=1ll*a[i].y*a[i].z,x-=a[i].z; else if (x) s+=1ll*a[i].y*x,s+=1ll*a[i].x*(a[i].z-x),y-=a[i].z-x,x=0; else s+=1ll*a[i].x*a[i].z,y-=a[i].z; } else { if (a[i].z<=y) s+=1ll*a[i].x*a[i].z,y-=a[i].z; else if (y) s+=1ll*a[i].x*y,s+=1ll*a[i].y*(a[i].z-y),x-=a[i].z-y,y=0; else s+=1ll*a[i].y*a[i].z,x-=a[i].z; } ans=max(ans,s); } signed main() { #ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif n=read(),m=read(); for (int i=1;i<=n;i++) t+=a[i].z=read(),a[i].x=read(),a[i].y=read(); sort(a+1,a+n+1); t=(t-1)/m+1; ll x=0;for (int i=1;i<=n;i++) if (a[i].y>a[i].x) x+=a[i].z;else break; sort(a+1,a+n+1,cmp); for (ll i=x/m-2;i<=x/m+2;i++) check(i); sort(a+1,a+n+1); x=0;for (int i=1;i<=n;i++) if (a[i].y>=a[i].x) x+=a[i].z;else break; sort(a+1,a+n+1,cmp); for (ll i=x/m-2;i<=x/m+2;i++) check(i); cout<<ans; return 0; //NOTICE LONG LONG!!!!! }
D:注意到在某一天卖出再买进对答案是没有影响的。于是维护一个小根堆,每次取出堆顶与当前天价格比较,若能赚钱则计入答案并将堆顶弹出,相当于在堆顶那天买进然后在当前天卖出,然后将当前天价格入堆。我们还需要支持一个反悔操作,即更换买进天的匹配对象,容易发现把当前天价格再入一次堆就可以了。大约是在模拟费用流。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> #include<queue> #include<vector> using namespace std; #define ll long long #define N 300010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,a[N]; ll ans=0; priority_queue<int,vector<int>,greater<int> > q; signed main() { #ifndef ONLINE_JUDGE freopen("d.in","r",stdin); freopen("d.out","w",stdout); #endif n=read(); for (int i=1;i<=n;i++) a[i]=read(); for (int i=1;i<=n;i++) { if (!q.empty()&&q.top()<a[i]) { ans+=a[i]-q.top(); int x=q.top();q.pop(); q.push(a[i]); } q.push(a[i]); } cout<<ans; return 0; //NOTICE LONG LONG!!!!! }
C:期望考虑倒推,但难以计算重置带来的影响。于是考虑先固定住重置的作用,即二分答案,这样重置之后就相当于可以用等于答案的时间直接通关。然后就可以dp了,即设f[i][j]为i~n关要用至多j时间通关的话最少要花多久。最后得到f[1][m],判断其和二分出的答案的大小,若比其大说明一开始就得重置,这显然说明二分出来的答案小了;反之亦然。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 110 #define inf ((double)100000000000) char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,m,a[N],b[N],p[N]; double f[N][N*N]; signed main() { #ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif n=read(),m=read(); for (int i=1;i<=n;i++) a[i]=read(),b[i]=read(),p[i]=read(); double l=0,r=inf,ans; for (int _=1;_<=100;_++) { double mid=(l+r)/2; for (int i=n;i>=1;i--) for (int j=0;j<=m;j++) { f[i][j]=0.01*((a[i]+(j>=a[i]?f[i+1][j-a[i]]:mid))*p[i]+(b[i]+(j>=b[i]?f[i+1][j-b[i]]:mid))*(100-p[i])); if (i>1) f[i][j]=min(f[i][j],mid); } if (f[1][m]>mid) ans=mid,l=mid;else r=mid; } printf("%.11f",ans); return 0; //NOTICE LONG LONG!!!!! }
标签:int,Codeforces,char,while,&&,437,Div,include,getchar 来源: https://www.cnblogs.com/Gloid/p/10488019.html