20190918CF训练
2024-04-26 17:26:40
A、Serial Time!
分层dfs大爆搜，每次往六个方向搜，搜一次答案加一即可
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 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;}
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 k,n,m;
char c[15][15][15];
int x,y;
int ans=0;bool ok(int x,int y,int z){if(x<=0||x>n||y<=0||y>m||z<=0||z>k||c[x][y][z]=='#') return 0;return 1;
}void dfs(int x,int y,int z,int cnt){if(!ok(x,y,z)) return;c[x][y][z]='#';ans++;dfs(x+1,y,z,cnt+1);dfs(x,y+1,z,cnt+1);dfs(x-1,y,z,cnt+1);dfs(x,y-1,z,cnt+1);dfs(x,y,z+1,cnt+1);dfs(x,y,z-1,cnt+1);
}signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>k>>n>>m;re(z,1,k){re(x,1,n){re(y,1,m){cin>>c[x][y][z];}}}cin>>x>>y;dfs(x,y,1,1);
// re(z,1,k) re(x,1,n) re(y,1,m) cout<<c[x][y][z];cout<<ans;return 0;
}B、Help Farmer
暴力枚举ab的值，在o(1)内把c算出来，看一下是否合法就行，最大值可以打表找规律发现是17+8*(n-1)
注意到最小值应该是ab很小的时候存在，把a的范围开小一点，b的范围开大一点，枚举即可
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 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;}
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;signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>n;int mx=(n-1)*8+17,mn=inf;int x,y,z;re(a,2,1000){re(b,3,100000){if(n%((a-1)*(b-2))!=0) continue;int c=n/((a-1)*(b-2))+2;if(a*b*c-n<mn){mn=a*b*c-n;x=a,y=b,z=c;}}}
// cout<<x<<' '<<y<<' '<<z<<endl;cout<<mn<<' '<<mx;return 0;
}
//999893227C、Rectangle and Square
给你八个点，问你能不能构成一个正方形和一个矩形，其中矩形可以为正方形
全排列暴力枚举，检查前四个是否为正方形，注意不仅要检查边长相等，还要检查对角线相等
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 scs(a) scanf("%s",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 ord[9];
pair<db,db> p[9];signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);re(i,1,8)cin>>p[i].fst>>p[i].snd,ord[i]=i;do{bool can=1;db l=len(p[ord[1]].fst,p[ord[1]].snd,p[ord[4]].fst,p[ord[4]].snd);re(i,1,3){if(!eql(l,len(p[ord[i]].fst,p[ord[i]].snd,p[ord[i+1]].fst,p[ord[i+1]].snd))){can=0;break;}}if(!eql(len(p[ord[1]].fst,p[ord[1]].snd,p[ord[3]].fst,p[ord[3]].snd),len(p[ord[2]].fst,p[ord[2]].snd,p[ord[4]].fst,p[ord[4]].snd))) can=0;if(!can) continue;vector<db> v;re(i,5,7)v.pub(len(p[ord[i]].fst,p[ord[i]].snd,p[ord[i+1]].fst,p[ord[i+1]].snd));v.pub(len(p[ord[5]].fst,p[ord[5]].snd,p[ord[8]].fst,p[ord[8]].snd));if(eql(v[0],v[2])&&eql(v[1],v[3])){if(!eql(len(p[ord[5]].fst,p[ord[5]].snd,p[ord[7]].fst,p[ord[7]].snd),len(p[ord[6]].fst,p[ord[6]].snd,p[ord[8]].fst,p[ord[8]].snd))) continue;yyesre(i,1,4) cout<<ord[i]<<' ';cout<<endl;re(i,5,8) cout<<ord[i]<<' ';return 0;}}while(next_permutation(ord+1,ord+9));nnoreturn 0;
}
/*
-1000 -736
1200 408
1728 12
188 -1000
1332 -516
-736 -208
452 -472
804 -120
*/D、 Logo Turtle
一个T和F的字符串，你可以把任意元素翻转任意次，一共有n次翻转机会，翻转就是T变F，F变T
F是前进一步，T是回头，问你n次翻转之后最终抵达的终点到原点的距离最大是多少
考虑三维dp[i][j][1/0]，第三维是朝向，前两维代表前i个字符翻转j次的答案
枚举当前字符的翻转次数k
如果翻成了T，那么dp[i][j][0]=dp[i-1][j-k][1],dp[i][j][1]=dp[i-1][j-k][0]
如果翻成了F，那么dp[i][[j][1]=dp[i-1][j-k][1]+1,dp[i][j][0]=dp[i-1][j-k][0]-1
这里很神秘，注意正方向上的前进等于负方向上的后退
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=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;}
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);}string s;
int l,n;
int dp[105][55][2];signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>s;l=s.length();s='#'+s;cin>>n;memmn(dp);dp[0][0][1]=dp[0][0][0]=0;re(i,1,l){re(j,0,n){re(k,0,j){if(s[i]=='F'){if(k%2==0){dp[i][j][1]=max(dp[i][j][1],dp[i-1][j-k][1]+1);dp[i][j][0]=max(dp[i][j][0],dp[i-1][j-k][0]-1);}else{dp[i][j][1]=max(dp[i][j][1],dp[i-1][j-k][0]);dp[i][j][0]=max(dp[i][j][0],dp[i-1][j-k][1]);}}else if(s[i]=='T'){if(k%2==0){dp[i][j][1]=max(dp[i][j][1],dp[i-1][j-k][0]);dp[i][j][0]=max(dp[i][j][0],dp[i-1][j-k][1]);}else{dp[i][j][1]=max(dp[i][j][1],dp[i-1][j-k][1]+1);dp[i][j][0]=max(dp[i][j][0],dp[i-1][j-k][0]-1);}}}}}
// re(i,1,l) re(j,0,n) cout<<dp[i][j][1]<<" \n"[j==n];cout<<max(dp[l][n][1],dp[l][n][0])<<endl;return 0;
}
/*
TFFTF
1FTFTFTFFFFTFTFTTTTTTFFTTTTFFTFFFTFTFTFFTFTFTFFFTTTFTTFTTTTTFFFFTTT
12
*/E、Hot Bath
两个水龙头各有流速，需要你按照公式在满足条件的情况下找到流速最大的解
那么如果当前温度大于目标温度就降低高温水龙头流速，否则降低低温水龙头流速
按照题意模拟就行，注意一下精度
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=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;}
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 t1,t2,v1,v2,t0;signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>t1>>t2>>v1>>v2>>t0;int a=v1,b=v2;int ansa=0,ansb=0;db dif=inf;while(a>=0&&b>=0){
// cout<<a<<' '<<b<<endl;db d=fabs(((db)t1*a+(db)t2*b)/((db)a+(db)b)-t0);int cnt=t1*a+t2*b;if(cnt>=t0*(a+b)&&smleql(d,dif)){if(a+b>ansa+ansb&&fabs(d-dif)<eps)ansa=a,ansb=b;if(d<dif)ansa=a,ansb=b;dif=d;}if(cnt>=t0*(a+b)) --b;else --a;}cout<<ansa<<' '<<ansb;return 0;
}
/*
10 70 100 100 25
*/转载于:https://www.cnblogs.com/oneman233/p/11562288.html
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