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牛客练习赛 59
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A.小喬和小灰灰
前幾天剛剛學(xué)了序列自動(dòng)機(jī)，這題直接也沒(méi)咋想暴力的做法，直接上序列自動(dòng)機(jī)匹配子序列即可。
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
using namespace std
;
const int N
=1010;
char s
[N
];
int ne
[N
][110],n
;
char p1
[]="XiaoQiao";
char p2
[]="XiaoHuiHui";
void build()
{for(int i
=n
;i
;i
--){for(int j
=0;j
<100;j
++)ne
[i
-1][j
]=ne
[i
][j
];ne
[i
-1][s
[i
]-'A']=i
;}
}
int main()
{IO
;int T
=1;while(T
--){cin
>>s
+1;n
=strlen(s
+1);build();bool ok1
=1,ok2
=1;for(int i
=0,now
=0;i
<8;i
++){now
=ne
[now
][p1
[i
]-'A'];if(!now
) {ok1
=0;break;}}for(int i
=0,now
=0;i
<10;i
++){now
=ne
[now
][p2
[i
]-'A'];if(!now
) {ok2
=0;break;}}if(ok1
&&ok2
) cout
<<"Happy\n";else cout
<<"emm\n";}return 0;
}
B.牛能和小鎮(zhèn)
∣xi2yi?xj2yj+yi2(yi?2xi)?yj2(yj?2xj)∣|x_i^2y_i-x_j^2y_j+y_i^2(y_i-2x_i)-y_j^2(y_j-2x_j)|∣xi2?yi??xj2?yj?+yi2?(yi??2xi?)?yj2?(yj??2xj?)∣
不難發(fā)現(xiàn)這個(gè)式子iii和jjj之間的坐標(biāo)沒(méi)有任何相關(guān)性我們做一步轉(zhuǎn)化
∣xi2yi+yi2(yi?2xi)?(xj2yj+yj2(yj?2xj))∣|x_i^2y_i+y_i^2(y_i-2x_i)-(x_j^2y_j+y_j^2(y_j-2x_j))|∣xi2?yi?+yi2?(yi??2xi?)?(xj2?yj?+yj2?(yj??2xj?))∣
發(fā)現(xiàn)要把每個(gè)點(diǎn)用一個(gè)坐標(biāo)表示即可xi2yi+yi2(yi?2xi)x_i^2y_i+y_i^2(y_i-2x_i)xi2?yi?+yi2?(yi??2xi?)，然后二維轉(zhuǎn)化成一維數(shù)軸上的點(diǎn)，連接就非常容易了
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
using namespace std
;
typedef long long ll
;
const int N
=100010;
ll p
[N
];
int n
;
int main()
{IO
;int T
=1;while(T
--){cin
>>n
;for(int i
=1;i
<=n
;i
++){ll x
,y
;cin
>>x
>>y
;p
[i
]=x
*x
*y
+y
*y
*(y
-2*x
);}sort(p
+1,p
+1+n
);cout
<<p
[n
]-p
[1]<<'\n';}return 0;
}
C.裝備合成
典型數(shù)學(xué)題目，直接像做高中線性規(guī)劃題做即可。
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
using namespace std
;
typedef long long ll
;
typedef pair
<int,int> pii
;
typedef unsigned long long ull
;
const int N
=100010;
ll x
,y
;
bool check(ll a
,ll b
)
{return (2*a
+4*b
<=x
)&&(3*a
+b
<=y
);
}
int main()
{IO
;int T
=1;cin
>>T
;while(T
--){cin
>>x
>>y
;ll a
=(4*y
-x
)/10;ll b
=(3*x
-2*y
)/10;if(a
>=0&&b
>=0){ll res
=0;if(check(a
,b
)) res
=max(res
,a
+b
);if(check(a
,b
+1)) res
=max(res
,a
+b
+1);if(check(a
+1,b
)) res
=max(res
,a
+b
+1);if(check(a
,b
-1)) res
=max(res
,a
+b
-1);if(check(a
-1,b
)) res
=max(res
,a
+b
-1);cout
<<res
<<'\n';}else if(a
<0)cout
<<min(x
/4,y
)<<'\n';else if(b
<0) cout
<<min(x
/2,y
/3)<<'\n';else cout
<<0<<'\n';}return 0;
}
D.取石子游戲
暴力打表+總結(jié)規(guī)律+瞎胡搞
畢竟我也看不出啥規(guī)律啊~~
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
using namespace std
;
typedef long long ll
;
typedef pair
<int,int> pii
;
typedef unsigned long long ull
;
const int N
=100010;
ll sg
[N
];
ll
dfs(ll u
)
{if(sg
[u
]!=-1) return sg
[u
];if(u
==1) return sg
[u
]=0;int a
=dfs(u
/2);int b
=dfs((u
+1)/2);for(int i
=0;;i
++)if(i
!=a
&&i
!=b
) return sg
[u
]=i
;
}ll f
[N
],s
[N
];
int sz
;
void init()
{f
[1]=1;f
[2]=2;s
[1]=1;s
[2]=3;for(int i
=3;;i
++){f
[i
]=f
[i
-1]+s
[i
-2];if(i
%2==0) f
[i
]++;s
[i
]=f
[i
]+s
[i
-1];if(s
[i
]>=1e18){sz
=i
;break;}}
}
int main()
{IO
;int T
=1;cin
>>T
;init();while(T
--){ll n
;cin
>>n
;int l
=1,r
=sz
;while(l
<r
){int mid
=l
+r
+1>>1;if(s
[mid
]<=n
) l
=mid
;else r
=mid
-1;}if(s
[l
]<n
) l
++;if(l
&1) cout
<<"XiaoQiao\n";else cout
<<"XiaoHuiHui\n";}return 0;
}
E.石子搬運(yùn)
O(qn2logn)O(qn^2logn)O(qn2logn)
首先考慮如果沒(méi)有修改操作，不難發(fā)現(xiàn)可以設(shè)計(jì)dp
狀態(tài)表示：f(i,j)f_{(i,j)}f(i,j)?表示考慮前iii堆石子，搬運(yùn)jjj次能夠搬完
狀態(tài)轉(zhuǎn)移：只需要考慮最后當(dāng)前這堆石子搬運(yùn)多少次能夠?qū)⑵浒嵬?/p>
由基本不等式可得，對(duì)于某一堆石子，搬運(yùn)次數(shù)一定的，那么均分搬運(yùn)石子代價(jià)一定最小。
如果考慮修改，我們用線段樹(shù)維護(hù)，對(duì)于區(qū)間合并直接暴力合并即可O(n2)O(n^2)O(n2)
單點(diǎn)修改時(shí)間復(fù)雜度n2log(n)n^2log(n)n2log(n)，總時(shí)間復(fù)雜度qn2log(n)qn^2log(n)qn2log(n)
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std
;
typedef long long ll
;
const int N
=410;
const ll INF
=0x3f3f3f3f3f3f3f3f;
int n
,m
,q
;
ll a
[N
];
struct node
{int l
,r
;ll f
[N
];
}tree
[N
*4];
void pushup(int u
)
{for(int i
=1;i
<=m
;i
++) tree
[u
].f
[i
]=INF
;for(int i
=1;i
<=m
;i
++)for(int j
=1;j
<=m
-i
;j
++){if(tree
[u
<<1].f
[i
]==INF
||tree
[u
<<1|1].f
[j
]==INF
) continue;tree
[u
].f
[i
+j
]=min(tree
[u
].f
[i
+j
],tree
[u
<<1].f
[i
]+tree
[u
<<1|1].f
[j
]);}
}
void build(int u
,int l
,int r
)
{tree
[u
]={l
,r
};memset(tree
[u
].f
,0x3f,8*N
);if(l
==r
){for(int k
=1;k
<=min(m
,(int)a
[l
]);k
++){int x
=a
[l
]/k
;int p
=a
[l
]%k
;tree
[u
].f
[k
]=1ll*(x
+1)*(x
+1)*p
+1ll*x
*x
*(k
-p
);}return;}int mid
=l
+r
>>1;build(u
<<1,l
,mid
);build(u
<<1|1,mid
+1,r
);pushup(u
);
}
void modify(int u
,int pos
,int val
)
{if(tree
[u
].l
==tree
[u
].r
){for(int k
=1;k
<=min(m
,val
);k
++){int x
=val
/k
;int p
=val
%k
;tree
[u
].f
[k
]=1ll*(x
+1)*(x
+1)*p
+1ll*x
*x
*(k
-p
);}return;}int mid
=tree
[u
].l
+tree
[u
].r
>>1;if(pos
<=mid
) modify(u
<<1,pos
,val
);else modify(u
<<1|1,pos
,val
);pushup(u
);
}
int main()
{IO
;int T
=1;while(T
--){cin
>>n
>>m
;for(int i
=1;i
<=n
;i
++) cin
>>a
[i
];build(1,1,n
);cin
>>q
;while(q
--){int x
,v
;cin
>>x
>>v
;modify(1,x
,v
);cout
<<tree
[1].f
[m
]<<'\n';}}return 0;
}
就會(huì)上面幾題~~
F-小松鼠吃松果
大佬題解
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std
;
typedef long long ll
;
typedef pair
<int,int> pii
;
typedef unsigned long long ull
;
const int N
=100010;
int n
,m
;
int pos
[N
],h
[N
];
int hs
[N
];
struct node
{int t
,p
,w
;bool operator<(const node
&o
) const{return t
==o
.t
?p
<o
.p
:t
<o
.t
;}
}q
[N
];
ll tree
[N
];
int lowbit(int x
)
{return x
&-x
;
}
void add(int k
,ll x
)
{for(;k
<=n
;k
+=lowbit(k
)) tree
[k
]=max(tree
[k
],x
);
}
ll
query(int k
)
{ll res
=0;for(;k
;k
-=lowbit(k
)) res
=max(res
,tree
[k
]);return res
;
}
int main()
{IO
;int T
=1;while(T
--){cin
>>n
>>m
;for(int i
=1;i
<=m
;i
++) cin
>>pos
[i
];for(int i
=1;i
<=m
;i
++) cin
>>h
[i
];for(int i
=1;i
<=n
;i
++){cin
>>q
[i
].t
>>q
[i
].p
>>q
[i
].w
;q
[i
].t
+=h
[q
[i
].p
];}for(int i
=1;i
<=n
;i
++){int x
=q
[i
].t
-pos
[q
[i
].p
];int y
=q
[i
].t
+pos
[q
[i
].p
];q
[i
].t
=x
,q
[i
].p
=y
;hs
[i
]=y
;}sort(hs
+1,hs
+1+n
);sort(q
+1,q
+1+n
);ll res
=0;for(int i
=1;i
<=n
;i
++){int k
=lower_bound(hs
+1,hs
+1+n
,q
[i
].p
)-hs
;ll dp
=query(k
)+q
[i
].w
;res
=max(res
,dp
);add(k
,dp
);}cout
<<res
<<'\n';}return 0;}
要加油哦~
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