A - Fence
把湊得那條邊當成最長的邊，如果a+b+c=d那么將會共線，只要稍微a+b+c大一點即可滿足四邊形。
#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
;
const int N
=100010;
const ll mod
=998244353;
int main()
{IO
;int T
=1;cin
>>T
;while(T
--){ll a
,b
,c
;cin
>>a
>>b
>>c
;cout
<<a
+b
+c
-1<<'\n';}return 0;}
B - Nice Matrix
不難發現由于需要滿足回文有以下關系：
a(i,j)=a(n?i+1,j)=a(i,m?j+1)=a(n?i+1,m?j+1)a_{(i,j)}=a_{(n-i+1,j)}=a_{(i,m-j+1)}=a_{(n-i+1,m-j+1)}a(i,j)?=a(n?i+1,j)?=a(i,m?j+1)?=a(n?i+1,m?j+1)?
單獨考慮這四元組，需要找到xxx使得∣x?a(i,j)∣+∣x?a(n?i+1,j)∣+∣x?a(i,m?j+1)+∣x?a(n?i+1,m?j+1)∣|x-a_{(i,j)}|+|x-a_{(n-i+1,j)}|+|x-a_{(i,m-j+1)}+|x-a_{(n-i+1,m-j+1)}|∣x?a(i,j)?∣+∣x?a(n?i+1,j)?∣+∣x?a(i,m?j+1)?+∣x?a(n?i+1,m?j+1)?∣最小，只需要找中位數即可。
#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
;
const int N
=110;
const ll mod
=998244353;
ll a
[N
][N
];
int n
,m
;
ll tmp
[5];
ll
calc(ll a
,ll b
,ll c
,ll d
)
{ll res
=0;tmp
[1]=a
,tmp
[2]=b
,tmp
[3]=c
,tmp
[4]=d
;sort(tmp
+1,tmp
+1+4);ll mid
=tmp
[2];return abs(a
-mid
)+abs(b
-mid
)+abs(c
-mid
)+abs(d
-mid
);}
int main()
{IO
;int T
=1;cin
>>T
;while(T
--){ cin
>>n
>>m
;for(int i
=1;i
<=n
;i
++)for(int j
=1;j
<=m
;j
++) cin
>>a
[i
][j
];ll res
=0;int rmid
=1+n
>>1,cmid
=1+m
>>1;if(n
&1){for(int j
=1;j
<=cmid
;j
++){ll mid
=a
[rmid
][j
]+a
[rmid
][m
-j
+1]>>1;res
+=abs(a
[rmid
][j
]-mid
)+abs(a
[rmid
][m
-j
+1]-mid
);}rmid
--;}if(m
&1){for(int i
=1;i
<=rmid
;i
++){ll mid
=a
[i
][cmid
]+a
[n
-i
+1][cmid
]>>1;res
+=abs(a
[i
][cmid
]-mid
)+abs(a
[n
-i
+1][cmid
]-mid
);}cmid
--;}for(int i
=1;i
<=rmid
;i
++)for(int j
=1;j
<=cmid
;j
++)res
+=calc(a
[i
][j
],a
[i
][m
-j
+1],a
[n
-i
+1][j
],a
[n
-i
+1][m
-j
+1]);cout
<<res
<<'\n';}return 0;
}
C - Bargain
單獨考慮每位的對答案的貢獻：
某位想要有貢獻那么首先該位不能被刪除，因而選擇刪除的子串只可能有兩種情況：①該位前面②該位后面
不妨設第iii位的數是ddd（iii從高位）
如果去掉的子串在該位前面，則貢獻為：i(i?1)2d×10n?i\frac{i(i-1)}{2}d×10^{n-i}2i(i?1)?d×10n?i
如果去掉的子串在該位后面，則貢獻為：d∑j=1n?jj×10j?1d\sum_{j=1}^{n-j}j×10^{j-1}d∑j=1n?j?j×10j?1
#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
;
const int N
=100010;
const ll mod
=1e9+7;
char s
[N
];
int n
;
int main()
{IO
;int T
=1;while(T
--){ cin
>>s
+1;n
=strlen(s
+1);ll p
=1,sum
=0,res
=0;for(int i
=n
;i
;i
--){ll d
=s
[i
]-'0';res
=(res
+1ll*i
*(i
-1)/2%mod
*d
%mod
*p
%mod
)%mod
;res
=(res
+1ll*sum
*d
)%mod
;sum
=(sum
+1ll*(n
-i
+1)*p
)%mod
;p
=p
*10%mod
;}cout
<<res
<<'\n';}return 0;}
D - Returning Home
Heltion大佬題解
第一次見這種建圖方式，行列建圖，如果(i,j)(i,j)(i,j)是一個特殊點那么在第iii行的任意位置即可0代價到達(i,j)(i,j)(i,j)即到達了第jjj列，同理第jjj列的任意位置即可0代價到達(i,j)(i,j)(i,j)即到達了第iii行。然后參考上述題解即可
#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
<ll
,int> pli
;
const int N
=200010,M
=2*N
;
const ll mod
=1e9+7;
const ll INF
=1e17;
int n
,m
;
int sx
,sy
,fx
,fy
;
vector
<int> dx
,dy
;
int x
[N
],y
[N
];
ll d
[N
];
int h
[N
],e
[M
],ne
[M
],idx
;
bool st
[N
];
void add(int a
,int b
)
{e
[idx
]=b
;ne
[idx
]=h
[a
];h
[a
]=idx
++;
}
int find(vector
<int> &mp
,int x
)
{return lower_bound(mp
.begin(),mp
.end(),x
)-mp
.begin()+1;
}
void dijkstra()
{priority_queue
<pli
,vector
<pli
>,greater
<pli
>> q
;for(int i
=1;i
<=2*m
;i
++)if(d
[i
]!=INF
)q
.push({d
[i
],i
});while(q
.size()){int t
=q
.top().second
;q
.pop();if(st
[t
]) continue;st
[t
]=1;for(int i
=h
[t
];i
!=-1;i
=ne
[i
]){int j
=e
[i
];if(d
[j
]>d
[t
])q
.push({d
[j
]=d
[t
],j
});}if(1<=t
&&t
<=dx
.size()){if(t
+1<=dx
.size()&&d
[t
+1]>d
[t
]+dx
[t
]-dx
[t
-1]){d
[t
+1]=d
[t
]+dx
[t
]-dx
[t
-1];q
.push({d
[t
+1],t
+1});}if(1<t
&&d
[t
-1]>d
[t
]+dx
[t
-1]-dx
[t
-2]){d
[t
-1]=d
[t
]+dx
[t
-1]-dx
[t
-2];q
.push({d
[t
-1],t
-1});}}if(m
<t
&&t
<=dy
.size()+m
){if(t
+1<=dy
.size()+m
&&d
[t
+1]>d
[t
]+dy
[t
-m
]-dy
[t
-1-m
]){d
[t
+1]=d
[t
]+dy
[t
-m
]-dy
[t
-1-m
];q
.push({d
[t
+1],t
+1});}if(m
+1<t
&&d
[t
-1]>d
[t
]+dy
[t
-1-m
]-dy
[t
-2-m
]){d
[t
-1]=d
[t
]+dy
[t
-1-m
]-dy
[t
-2-m
];q
.push({d
[t
-1],t
-1});}}}
}
int main()
{IO
;int T
=1;while(T
--){ cin
>>n
>>m
;cin
>>sx
>>sy
>>fx
>>fy
;memset(h
,-1,sizeof h
);for(int i
=1;i
<=m
;i
++){d
[i
]=d
[i
+m
]=INF
;cin
>>x
[i
]>>y
[i
];dx
.push_back(x
[i
]);dy
.push_back(y
[i
]);}sort(dx
.begin(),dx
.end());sort(dy
.begin(),dy
.end());dx
.erase(unique(dx
.begin(),dx
.end()),dx
.end());dy
.erase(unique(dy
.begin(),dy
.end()),dy
.end());for(int i
=1;i
<=m
;i
++){int u
=find(dx
,x
[i
]);int v
=find(dy
,y
[i
]);add(u
,m
+v
);add(m
+v
,u
);d
[u
]=min(d
[u
],(ll
)abs(sx
-x
[i
]));d
[m
+v
]=min(d
[m
+v
],(ll
)abs(sy
-y
[i
]));}dijkstra();ll res
=abs(sx
-fx
)+abs(sy
-fy
);for(int i
=1;i
<=m
;i
++){int u
=find(dx
,x
[i
]);res
=min(res
,abs(fx
-x
[i
])+abs(fy
-y
[i
])+d
[u
]);}cout
<<res
<<'\n';}return 0;}
F - Boring Queries
靜態版本題解都看不懂的渣渣，先標記一下，以后會了補！
最近數電信號要頂不住了啊啊啊啊啊
要加油哦！！！
總結
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