题目链接：https://loj.ac/problems/search?keyword=NOIP2013
\(ans=(m*10^k+x)\% n\)，快速幂求之
#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=10000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define maxd 998244353
#define eps 1e-8
int n,m,k,x;
int read()
{
int x=0,f=1;char ch=getchar();
while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
ll qpow(ll x,int y,int p)
{
ll ans=1;
while (y)
{
if (y&1) ans=ans*x%p;
x=x*x%p;y>>=1;
}
return ans;
}
int main()
{
n=read();m=read();k=read();x=read();
ll ans=m*qpow(10,k,n)%n;
ans=(ans+x)%n;
printf("%lld",ans);
return 0;
}原问题等价于最大化\(\sum_{i=1}^na_ib_i\)，根据排序不等式可得当\(\{a_i\}\)与\(\{b_i\}\)均按照从小到大的顺序排列时两列火柴的距离最小。原序列离散化后构造一个新序列\(\{c_i\}\)表示\(\{b_i\}\)中值为\(i\)的数将要移动到的位置。于是转化交换\(\{c_i\}\)中的相邻元素以排序，也就是求\(c_i\)的逆序对。
#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=10000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define maxd 99999997
#define eps 1e-8
int n,a[100100],b[100100],id1[100100],id2[100100],pos[100100],t[100100];
int read()
{
int x=0,f=1;char ch=getchar();
while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
void modify(int p)
{
for (int i=p;i<=n;i+=lowbit(i)) t[i]++;
}
int query(int p)
{
int ans=0;
for (int i=p;i;i-=lowbit(i)) ans+=t[i];
return ans;
}
bool cmp1(int x,int y) {return a[x]<a[y];}
bool cmp2(int x,int y) {return b[x]<b[y];}
int main()
{
n=read();
rep(i,1,n) a[i]=read();
rep(i,1,n) b[i]=read();
rep(i,1,n) id1[i]=id2[i]=i;
sort(id1+1,id1+1+n,cmp1);
sort(id2+1,id2+1+n,cmp2);
rep(i,1,n) pos[id2[i]]=id1[i];
int ans=0;
rep(i,1,n)
{
modify(pos[i]);
ans=(ans+i-query(pos[i]))%maxd;
}
printf("%d",ans);
return 0;
}符合要求的边一定在图的最大生成树上，倍增维护最小值
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=10000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define maxd 998244353
#define eps 1e-8
struct node{int to,nxt,cost;}sq[100100];
int all=0,head[100100];
int n,m,q,col[10010],fa[10010][16],dis[10010][16],dep[10010];
struct edgenode{int u,v,w;}edge[50050];
bool operator<(edgenode p,edgenode q) {return p.w>q.w;}
int read()
{
int x=0,f=1;char ch=getchar();
while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
void add(int u,int v,int w)
{
all++;sq[all].to=v;sq[all].nxt=head[u];sq[all].cost=w;head[u]=all;
}
int find(int x) {if (col[x]==x) return col[x];col[x]=find(col[x]);return col[x];}
void dfs(int u,int fu)
{
dep[u]=dep[fu]+1;
rep(i,1,15)
{
fa[u][i]=fa[fa[u][i-1]][i-1];
dis[u][i]=min(dis[u][i-1],dis[fa[u][i-1]][i-1]);
}
for (int i=head[u];i;i=sq[i].nxt)
{
int v=sq[i].to;
if (v==fu) continue;
fa[v][0]=u;dis[v][0]=sq[i].cost;
dfs(v,u);
}
}
int query(int u,int v)
{
if (dep[u]<dep[v]) swap(u,v);
int tmp=dep[u]-dep[v],ans=maxd;
per(i,15,0)
if ((tmp>>i)&1) {ans=min(ans,dis[u][i]);u=fa[u][i];}
if (u==v) return ans;
per(i,15,0)
if (fa[u][i]!=fa[v][i])
{
ans=min(ans,min(dis[u][i],dis[v][i]));
u=fa[u][i];v=fa[v][i];
}
ans=min(ans,min(dis[u][0],dis[v][0]));
return ans;
}
int main()
{
n=read();m=read();
rep(i,1,m)
{
edge[i].u=read();edge[i].v=read();edge[i].w=read();
}
sort(edge+1,edge+1+m);
rep(i,1,n) col[i]=i;
rep(i,1,m)
{
int u=edge[i].u,v=edge[i].v,w=edge[i].w;
int fx=find(u),fy=find(v);
if (fx!=fy)
{
add(u,v,w);add(v,u,w);
col[fx]=fy;
}
}
memset(dis,0x3f,sizeof(dis));
rep(i,1,n)
if (!dep[i]) dfs(i,0);
q=read();
while (q--)
{
int u=read(),v=read();
if (find(u)!=find(v)) {puts("-1");continue;}
printf("%d\n",query(u,v));
}
return 0;
}考虑\(h_{i-1}\)对\(h_i\)的影响
为了覆盖\(h_{i-1}\),我们有\(h_{i-1}\)个以\(i-1\)为终点的区间。接下来覆盖\(h_i\)时，若\(h_{i-1}\geq h_i\)，我们可以将之前的某些以\(i-1\)为终点的区间向右拓展一位以覆盖\(h_i\)。反之我们则需要新开\(h_{i-1}-h_i\)个区间来覆盖\(h_i\)。直接累加即可
#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=10000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define maxd 998244353
#define eps 1e-8
int n,h[100100];
int read()
{
int x=0,f=1;char ch=getchar();
while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
int main()
{
n=read();
rep(i,1,n) h[i]=read();
int ans=0;
rep(i,1,n)
if (h[i]>h[i-1]) ans+=(h[i]-h[i-1]);
printf("%d",ans);
return 0;
}蜜汁easy的dp
记\(f_{i,0/1}\)表示\(1-i\)中，\(h_i\)作为较大值/较小值的最大合法序列长度。
当\(h_i>h_{i-1}\)时，\(f_{i,0}=f_{i-1,1}+1,f_{i,1}=f_{i-1,1}\)。其它情况同理
#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=10000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define maxd 998244353
#define eps 1e-8
int n,h[2000100],f[2000100][2];
int read()
{
int x=0,f=1;char ch=getchar();
while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
int main()
{
n=read();
rep(i,1,n) h[i]=read();
f[1][0]=f[1][1]=1;
rep(i,2,n)
{
if (h[i]>h[i-1]) f[i][0]=f[i-1][1]+1;
else f[i][0]=f[i-1][0];
if (h[i]<h[i-1]) f[i][1]=f[i-1][0]+1;
else f[i][1]=f[i-1][1];
}
//rep(i,1,n) cout << f[i][0] << " " << f[i][1] << endl;
printf("%d",max(f[n][0],f[n][1]));
return 0;
}暴力的话考虑记一个四元组\((x_1,y_1,x_2,y_2)\)表示空格的位置和棋子的位置，大力\(bfs\)就行了，每次转移空格的位置，当空格的位置和棋子的位置相同时就移动棋子即可。
这个东西可以通过单组数据，但是无法应对多组数据。同时注意到多组数据对应的网格是相同的，于是可以考虑抠出关键状态建图然后最短路。
第一个问题是关键状态是什么呢？注意到终止的时候空格一定是在棋子旁边的，并且如果空格跑的离棋子很远也没有什么意义。于是可以将所有的空格在棋子周围的状态作为关键状态，具体的，关键状态可以被刻画为\((i,j,k)\)表示棋子在\((i,j)\)且空格在其的上/下/左/右。
第二个问题是如何连边？这里还需要考虑的是求边的长度时注意时间。首先显然的是：如果一个棋子在格子的上方，那么可以花一步把它转移到棋子的下方。但是我们发现这么简单的建边无法使这个图联通。接下来还有的是\((i,j,k)\)向\((i,j,p)(k\neq p)\)连边。这个长度由于\((i,j)\)恒定可以直接对一个点bfs。之后这个图就联通了，可以欢快的跑最短路了
#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=10000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define maxd 998244353
#define eps 1e-8
const int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
struct node{int x,y,len;};
struct sqnode{int to,nxt,cost;}sq[200200];
int all=0,head[4040];
int n,m,ask,id[31][31][4],dis[4040],a[31][31],tot=0;
bool in[4040],vis[31][31];
int read()
{
int x=0,f=1;char ch=getchar();
while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
void add(int u,int v,int w)
{
//cout << u << " " << v << " " << w << endl;
all++;sq[all].to=v;sq[all].nxt=head[u];sq[all].cost=w;head[u]=all;
}
int bfs(int sx,int sy,int ex,int ey,int bx,int by)
{
if ((sx==ex) && (sy==ey)) return 0;
queue<node> q;
while (!q.empty()) q.pop();
q.push((node){sx,sy,0});
memset(vis,0,sizeof(vis));
vis[sx][sy]=1;
while (!q.empty())
{
node now=q.front();q.pop();
if ((now.x==ex) && (now.y==ey)) return now.len;
rep(i,0,3)
{
int nx=now.x+dx[i],ny=now.y+dy[i];
if ((nx==bx) && (ny==by)) continue;
if ((a[nx][ny]) && (!vis[nx][ny]))
{
vis[nx][ny]=1;
if ((nx==ex) && (ny==ey)) return now.len+1;
q.push((node){nx,ny,now.len+1});
}
}
}
return maxd;
}
int spfa(int sx,int sy,int ex,int ey,int bx,int by)
{
if ((sx==ex) && (sy==ey)) return 0;
queue<int> q;
memset(dis,0x3f,sizeof(dis));
memset(in,0,sizeof(in));
rep(i,0,3)
if (id[sx][sy][i])
{
dis[id[sx][sy][i]]=bfs(bx,by,sx+dx[i],sy+dy[i],sx,sy);
q.push(id[sx][sy][i]);
in[id[sx][sy][i]]=1;
}
while (!q.empty())
{
int u=q.front();q.pop();in[u]=0;
for (int i=head[u];i;i=sq[i].nxt)
{
int v=sq[i].to;
if (dis[v]>dis[u]+sq[i].cost)
{
dis[v]=dis[u]+sq[i].cost;
if (!in[v]) {q.push(v);in[v]=1;}
}
}
}
int ans=maxd;
rep(i,0,3)
if (id[ex][ey][i])
ans=min(ans,dis[id[ex][ey][i]]);
return ans;
}
void init()
{
n=read();m=read();ask=read();
rep(i,1,n) rep(j,1,m)
{
a[i][j]=read();
}
rep(i,1,n) rep(j,1,m)
rep(k,0,3)
if ((a[i][j]) && (a[i+dx[k]][j+dy[k]]))
id[i][j][k]=(++tot);
rep(i,1,n)
rep(j,1,m)
rep(k,0,3)
if (id[i][j][k])
add(id[i][j][k],id[i+dx[k]][j+dy[k]][k^1],1);
rep(i,1,n)
rep(j,1,m)
rep(p,0,3)
rep(q,0,3)
if ((p!=q) && (id[i][j][p]) && (id[i][j][q]))
{
int tmp=bfs(i+dx[p],j+dy[p],i+dx[q],j+dy[q],i,j);
add(id[i][j][p],id[i][j][q],tmp);
}
}
void work()
{
while (ask--)
{
int bx=read(),by=read(),sx=read(),sy=read(),ex=read(),ey=read();
int ans=spfa(sx,sy,ex,ey,bx,by);
if (ans>=maxd) ans=-1;
printf("%d\n",ans);
}
}
int main()
{
init();
work();
return 0;
}原文：https://www.cnblogs.com/encodetalker/p/11749733.html