【Codeforces Round 431 (Div. 2) A B C D E五个题】

#include<stdio.h>
int n,a[103],sum,m;
int main()
{
scanf("%d",&n);m=n;
while(n)scanf("%d",a+n),sum+=a
,n--;
puts(!(m&1)||!(a[1]&1)||!(a[m]&1)?"No":"Yes");return 0;
}//Paul_Guderian

列出一些样例，画画图，或者思考思考：

#include<stdio.h>
#include<math.h>
#define go(i,a,b) for(int i=a;i<=b;i++)
int n,t,cnt,tot,I,use[1003];double x[1003],y[1003],K[1003],k;
int main()
{
scanf("%d",&n);
go(i,1,n)x[i]=i,scanf("%lf",&y[i]);
go(i,1,2)go(j,i+1,3)K[++t]=(y[i]-y[j])/(x[i]-x[j]);
go(j,1,t)
{
k=K[j];
go(i,1,n)use[i]=0;tot=cnt=use[1]=1;
go(i,2,n)if(fabs((y[i]-y[1])/(x[i]-x[1])-k)<1e-6)cnt+=(use[i]=1);
go(i,1,n)if(!use[i])I=i,i=n;if(cnt==n)continue;
go(i,I+1,n)if(fabs((y[i]-y[I])/(x[i]-x[I])-k)<1e-6)tot++;
if(!tot||tot+cnt<n)continue;puts("Yes");return 0;
}
puts("No");return 0;
}//Paul_Guderian

#include<stdio.h>
#define sum(j) (j*(j-1)/2)
#define go(i,a,b) for(int i=a;i<=b;i++)
int main()
{
int k;scanf("%d",&k);
if(!k){puts("z");return 0;}
go(i,0,25){int j=2;while(k-sum(j)>=0)j++;j--;
go(t,1,j)printf("%c",'a'+i);if(!(k-=sum(j)))return 0;}return 0;
}//Paul_Guderian

#include<stdio.h>
#include<algorithm>
#define go(i,a,b) for(int i=a;i<=b;i++)
struct info{int Type,tot,x,y,id;};info a[100003],tmp[100003];
bool cmp1(info A,info B){return A.tot==B.tot?(A.Type==B.Type?
(A.Type==2?(A.y>B.y):(A.x<B.x)):(A.Type<B.Type)):(A.tot<B.tot);}
bool cmp2(info A,info B){return A.Type==B.Type?(A.Type==2?
(A.y>B.y):(A.x<B.x)):(A.Type>B.Type);}
bool cmp3(info A,info B){return A.id<B.id;}int n,w,h,t;
int main()
{
scanf("%d%d%d",&n,&w,&h);
go(i,1,n)
{
int Type,p,t;scanf("%d%d%d",&Type,&p,&t);
if(Type==1)a[i]=(info){Type,p-t,p,h,i};
if(Type==2)a[i]=(info){Type,p-t,w,p,i};
}
std::sort(a+1,a+n+1,cmp1);int l=1,r=1;
while(l<=n)
{
while(a[l].tot==a[r+1].tot&&r+1<=n)r++;
t=0;go(i,l,r)tmp[++t]=a[i];std::sort(tmp+1,tmp+t+1,cmp2);
go(i,l,r)a[i].id=tmp[i-l+1].id;l=r=r+1;
}
std::sort(a+1,a+n+1,cmp3);go(i,1,n)printf("%d %d\n",a[i].x,a[i].y);return 0;
}//Paul_Guderian

我们发现修改+询问很难想。先试着去掉修改，只询问，那么应该怎么做呢？第一种方法一定是对于每种颜色维护区间最早位置和最晚位置，询问的时候相减就是答案了。另一种解法是为将来处理修改操作而考虑的——由于修改仅仅每次修改一个值，那么这一个值可能会影响答案，也可能不会，所以既然修改一个点，那么我们可以将答案这样统计：

#include<set>
#include<cstdio>
#include<algorithm>
#define ll long long
#define go(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N=100010;
int k,n,m,q,color
,x,y;
ll ans
,c
;bool yes
;
set<int>S
;set<int>::iterator it;
struct info{int p,x,w,id;}Q[N*20],T[N*20];
void Add(int x,int d){while(x<=n)c[x]+=d,x+=x&-x;}
ll Sum(int x){ll R=0;while(x)R+=c[x],x-=x&-x;return R;}

void Revise(int x,int y)
{
if(color[x]==y)return;it=S[color[x]].find(x);int left=0;

if(it!=S[color[x]].begin()){it--;left=(*it);it++;
Q[++m]=(info){x,left,-(x-left),0};}it++;
if(it!=S[color[x]].end()){Q[++m]=(info){*it,x,-((*it)-x),0};
if(left)Q[++m]=(info){*it,left,(*it)-left,0};}

S[color[x]].erase(x);color[x]=y;
S[y].insert(x);it=S[y].find(x);left=0;

if(it!=S[y].begin()){it--;left=(*it);it++;
Q[++m]=(info){x,left,x-left,0};}it++;
if(it!=S[y].end()){if(left)Q[++m]=(info){*it,left,left-(*it),0};
Q[++m]=(info){*it,x,(*it)-x,0};}
}

void CDQ(int l,int r)
{
if(l==r)return;int M=l+r>>1;CDQ(l,M);CDQ(M+1,r);
int p1=l-1,p2=M,t=l-1;go(i,l,r)T[i]=Q[i];

while(p1<M||p2<r)if(p2>=r||(p1<M&&T[p1+1].p<=T[p2+1].p))
Q[++t]=T[++p1],!Q[t].id?Add(n-Q[t].x+1,Q[t].w),1:1;
else Q[++t]=T[++p2],Q[t].id?ans[Q[t].id]+=Sum(n-Q[t].x+1),1:1;

go(i,l,p1)if(!T[i].id)Add(n-T[i].x+1,-T[i].w);
}
int main()
{
scanf("%d%d",&n,&q);
go(i,1,n)scanf("%d",&color[i]),S[color[i]].insert(i);
go(i,1,n){it=S[color[i]].find(i);if(it!=S[color[i]].begin())it--,
Q[++m]=(info){i,*it,i-(*it),0};}
go(i,1,q)scanf("%d%d%d",&k,&x,&y),
k==1?Revise(x,y),1:(Q[++m]=(info){y,x,0,i},yes[i]=1);
CDQ(1,m);go(i,1,q)if(yes[i])printf("%I64d\n",ans[i]);return 0;
}