Construction ...... G to be filled pit, AB did not see estimates will not make up ^ _ ^.
Somehow it seems, in addition to the G looks a bit should not make it difficult or div3 ......
C. Exam in BerSU
The meaning of problems
To give you a length \ (n-\) sequence \ (a_i \) . For each \ (i \ in [1, N] \) seeking \ ([1, i-1 ] \) at least satisfy the remaining omitted by how many \ ([1, i] \ ) number and less than \ (M \) .
\ (n \ 2 \ cdot 10 ^ 5, M \ 2 \ cdot 10 ^ 7, a_i \ 100 \) .
Solution
See range \ (\ Le 100 \) , appears directly in buckets remember how many times each number, then scan again descending violence barrel finished it. . .
code
#include<bits/stdc++.h>
using namespace std;
priority_queue<int> q;
const int N=2e5+5;
int a[N],t[233],n,m,sum;
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;++i) scanf("%d",&a[i]);
for(int i=1;i<=n;++i)
{
sum+=a[i];
if(sum<=m)
{
++t[a[i]];
printf("0 ");
continue;
}
int ans=0,tmp=sum;
for(int j=100;j&&tmp>m;--j)
{
if(tmp-t[j]*j<=m)
{
printf("%d ",ans+(tmp-m-1)/j+1);
break;
}
else tmp-=t[j]*j,ans+=t[j];
}
++t[a[i]];
}
}D. Extra Element
The meaning of problems
To a length \ (n-\) is the number of columns, by deleting a number such that the number of columns of the original rearrangement of arithmetic sequence.
Solution
...... I do not know what to say, if the difference ranging from a deleted directly on the line. The first three special number may be sentenced to it.
code
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=2e5+5;
int a[N],id[N];
bool cmp(int x, int y) {
return a[x]<a[y];
}
int main()
{
int n; scanf("%d",&n);
for(int i=1;i<=n;++i) scanf("%d",&a[i]),id[i]=i;
sort(id+1,id+1+n,cmp);
sort(a+1,a+1+n);
if(n<=3)
{
puts("1");
return 0;
}
int del=0,cha;
if(a[3]-a[2]!=a[2]-a[1])
{
if(a[3]-a[2]==a[4]-a[3])
{
del=1;
cha=a[3]-a[2];
}
else if(a[3]-a[1]==a[4]-a[3])
{
del=2;
cha=a[3]-a[1];
}
else if(a[2]-a[1]==a[4]-a[2])
{
del=3;
cha=a[2]-a[1];
}
else
{
puts("-1");
return 0;
}
}
else cha=a[2]-a[1];
for(int i=del?4:3;i<=n;++i)
{
int l=(del==i-1?i-2:i-1);
if(a[i]-a[l]!=cha)
{
if(del)
{
puts("-1");
return 0;
}
del=i;
}
}
printf("%d",!del?id[1]:id[del]);
}E. Polycarp and snakes
The meaning of problems
...... redundancy complex topic, or to directly see the original title face it ......
Solution
I do not know what to say, according to the meaning of the questions you can direct simulation. Note that some of the details.
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=2005;
char s[N][N];
int x1[N],x2[N],y1[N],y2[N];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,m,mx=0;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;++i) scanf("%s",s[i]+1);
bool flag=true;
for(int a=0;flag&&a<26;++a)
{
for(int i=1;i<=n;++i)
for(int j=1;j<=m;++j)
{
if(s[i][j]!=a+'a') continue;
if(!x1[a]&&!y1[a]) x1[a]=x2[a]=i,y1[a]=y2[a]=j;
else
{
x1[a]=min(x1[a],i),x2[a]=max(x2[a],i);
y1[a]=min(y1[a],j),y2[a]=max(y2[a],j);
}
}
if(x1[a]!=x2[a]&&y1[a]!=y2[a])
{
flag=false;
break;
}
if(x1[a]) mx=a+1;
else continue;
if(x1[a]==x2[a])
{
for(int j=y1[a];j<=y2[a];++j)
{
if(s[x1[a]][j]<a+'a') {
flag=false;
continue;
}
}
}
else
{
for(int j=x1[a];j<=x2[a];++j)
{
if(s[j][y1[a]]<a+'a') {
flag=false;
continue;
}
}
}
}
if(!flag)
{
puts("NO");
for(int i=0;i<26;++i) x1[i]=y1[i]=x2[i]=y2[i]=0;
continue;
}
printf("YES\n%d\n",mx);
for(int a=0;a<mx;++a)
{
if(!x1[a])
{
bool f=true;
for(int i=1;f&&i<=n;++i)
for(int j=1;f&&j<=m;++j)
{
if(s[i][j]>a+'a')
{
printf("%d %d %d %d\n",i,j,i,j);
f=false;
}
}
continue;
}
printf("%d %d %d %d\n",x1[a],y1[a],x2[a],y2[a]);
x1[a]=y1[a]=x2[a]=y2[a]=0;
}
}
}F. Two Pizzas
The meaning of problems
There \ (n \) personal, everyone has some favorite ingredients; there \ (m \) pizzas, each pizza has a price and a number of ingredients. As little as possible at the price now want you to buy two pizzas, so as to satisfy the people as much as possible premise. A person can be satisfied if and only if all of his favorite ingredients appeared on two pizza.
Several ingredients does not exceed \ (9 \) . \ (n-, m \ Le 2 \ ^. 5 CDOT 10 \) .
Solution
Obviously, several kinds of pizza and people of no more than \ (2 ^ 9 \) , the direct use of barrels remember what each person how many each pizza minimum / second smallest price and its number. Then \ ((2 ^ 9) ^ 3 \) enumeration of pizza and two people can be. Note that some of the details.
code
#include<cstdio>
#include<cstring>
const int N=515;
int a[N],b[N],id[N],b2[N],xid[N],id1,id2,ans,cost;
inline int gi()
{
char c=getchar(); int x=0;
for(;c<'0'||c>'9';c=getchar());
for(;c>='0'&&c<='9';c=getchar())x=(x<<1)+(x<<3)+c-'0';
return x;
}
int main()
{
int n=gi(),m=gi();
for(int i=1;i<=n;++i)
{
int x=gi(),s=0;
while(x--) s|=(1<<gi()-1);
++a[s];
}
memset(b,0x3f,sizeof(b));
memset(b2,0x3f,sizeof(b2));
for(int i=1;i<=m;++i)
{
int c=gi(),x=gi(),s=0;
while(x--) s|=(1<<gi()-1);
if(c<b[s]) b[s]=c,id[s]=i;
else if(c<b2[s]) b2[s]=c,xid[s]=i;
}
cost=2e9+5;
for(int s1=0;s1<512;++s1)
for(int s2=0;s2<512;++s2)
{
if(!id[s1]||!id[s2]) continue;
int tid1,tid2,tcost;
if(s1==s2) tid1=id[s1],tid2=xid[s1],tcost=b[s1]+b2[s1];
else tid1=id[s1],tid2=id[s2],tcost=b[s1]+b[s2];
int ret=0;
for(int s3=0;s3<512;++s3)
if(((s1|s2)&s3)==s3) ret+=a[s3];
if(ret>ans||(ret==ans&&cost>tcost)) cost=tcost,ans=ret,id1=tid1,id2=tid2;
}
printf("%d %d",id1,id2);
}