K. Video Posts
Input
The first line contains two integers n and k (1≤k≤n≤105). The next line contains n positive integer numbers a1,a2,…,an (1≤ai≤104), where ai is the duration of the i-th video.
Output
If solution exists, print "Yes" in the first line. Print k positive integers s1,s2,…,sk (s1+s2+⋯+sk=n) in the second line. The total duration of videos in each post should be the same. It can be easily proven that the answer is unique (if it exists).
If there is no solution, print a single line "No".
Examples
Input
6 3
3 3 1 4 1 6Output
Yes
2 3 1 Input
3 3
1 1 1
Output
Yes
1 1 1 Input
3 3
1 1 2Output
NoInput
3 1
1 10 100Output
Yes
3 题意：就是看能否均分，是的话就Yes,然后输出可以由几个数组成，不是就No
#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<math.h>
#include<map>
#include<string>
using namespace std;
typedef long long ll;
map<int,int>mp;
map<int,int>::iterator it;
ll c[100010],a[101010];
int main()
{
ll n,k,i,j,s=0,ss;
cin>>n>>k;
for(i=0; i<n; i++)
{
cin>>a[i];
s+=a[i];
}
if(s%k)
printf("No\n");
else
{
ss=s/k;
ll e=0,ans=0,p=0;
memset(c,0,sizeof(c));
for(i=0; i<n; i++)
{
ans+=a[i];
p++;
if(ans==ss)
{
c[e++]=p;
ans=0;
p=0;
}
}
//cout<<e<<endl;
if(e==k)
{
cout<<"Yes"<<endl;
cout<<c[0];
for(i=1; i<e; i++)
cout<<" "<<c[i];
cout<<endl;
}
else
cout<<"No"<<endl;
}
return 0;
}D. Garbage Disposal
Examples
Input
3 2
3 2 1Output
3
Input
5 1
1000000000 1000000000 1000000000 1000000000 1000000000Output
5000000000
Input
3 2
1 0 1Output
2
Input
4 4
2 8 4 1Output
4题意：扔垃圾，垃圾最多隔一天扔，问最少需要几个垃圾袋
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
const int A=1e9+7;
ll a[201010];
int main()
{
ll n,k,i,m,ans=0,j,x=0;
cin>>n>>m;
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<n-1;i++)
{
if(a[i]+x>=m)
{
ans+=(a[i]+x)/m;
x=(a[i]+x)%m;
}
else
{
if(x==0)
x=a[i];
else
{
x=0;
ans++;
}
}
}
if((a[n-1]+x)%m==0)
ans+=(a[n-1]+x)/m;
else
ans+=(a[n-1]+x)/m+1;
cout<<ans<<endl;
return 0;
}
F. Debate
Elections in Berland are coming. There are only two candidates — Alice and Bob.
The main Berland TV channel plans to show political debates. There are n people who want to take part in the debate as a spectator. Each person is described by their influence and political views. There are four kinds of political views:
- supporting none of candidates (this kind is denoted as "00"),
- supporting Alice but not Bob (this kind is denoted as "10"),
- supporting Bob but not Alice (this kind is denoted as "01"),
- supporting both candidates (this kind is denoted as "11").
The direction of the TV channel wants to invite some of these people to the debate. The set of invited spectators should satisfy three conditions:
- at least half of spectators support Alice (i.e. 2⋅a≥m, where a is number of spectators supporting Alice and mis the total number of spectators),
- at least half of spectators support Bob (i.e. 2⋅b≥m, where b is number of spectators supporting Bob and m is the total number of spectators),
- the total influence of spectators is maximal possible.
Help the TV channel direction to select such non-empty set of spectators, or tell that this is impossible.
Input
The first line contains integer n(1≤n≤4⋅105) — the number of people who want to take part in the debate as a spectator.
These people are described on the next n lines. Each line describes a single person and contains the string si and integer ai separated by space (1≤ai≤5000), where si denotes person's political views (possible values — "00", "10", "01", "11") and ai — the influence of the i-th person.
Output
Print a single integer — maximal possible total influence of a set of spectators so that at least half of them support Alice and at least half of them support Bob. If it is impossible print 0 instead.
-
Output6 11 6 10 4 01 3 00 3 00 7 00 9
Input22
Output5 11 1 01 1 00 100 10 1 01 1
Input103
Output6 11 19 10 22 00 18 00 29 11 29 10 28
Input105
Output3 00 5000 00 5000 00 50000Note
In the first example 4
spectators can be invited to maximize total influence: 1, 2, 3 and 6. Their political views are: "11", "10", "01" and "00". So in total 2 out of 4 spectators support Alice and 2 out of 4 spectators support Bob. The total influence is 6+4+3+9=22.In the second example the direction can select all the people except the 5-th person.
In the third example the direction can select people with indices: 1, 4, 5 and 6.
In the fourth example it is impossible to select any non-empty set of spectators.
题意：两个人要进行辩论赛，剧场要选观众，所选的观众要满足3个条件，至少有一半的人支持第一个人，至少有一半的人支持另一个人，所选观众影响力最大，问你最大的影响力是多少。
思路：11的肯定要，在01跟10的情况下找最小的是都要的 ，然后在剩下的一些中找最大的，先判断还需要几个人
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
const int AA=4e5+7;
struct A
{
ll p;
ll q;
}s1[AA],s2[AA],s3[AA],s4[AA];
bool cmp(struct A l,struct A ll)
{
if(l.p==ll.p)
return l.q>ll.q;
else
return l.p>ll.p;
}
bool cmp1(struct A l,struct A ll)
{
return l.q>ll.q;
}
int main()
{
ll a,ans=0,b,i,j=0,k=0,t=0,tt=0,m=0,sum=0,n;
cin>>n;
for(i=0;i<n;i++)
{
cin>>a>>b;
if(a==11)
{
s1[j].p=a;
s1[j].q=b;
j++;
}
if(a==10)
{
s2[k].p=a;
s2[k].q=b;
k++;
}
if(a==1)
{
s3[t].p=a;
s3[t].q=b;
t++;
}
if(a==0)
{
s4[tt].p=a;
s4[tt].q=b;
tt++;
}
}
for(i=0;i<j;i++)
{
ans+=s1[i].q;
}
sort(s2,s2+k,cmp);
sort(s3,s3+t,cmp);
sort(s4,s4+tt,cmp);
//cout<<j<<" "<<k<<" "<<t<<" "<<tt<<endl;
a=b=j;
sum+=j;
//cout<<a<<" "<<b<<" "<<sum<<" "<<ans<<endl;
if(k==t)
{
for(i=0;i<k;i++)
{
ans+=s2[i].q;
ans+=s3[i].q;
}
a+=k;
b+=k;
sum+=k*2;
//cout<<a<<" "<<b<<" "<<sum<<" "<<ans<<endl;
sort(s4,s4+tt,cmp);
for(i=0;i<tt;i++)
{
if(2*a-sum<=0)
break;
ans+=s4[i].q;
sum++;
}
}
else
if(k>t)
{
//cout<<t<<endl;
//cout<<s2[0].q<<" "<<s3[0].q<<endl;
for(i=0;i<t;i++)
{
ans+=s2[i].q;
ans+=s3[i].q;
}
a+=t;
b+=t;
sum+=t*2;
//cout<<a<<" "<<b<<" "<<sum<<" "<<ans<<endl;
if(2*a-sum==0)
{
cout<<ans<<endl;
return 0;
}
for(i=t;i<k;i++)
{
s4[tt].p=10;
s4[tt].q=s2[i].q;
tt++;
}
sort(s4,s4+tt,cmp1);
for(i=0;i<tt;i++)
{
if(2*a-sum<=0)
break;
ans+=s4[i].q;
sum++;
}
}
else
if(k<t)
{
for(i=0;i<k;i++)
{
ans+=s2[i].q;
ans+=s3[i].q;
}
a+=k;
b+=k;
sum+=k*2;
if(2*a-sum==0)
{
cout<<ans<<endl;
return 0;
}
//cout<<a<<" "<<b<<" "<<sum<<" "<<ans<<endl;
for(i=k;i<t;i++)
{
s4[tt].p=01;
s4[tt].q=s3[i].q;
tt++;
}
sort(s4,s4+tt,cmp1);
for(i=0;i<tt;i++)
{
if(2*a-sum<=0)
break;
ans+=s4[i].q;
sum++;
}
}
cout<<ans<<endl;
return 0;
}
H - BerOS File Suggestion
Input
4
test
contests
test.
.test
6
ts
.
st.
.test
contes.
st
Output
1 contests
2 test.
1 test.
1 .test
0 -
4 .test题意：输出出现过几次然后随便输出一个满足的主串
思路：直接找的话，会时间超限，所以可以用map把上面输入的串的所有字串都记下来，用两个map，一个记录有几个串有匹配的，一个记录匹配的串是哪个。单个去重，总的不去重。
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
const int AA=4e5+7;
map<string,ll> M;
map<string,string> N;
int main()
{
ll n,ans=0,i,j,m,k;
string s;
cin>>n;
for(i=0;i<n;i++)
{
cin>>s;
ll len=s.size();
map<string,ll> qc;
for(j=0;j<len;j++)
{
for(k=1;k<=len;k++)
{
string t;
t=s.substr(j,k);
if(qc[t]==0)
{
M[t]++;
N[t]=s;
qc[t]=1;
}
}
}
}
cin>>m;
string ss;
for(i=0;i<m;i++)
{
cin>>ss;
if(M[ss]==0)
cout<<"0 -"<<endl;
else
cout<<M[ss]<<" "<<N[ss]<<endl;
}
return 0;
}