1 second
512 megabytes
standard input
standard output
A conglomerate consists of nn companies. To make managing easier, their owners have decided to merge all companies into one. By law, it is only possible to merge two companies, so the owners plan to select two companies, merge them into one, and continue doing so until there is only one company left.
But anti-monopoly service forbids to merge companies if they suspect unfriendly absorption. The criterion they use is the difference in maximum salaries between two companies. Merging is allowed only if the maximum salaries are equal.
To fulfill the anti-monopoly requirements, the owners can change salaries in their companies before merging. But the labor union insists on two conditions: it is only allowed to increase salaries, moreover all the employees in one company must get the same increase.
Sure enough, the owners want to minimize the total increase of all salaries in all companies. Help them find the minimal possible increase that will allow them to merge companies into one.
The first line contains a single integer nn — the number of companies in the conglomerate (1≤n≤2⋅1051≤n≤2⋅105). Each of the next nn lines describes a company.
A company description start with an integer mimi — the number of its employees (1≤mi≤2⋅1051≤mi≤2⋅105). Then mimi integers follow: the salaries of the employees. All salaries are positive and do not exceed 109109.
The total number of employees in all companies does not exceed 2⋅1052⋅105.
Output a single integer — the minimal total increase of all employees that allows to merge all companies.
3
2 4 3
2 2 1
3 1 1 1
13
One of the optimal merging strategies is the following. First increase all salaries in the second company by 22, and merge the first and the second companies. Now the conglomerate consists of two companies with salaries [4,3,4,3][4,3,4,3] and [1,1,1][1,1,1]. To merge them, increase the salaries in the second of those by 33. The total increase is 2+2+3+3+3=132+2+3+3+3=13.
思路:一个大集团下面有n个公司,每个公司m个员工,每个员工的工资mi,现在要合并公司.两两合并,最后只剩一个才行,合并要求是,两个公司之间的员工最大工资相等才行.
只要记录每个公司的最大工资,然后排序,再开始从小到大挨着两两合并.
#include <iostream>
#include <algorithm>
#include <cstdlib> using namespace std; struct everycom{
int maxx;
int num;
}; struct everycom ec[]; int cmp(struct everycom a,struct everycom b){
return a.maxx<b.maxx;
} int main(){
int n,m;
int now=;
int maxx=;
//while(scanf("%d",&n)){
scanf("%d",&n);
long long sum=;
long long nownum=;
for(int i=;i<n;i++){
cin>>m;
ec[i].num=m;
maxx=;
for(int j=;j<m;j++){
cin>>now;
maxx=now>maxx?now:maxx;
}
ec[i].maxx=maxx;
}
sort(ec,ec+n,cmp);
for(int i=;i<n-;i++){
long long tmp=ec[i+].maxx-ec[i].maxx;
nownum+=ec[i].num;
sum+=tmp*nownum;
}
cout<<sum<<endl;
}
M. The Pleasant Walk
1 second
512 megabytes
standard input
standard output
There are nn houses along the road where Anya lives, each one is painted in one of kk possible colors.
Anya likes walking along this road, but she doesn't like when two adjacent houses at the road have the same color. She wants to select a long segment of the road such that no two adjacent houses have the same color.
Help Anya find the longest segment with this property.
The first line contains two integers nn and kk — the number of houses and the number of colors (1≤n≤1000001≤n≤100000, 1≤k≤1000001≤k≤100000).
The next line contains nn integers a1,a2,…,ana1,a2,…,an — the colors of the houses along the road (1≤ai≤k1≤ai≤k).
Output a single integer — the maximum number of houses on the road segment having no two adjacent houses of the same color.
8 3
1 2 3 3 2 1 2 2
4
In the example, the longest segment without neighboring houses of the same color is from the house 4 to the house 7. The colors of the houses are [3,2,1,2][3,2,1,2] and its length is 4 houses.
思路:让找最长的序列,序列中任何两个相邻的元素不能相等.
#include <iostream>
using namespace std;
int main(){
int n,k;
//int a[100005];
int ai;
while(cin>>n>>k){
int la=;
int now=;
int maxx=;
cin>>ai;
la=ai;
for(int i=;i<n;i++){
//cin>>a[i];
cin>>ai;
if(ai!=la){
now++;
}else{
now=;
}
la=ai;
maxx= now>maxx?now:maxx;
}
cout<<maxx<<endl;
}
}