Fight Against Monsters （2018-2019 ACM-ICPC, China Multi-Provincial Collegiate Programming Contest）（The 2019 Asia Yinchuan First Round Online Programming）

It is my great honour to introduce myself to you here. My name is Aloysius Benjy Cobweb Dartagnan Egbert Felix Gaspar Humbert Ignatius Jayden Kasper Leroy Maximilian. As a storyteller, today I decide to tell you and others a story about the hero Huriyyah, and the monsters.

Once upon a time, citizens in the city were suffering from

For the good of these unfortunate citizens, Huriyyah set off to the forest which was the main lair of monsters and fought with

They fought in a cave through a turn-based battle. During each second, the hero Huriyyah was attacked by monsters at first, and the damage was the sum of attack values of all alive monsters. Then he selected a monster and attacked it. The monster would suffer the damage of

Now, my smart audience, can you calculate the minimum amount of total damages our hero should suffer before he won the battle?

Input

The input contains several test cases, and the first line is a positive integer

For each test case, the first line contains an integers

The

We guarantee that the sum of

Output

For each test case, output a line containing Case #x: y, where x is the test case number starting from

Example

input

Copy

output

Copy

Case #1: 19
Case #2: 14


分析：要是受到的伤害最低，那么只要Kill小怪兽的顺序最优就可以啦，怎样最优呢？最优的影响因素是 kill 小怪兽的时间，而时间又受到伤害的影响
　　  所以呢，我们排序的时候就判断先kill A怪兽再kill B怪兽所受到的伤害高还是先kill B怪兽再kill A怪兽受到的伤害高
PS：有队友提出优先队列做，这里提供一下思路，有兴趣的可以搞一下

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
#define pq priority_queue<int>
#define pql priority_queue<ll>
#define pqn priority_queue<node>
#define v vector<int>
#define vl vector<ll>
#define lson rt<<1, l, m  
#define rson rt<<1|1, m+1, r
#define read(x) scanf("%d",&x)
#define lread(x) scanf("%lld",&x);
#define pt(x) printf("%d\n",(x))
#define yes printf("YES\n");
#define no printf("NO\n");
#define gcd __gcd
#define cn cin>>
#define ct cout<<
#define ed <<endl;
#define ok return 0 ;
#define over cout<<endl;
#define rep(j,k) for (int i = (int)(j); i <= (int)(k); i++)
#define input(k) for (int i = 1; i <= (int)(k); i++)  {cin>>a[i] ; }
#define mem(s,t) memset(s,t,sizeof(s))
#define re return 0;
#define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define mod(x) ((x)%9973)
#define test cout<<"     ++++++      "<<endl; 
typedef long long ll;
const int INT=1e6+5;
const int maxn=10000+5;
const int len = 1e5+5;

typedef struct node
{
    ll x,y,ankle,T;
}node;
//double getlen(node xx,node yy) { return sqrt( (xx.x-yy.x)*(xx.x-yy.x) +(xx.y-yy.y)*(xx.y-yy.y) ); } //计算两点间距离 
node dp[len]; 
int cmp(node a,node b)
{
　　/*  可以加可以不加
    if(b.y*a.T == a.y*b.T)
        return a.y>b.y;
　　*/
    return a.y*b.T > b.y*a.T ; 
}
int runtime(int num)
{
    int k=1,t=0;
    while(num>0)
    {
        t++;
        num-=k;
        k++;
    }
    return t;
}
int main()
{
    int t,n,k;
    cin>>t;
    for(int kk=1;kk<=t;kk++)
    {
        cin>>n;
        ll cnt=0,sum=0;
        rep(0,n-1)
        {
            cin>>dp[i].x>>dp[i].y;
            dp[i].T = runtime(dp[i].x) ;
            cnt+=dp[i].y;
        }
        sort(dp,dp+n,cmp);
        rep(0,n-1)
        {
            sum+=cnt*dp[i].T;
            cnt-=dp[i].y;
        }
        cout<<"Case #"<<kk<<": "<<sum<<endl;
    }
    ok;
}

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Fight Against Monsters （2018-2019 ACM-ICPC, China Multi-Provincial Collegiate Programming Contest）（The 2019 Asia Yinchuan First Round Online Programming）

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