2019 ICPC 银川网络赛 H. Fight Against Monsters

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 nn powerful monsters. They ate small children who went out alone and even killed innocent persons. Before the hero appeared, the apprehension had overwhelmed the people for several decades. For the good of these unfortunate citizens, Huriyyah set off to the forest which was the main lair of monsters and fought with nn fierce and cruel monsters. The health point of the ii-th monster was HP_iHPi, and its attack value was ATK_iATKi.

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 kk (its health point would decrease by kk) which was the times of attacks it had been came under. That is to say, for each monster, the damage of the first time that Huriyyah attacked it was 11, and the damage of Huriyyah's second attack to this monster was 22, the third time to this monster was 33, and so on. If at some time, the health point of a monster was less than or equal to zero, it died. The hero won if all monsters were killed.

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 TT indicating the number of test cases which is up to 10^3103.

For each test case, the first line contains an integers n (1 le n le 10^5)n(1≤n≤105) which is the number of monsters. The ii-th line of the following nn lines contains two integers HP_iHPi and ATK_i (1 le HP_i, ATK_i le 10^5)ATKi(1≤HPi,ATKi≤105) which describe a monster.

We guarantee that the sum of nn in all test cases is up to 10^6106.

Output

For each test case, output a line containing Case #x: y, where xx is the test case number starting from 11, and yy is the minimum amount of total damages the hero should suffer.

输出时每行末尾的多余空格，不影响答案正确性

样例输入复制

样例输出复制

Case #1: 19
Case #2: 14

这个题是一个贪心题，一开始使用的攻击力/血量作为贪心准则，然后样例能过，就在我要交的时候，队友说不同的HP杀死用的时间是一样的，所以这个题，确实是没读太懂题，在读了一遍题目，然后1A过了，题意举个生动的例子，时下最火的自走棋，对方上了一群大汉，现在你的德莱文畏畏缩缩的缩在角落，敌人的血量与攻击力不同，你攻击别人一下，然后所有人都打你一下，现在让你求怎样杀死所有人使得自己掉血最少。

#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include<cstdio>
#include<cstring>
//---------------------------------Sexy operation--------------------------//

#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define cins(s) scanf("%s",s)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define debug(n) printf("%d_________________________________n",n);
#define speed ios_base::sync_with_stdio(0)
#define file  freopen("input.txt","r",stdin);freopen("output.txt","w",stdout)
//-------------------------------Actual option------------------------------//
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define Swap(a,b) a^=b^=a^=b
#define Max(a,b) (a>b?a:b)
#define Min(a,b) a<b?a:b
#define mem(n,x) memset(n,x,sizeof(n))
#define mp(a,b) make_pair(a,b)
#define pb(n)  push_back(n)
#define dis(a,b,c,d) ((double)sqrt((a-c)*(a-c)+(b-d)*(b-d)))
//--------------------------------constant----------------------------------//

#define INF  0x3f3f3f3f
#define esp  1e-9
#define PI   acos(-1)
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef  long long ll;
//___________________________Dividing Line__________________________________/
struct PPP
{
    int HP;
    int GJ;
    int Time;
}a[100010];
int num(int a)
{
    int sum=0,cnt=1;
    while(1)
    {
        sum++;
        a=a-cnt++;
        if(a<=0)  return sum;
    }
}
bool cmp(PPP a,PPP b)
{
    return a.Time*b.GJ<b.Time*a.GJ;
}
int main()
{
    int t,k=0;
    scanf("%d",&t);
    while(k++<t)
    {
        int n,i;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            cini(a[i].HP),cini(a[i].GJ);
            a[i].Time=num(a[i].HP);
        }
        sort(a,a+n,cmp);
        ll sum=0,cnt=0;
        for(i=0;i<n;i++)
        {
            cnt+=a[i].Time;
            sum+=cnt*a[i].GJ;
        }
        printf("Case #%d: %lldn",k,sum);

    }
    return 0;
}