SPOJ 18937 Touch of Venom

Description

Sometimes you have to try fighting even though you know that your enemy is very powerful than you. Your hero with initial health H is about to fight against a venomous enemy who has a poisonous value of P. The enemy's poison deals i*P damage at it's ith attacking chance(i>=1). The hero dies when his health becomes <=0. After enemy's attack, if the hero survives, he heals himself with a health of A by using his skills. Then the enemy gets the chance again and the cycle continues till the hero dies. Find the survival time of the hero. You can safely assume that the hero is mortal.

Example Scenario:

Initial Health(H) = 10, Poison (P) = 2, Heal value(A) = 1

At time 1, enemy does 1*2 damage reducing the hero's health to 8

At time 2, hero heals himself by 1 increasing his health to 9

At time 3, enemy does 2*2 damage reducing the hero's health to 5

At time 4, hero heals himself by 1 increasing his health to 6

At time 5, enemy does 3*2 damage and kill the hero.

The hero survived 5 units of time.

Input:

The first line consists of an integer t, the number of test cases. For each test case there is a line with 3 integers H, P and A.

Output:

For each test case, find the survival time of the hero.

Input Constraints:

1<=t<=10^6

1<=H<=10^6

1<=P<=10^6

0<=A<P

Sample Input:

3

3 7 2

81 4 1

87 8 4

Sample Output:

1

13

9

二分題，如果有人把它當做暴力題來做，而且執迷不悟，那麼歡迎前來和聰哥組隊去跑馬拉松了~~~

直接二分英雄死亡的時間，然後計算判斷即可~~~

還剩半個小時開的這道題，乍一看感覺很水，直接模擬，然後果斷TLE(10^6的數據，怎麼看都不會超時啊！後來看到了喪心病狂的測試組數，T<=10^6,Orz)

#include <cstdio>
typedef long long LL;
int H,P,A;
bool judge(int n){
    LL ans = (LL)(n+1)*n/2*P-(LL)(n-1)*A-H;
    if(ans >= 0) return 1;
    return 0;
}

int main(){
    int kase;
    scanf("%d",&kase);
    while(kase--){
        scanf("%d%d%d",&H,&P,&A);
        int l = 1,r = 1000000,ans = l;
        while(l <= r){
            int mid = (l+r)>>1;
            if(judge(mid))  r = mid-1,ans = mid;
            else    l = mid+1;
        }
        printf("%d\n",ans*2-1);
    }
    return 0;
}