题目链接:http://abc063.contest.atcoder.jp/tasks/arc075_b
Time limit : 2sec / Memory limit : 256MB
Score : 400 points
You are going out for a walk, when you suddenly encounter N monsters. Each monster has a parameter called health, and the health of the i-th monster is hi at the moment of encounter. A monster will vanish immediately when its health drops to 0 or below.
Fortunately, you are a skilled magician, capable of causing explosions that damage monsters. In one explosion, you can damage monsters as follows:
Select an alive monster, and cause an explosion centered at that monster. The health of the monster at the center of the explosion will decrease by A, and the health of each of the other monsters will decrease by B. Here, A and B are predetermined parameters, and A>B holds.At least how many explosions do you need to cause in order to vanish all the monsters?
Input is given from Standard Input in the following format:
N A B h1 h2 : hNPrint the minimum number of explosions that needs to be caused in order to vanish all the monsters.
You can vanish all the monsters in two explosion, as follows:
First, cause an explosion centered at the monster with 8 health. The healths of the four monsters become 3, 4, 1 and −1, respectively, and the last monster vanishes.Second, cause an explosion centered at the monster with 4 health remaining. The healths of the three remaining monsters become 0, −1 and −2, respectively, and all the monsters are now vanished.You need to cause two explosions centered at each monster, for a total of four.
Submit
解析:二分选择A操作的次数
代码:
#include<bits/stdc++.h> #define N 100009 using namespace std; typedef long long LL; const LL INF = 0x3f3f3f3f; LL h[N]; LL n, a, b; bool judge(LL mid) { LL ans = mid; for(int i = 1; i <= n; i++) { LL tm = h[i] - b*mid; if(tm > 0) { LL cnt = tm / (a - b) + (tm % (a - b) == 0 ? 0 : 1); ans -= cnt; if(ans < 0) return false; } } return true; } int main() { scanf("%lld%lld%lld", &n, &a, &b); for(int i = 1; i <= n; i++) scanf("%lld", &h[i]); LL l = 0, r = INF; LL ans = INF; while(l <= r) { LL mid = (l + r)>>1; if(judge(mid)) ans = mid, r = mid - 1; else l = mid + 1; } printf("%lld\n", ans); return 0; }