HDOJ 4950 Monster
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题目:
Monster
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 198 Accepted Submission(s): 81
Problem Description
Teacher Mai has a kingdom. A monster has invaded this kingdom, and Teacher Mai wants to kill it.
Monster initially has h HP. And it will die if HP is less than 1.
Teacher Mai and monster take turns to do their action. In one round, Teacher Mai can attack the monster so that the HP of the monster will be reduced by a. At the end of this round, the HP of monster will be increased by b.
After k consecutive round's attack, Teacher Mai must take a rest in this round. However, he can also choose to take a rest in any round.
Output "YES" if Teacher Mai can kill this monster, else output "NO".
Monster initially has h HP. And it will die if HP is less than 1.
Teacher Mai and monster take turns to do their action. In one round, Teacher Mai can attack the monster so that the HP of the monster will be reduced by a. At the end of this round, the HP of monster will be increased by b.
After k consecutive round's attack, Teacher Mai must take a rest in this round. However, he can also choose to take a rest in any round.
Output "YES" if Teacher Mai can kill this monster, else output "NO".
Input
There are multiple test cases, terminated by a line "0 0 0 0".
For each test case, the first line contains four integers h,a,b,k(1<=h,a,b,k <=10^9).
For each test case, the first line contains four integers h,a,b,k(1<=h,a,b,k <=10^9).
Output
For each case, output "Case #k: " first, where k is the case number counting from 1. Then output "YES" if Teacher Mai can kill this monster, else output "NO".
Sample Input
5 3 2 20 0 0 0
Sample Output
Case #1: NO
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如果a>=h,直接可以打死怪兽,否则,只有在第k次打死怪兽或者是没k+1次都可以对怪兽造成伤害,才能打死怪兽。
代码:
#include <cstdio>#include <cstring>typedef long long lint;int main(){ int h, a, b, k, icase = 1; while(~scanf("%d%d%d%d", &h, &a, &b, &k)) { if(0==h && 0==a && 0== b && 0==k) break; bool flag = false; if(a >= h) flag = true; else { lint t = 1ll*h - 1ll*(k-1)*(a-b) - 1ll*a; if(t <= 0) flag = true; t = 1ll*a*k - 1ll*b*(k+1); if(t > 0) flag = true; } printf("Case #%d: %s\n", icase++, flag ? "YES" : "NO"); } return 0;}
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