Codeforces Round #278 (Div. 2) C. Fight the Monster 二分+枚举

二分所需的花费，对给定花费枚举hp、atk、def的购买数量。

代码如下

#include <cstdio>using namespace std;#define INF 30100#define max(a, b) (a) > (b) ? (a) : (b)int hp[2], atk[2], def[2];int h, a, d;bool check(int x){int bound_hp = x / h, rest = x;bool found = false;for(int i = 0; i <= bound_hp && !found; ++i){int bound_atk = rest / a, rest_d = rest, new_hp = hp[0] + i;for(int j = 0; j <= bound_atk && !found; ++j){int new_atk = atk[0] + j;int new_def = def[0] + rest_d / d;rest_d -= a;int dec_y = max(0, atk[1] - new_def), dec_m = max(0, new_atk - def[1]);if(dec_m == 0)continue;else if(dec_y == 0)found = true;else{int time = hp[1] % dec_m ? hp[1] / dec_m + 1 : hp[1] / dec_m;if(dec_y * time < new_hp)found = true;}}rest -= h;}return found;}int main(){scanf("%d %d %d", &hp[0], &atk[0], &def[0]);scanf("%d %d %d", &hp[1], &atk[1], &def[1]);scanf("%d %d %d", &h, &a, &d);int l = 0, r = INF;while(l < r){int mid = l + ((r - l) >> 1);if(check(mid))r = mid;elsel = mid + 1;}printf("%d\n", l);return 0;}