codeforces Round #413 Div. 1 + Div. 2 A Carrot Cakes
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增加一个炉后 ,相同时间 原来可以制造一份的,现在可以制造两份,,
只要把最后的1份拿出来,看看之前制造那些蛋糕的同时能不能建造成新的炉,
如果可以的话,那倒数第二份可以拿出来跟最后一份合起来 花费 一个 t 的时间的到 n 个蛋糕(这样判断时间会减少)
#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;typedef long long ll;const int maxn = 30 + 7, INF = 0x3f3f3f3f, mod = 1e9 + 7;int n, t, k, d;int main() { scanf("%d %d %d %d", &n, &t, &k, &d); int cnt = n / k + (n % k == 0 ? 0 : 1); if(cnt == 1) return 0*puts("NO"); if((cnt-1)*t > d) return 0*puts("YES"); puts("NO"); return 0;}
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