CodeForces-842A
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感觉自己傻了,这么一道题竟然用n2的循环嵌套来做(我承认刚开始没有想到化简复杂度。。。),言归正题a/b可能的最大值是r/x, 可能的最小值是l/y即l<=k*j且r>=k*x为基本条件
Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number.
For each two integer numbers a and b such that l ≤ a ≤ r andx ≤ b ≤ y there is a potion with experiencea and costb in the store (that is, there are(r - l + 1)·(y - x + 1) potions).
Kirill wants to buy a potion which has efficiency k. Will he be able to do this?
First string contains five integer numbers l,r,x,y,k (1 ≤ l ≤ r ≤ 107,1 ≤ x ≤ y ≤ 107,1 ≤ k ≤ 107).
Print "YES" without quotes if a potion with efficiency exactlyk can be bought in the store and "NO" without quotes otherwise.
You can output each of the letters in any register.
1 10 1 10 1
YES
1 5 6 10 1
NO#include#include #include #include using namespace std; int main() {long long l, r, x, y, k;cin >>l>>r>>x>>y>>k;int flag = 0;if (l > y*k || r < x*k){cout<<"NO"< = l && j*k <= r)//判断j*k是否在区间(l, r)内 {flag = 1;break;}}if(flag == 1){cout<<"YES"<
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