Codeforces Round #430 (Div. 2) 签到题
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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 and x ≤ b ≤ y there is a potion with experience a and cost b 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 exactly k 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
代码实现:
#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<queue>#include<cstdio>#define ll long long#define mset(a,x) memset(a,x,sizeof(a))using namespace std;const double PI=acos(-1);const int inf=0x3f3f3f3f;const double esp=1e-6;const int maxn=100005;const int mod=1e9+7;int dir[4][2]={0,1,1,0,0,-1,-1,0};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}int main(){int l,r,a,b,ans;double x,y,z,k;cin>>l>>r>>a>>b>>k;int flag=1;int l1=l,r1=b;for(int i=a;i<=y;i++){if(k*i>=l&&k*i<=r){cout<<"YES"<<endl;return 0;}}cout<<"NO"<<endl;return 0;}
依旧水。
代码实现:
#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<queue>#include<cstdio>#define ll long long#define mset(a,x) memset(a,x,sizeof(a))using namespace std;const double PI=acos(-1);const int inf=0x3f3f3f3f;const double esp=1e-6;const int maxn=100005;const int mod=1e9+7;int dir[4][2]={0,1,1,0,0,-1,-1,0};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}struct node{double x;double y;double r;}p[maxn];double dis(double x,double y){return sqrt(x*x+y*y);}int main(){double d,r,n;int i,j,k;while(cin>>r>>d){cin>>n;for(i=0;i<n;i++){cin>>p[i].x>>p[i].y>>p[i].r;}int ans=0;for(i=0;i<n;i++){double ans1=dis(p[i].x,p[i].y);double ans2=ans1-p[i].r;if(ans1>=(r-d)&&ans1<=r&&ans2>=(r-d)&&(ans1+p[i].r<=r))ans++; }cout<<ans<<endl;}return 0;}
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