题库链接

https://codeforces.com/contest/1282

A. Temporarily unavailable

x轴,三个点a,b,c,在c处有一个半径为r的圈,求[a,b]段不在圈内的点的个数

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 200100;
const int M = 1e9+7;
 
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    int t;
    cin>>t;
    while(t--)
    {   
        int a,b,c,d;
        cin>>a>>b>>c>>d;
        if(a > b) swap(a,b);
        int l = c-d;
        int r = c+d;
        int ans = 0;
        if(r < a || l > b)
        {
            cout<<b-a<<endl;
            continue;
        }
        if(l > a) ans += l-a;
        if(r < b) ans += b-r;
        cout<<ans<<endl;
    }
    return 0;
}

B. K for the Price of One

n个物品,买k个物品只需要花k个里面的最贵的那个的钱就行了(可以使用无数次,有没有被坑到了的),p块钱最多可以买多少个物品?
按价格排个序,然后就按k求出k个序列

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 210000;
const int M = 1e9+7;
 
ll a[maxn];
ll b[maxn];
 
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    int n,k,t;
    ll p;
    cin>>t;
    while(t--)
    {
        cin>>n>>p>>k;
        mem(a,0);mem(b,0);
        for(int i = 1; i <= n; i++) 
        {
            cin>>a[i];
        }
        sort(a+1,a+n+1);
        ll res = 0,ans = 0;
        for(int i = 1; i < k; i++) 
        {
            b[i] = b[i-1]+a[i];
            if(b[i] <= p) ans = i;
        }
        for(int i = 0; i < k; i++)
        {
            for(int j = i+k; j <= n; j+=k) 
            {   
                b[i] += a[j];
                //cout<<b[i]<<endl;
                if(b[i] <= p && j > ans) ans = j;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

C. Petya and Exam

参加考试,0代表容易题,解出需要a分钟,1代表难题,解出需要b分钟,可以从任意时间s退出考试,n个题目,每个题目有个限制时间ti,如果要从s时间退出考试则必须把ti<=s的全部题目做完,求最多可以做几个题目;
贪心,先对t排序,当做完i题,总时间小于\(t_{i+1}-1\),是合法的,并且可以在多余的时间里面多做一些简单题
会爆int

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 200100;
const int M = 1e9+7;
int n,m,k,t,x,y;
 
ll sum[maxn];
 
struct node
{
    ll ty,t;
}a[maxn];
 
bool cmp(node a,node b)
{   
    return a.t < b.t;
}
 
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    cin(t);
    while(t--)
    {
        cin(n);cin(m);cin(x);cin(y);
        for(int i = 0; i < n; i++) 
        {
            cin>>a[i].ty;
        }
        for(int i = 0; i < n; i++) 
        {
            cin>>a[i].t;
        }
        sort(a,a+n,cmp);
        a[n].t = m+1;
        ll time = 0;
        int res = 0;
 
        for(int i = n-1; i >= 0; i--)
        {   
            sum[i] = res;
            if(a[i].ty == 0) res++;
        }
        int ans = 0;
        if(a[0].t) ans = min((a[0].t-1)/x,sum[0]);      //在t[0]前面做题
        res = 0;
        for(int i = 0; i < n; i++) 
        {
            if(a[i].ty == 0) time += x;
            else time += y;
            if(time < a[i+1].t)
            {
                res = i+1;
                res += min((a[i+1].t-time-1)/x,sum[i]);
                ans = max(ans,res);
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}