【Codeforces Round #523 (Div. 2)】 A B C D

给你无限个 1-n的硬币问你怎么最少的组成m

那么我们贪心的去组成m即可所以直接除以n 判断是否除的尽即可

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}

int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    long long n,m;
    scanf("%lld%lld",&n,&m);
    printf("%lld\n",m/n+(m%n==0?0:1));
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

给你一个由n个快组成的图形有俯视图和侧视图问你最多去几个小格子使得侧视图和俯视图不变

我们给块排个序然后贪心的每个块都留高度为之前块加一的块如果高度超过块本身高度就留块本身高度最后不够的用最后一个高块去补全即可

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
long long arr[1000025];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n;
    long long m;
    scanf("%d%lld",&n,&m);
    for(int i = 1;i<=n;++i) scanf("%lld",&arr[i]);
    sort(arr+1,arr+1+n);
    long long ans = 0,maxx = arr[n],tmp = 0;
    for(int i = 1;i<n;++i)
    {
        ans+=arr[i] - 1;
        tmp=min(arr[i],tmp+1);
    }
    if(maxx>tmp) ans+=arr[n]-(maxx-tmp);
    else ans+=arr[n] - 1;
    printf("%lld\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

n是 1e5 n^2的做法过不了一开始很难处理

用dp做定义 dp[i] 表示 i 长度的合法方案数那么我们知道对每一个数的所有因子 x 可以让 dp[x] += dp[x-1] 因为代表可以放x-1这个位置后面的x位置上从大往小枚举因子使得这次放的方案不受自己因子的影响

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
#define MOD 1000000007
vector<int > vt[1000025];
const int MAX_N = 1000025;
void init()
{
    for(int i = 1;i<=1000000;i++)
    {
        for(int j = i;j<=1000000;j+=i)
        {
            vt[j].push_back(i);
        }
    }
}
int arr[100025];
long long dp[MAX_N];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n;
    init();
    scanf("%d",&n);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
    dp[0] = 1;
    for(int i = 1;i<=n;++i)
    {
        for(int j = vt[arr[i]].size()-1;j>=0;j--)
        {
            int x = vt[arr[i]][j];
            dp[x] = (dp[x] + dp[x-1])%MOD;
        }
    }
    long long ans = 0;
    for(int i = 1;i<=1000000;i++) ans = (ans+dp[i])%MOD;
    printf("%lld\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

题意给你n个电视节目以及x y 开一个新电视的贡献是 x + 时间 *y 否则是上一个电视延续到结束时间 * y

所以我们把电视时间二维排序按照时间从小到大然后用优先队列去判断哪个时间更优取更优的贪心即可

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
#define MOD 1000000007
const int MAX_N = 100025;
struct node
{
    int l,r,id;
    bool operator<(const node other) const
    {
        if(l==other.l) return id < other.id;
        return l < other.l;
    }
}arr[MAX_N<<1];
priority_queue<int > q;
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,l,r;
    long long x,y,ans = 0;
    scanf("%d%lld%lld",&n,&x,&y);
    for(int i = 1;i<=n;++i)
    {
        scanf("%d%d",&l,&r);
        arr[i*2 - 1].l = l,arr[i*2-1].r = r,arr[i*2-1].id = 0;
        arr[i*2].l = r,arr[i*2].id = 1;
    }
    n*=2;
    sort(arr+1,arr+1+n);
    for(int i = 1;i<=n;++i)
    {
        if(arr[i].id==1)
            q.push(arr[i].l);
        else
        {
            if(!q.empty())
            {
                int top = q.top();
                ll tmp1 = (arr[i].r - top) * y;
                ll tmp2 = (arr[i].r - arr[i].l) * y + x;
                if(tmp1<tmp2)
                {
                    ans+=tmp1;
                    ans%=MOD;
                    q.pop();
                }
                else ans+=tmp2,ans%=MOD;
            }
            else ans = ans+x+(arr[i].r - arr[i].l)*y,ans%=MOD;
        }
    }
    printf("%lld\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

【Codeforces Round #523 (Div. 2)】 A B C D

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