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HZNU 2019 Summer training 7

2019-07-12 19:51:51 作者：互联网

A - Coins

CodeForces - 1061A 题意：需要几个硬币使得比S大题解：除一下向上取整

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include<stack>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;

const int maxn = 8e4 + 10;

int main()
{
    ll n,s;
    cin >> n >> s;
    ll tmp = ceil(s * 1.0 / (n * 1.0 ));
    printf("%lld\n",tmp);
}

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B - Views Matter

CodeForces - 1061B

题意：n列，原来每列都放有a[ i ] 个盒子，问最多能移除多少个盒子后，使移除后的俯视图和右视图与移除前的相同

题解：每一列尽可能的只拿最上面的一个，接下来的剩下的在最高的一列拿，遍历一下就好了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include<stack>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;

const int maxn = 1e5 + 10;

ll a[maxn];
int main()
{
    ll n,m;
    scanf("%lld %lld",&n,&m);
    ll maxx = -1;

    ll sum = 0;
    for(int i=0;i<n;i++)
    {
        scanf("%lld",&a[i]);
        maxx = max(maxx,a[i]);
        sum += a[i];
    }
    sort(a,a+n);

    int ans = 0;
    ll pos = 1;
    for(int i=0;i<n;i++)
    {
        if(a[i] >= pos)
        {
            ans++;
            pos++;
        }
        else
        {
            ans++;
        }
    }
    pos--;
    maxx = max(0LL,maxx - pos);
   // cout<<maxx<<endl;
    printf("%lld\n",sum - (ans + maxx));

}

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C - Multiplicity

CodeForces - 1061C

题意：给出序列ai，询问有多少个 a 的子序列 b 满足，对于任意的b[i]，b[i] mod i = 0

题解：先预处理因子，将每一个的倍数处理出来。之后开始dp累加前一个的方案数

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include<stack>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;

const int maxn = 1e5 + 10;

const int mod = 1e9 + 7;
int n,a[maxn],dp[maxn * 10],ans;
vector<int>fact[maxn * 10];

int maxx = -1;
int main()
{
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d",&a[i]);
        maxx = max(maxx,a[i]);
    }
    for(int i  = 1; i <= maxx; i++)
        for(int j = 1; j <= maxx / i; j++)
            fact[i * j].push_back(i);
    for(int i = 1; i <= n; i++)
        sort(fact[a[i]].begin(),fact[a[i]].end());

    dp[0] = 1;

    for(int i = 1; i <= n; i++)
    {
        for(int j = fact[a[i]].size() - 1; j >= 0; j--)
        {
            int x = fact[a[i]][j];
            dp[x] += dp[x - 1];
            dp[x] %= mod;
            //printf("i = %d dp[%d] = %d\n",i,x,dp[x]);
        }
    }

    for(int i = 1; i <= n; i++)
    {
        ans += dp[i];
        ans %= mod;
    }
    printf("%d\n",ans);
}

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D - Sleepy Game

CodeForces - 937D 题意：Petya and Vasya 在玩移动旗子的游戏，谁不能移动就输了。 Vasya在订移动计划的时候睡着了，然后Petya 就想趁着Vasya睡着的时候同时定下策略，如果可以赢得话输出Win 并输出路径，如果步数在达到1e6的情况下，就认定为平局，输出Draw，如果输的话就输出lost。题解：需要找到一条奇数路线的路径，dfs判环，一个奇数的环可以改变到达终点的奇偶步数

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include<stack>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;

const int maxn = 1e5 + 10;

vector<int>g[maxn];
vector<int>path;
int vis[maxn][2];
int st[maxn];
int n,m;
bool circle = false;
void dfs(int x,int dep)
{
    if(vis[x][dep & 1]) //vis的后面：1代表奇数步可达，0代表偶数不可达
        return;
    vis[x][dep & 1] = 1;
    if(g[x].size() == 0 && (dep & 1))
    {
        puts("Win");
        for(int i = 0; i < path.size(); i++)
            printf("%d%c",path[i],i == path.size() - 1 ? '\n' : ' ');
        exit(0);
    }
    for(int i = 0; i < g[x].size(); i++)
    {
        int next = g[x][i];
        if(st[next])
            circle = true;
        path.push_back(next);
        st[next] = true;
        dfs(next,dep + 1);
        path.pop_back();
        st[next] = false;
    }
}
int main()
{
    scanf("%d %d",&n,&m);
    int tmp,u;
    for(int i = 1; i <= n; i++)
    {
        scanf("%d",&tmp);
        while(tmp--)
        {
            scanf("%d",&u);
            g[i].push_back(u);
        }
    }
    int start;
    scanf("%d",&start);
    st[start] = true;
    path.push_back(start);
    dfs(start,0);
    if(circle)
        puts("Draw");
    else
        puts("Lose");
}

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E - Tree Reconstruction

CodeForces - 1041E

标签：training,int,HZNU,ll,long,2019,maxn,include,define
来源： https://www.cnblogs.com/smallhester/p/11178065.html