8VC Venture Cup 2017 - Elimination Round

传送门：http://codeforces.com/contest/755

A题题意是给你一个数字n，让你找到一个数字m，使得n*m+1为合数，范围比较小，直接线性筛出1e6的质数，然后暴力枚举一下就好了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#define X first
#define Y second
#define clr(u,v); memset(u,v,sizeof(u));
#define in() freopen("data","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e6 + ;
const int INF = 0x3f3f3f3f;
bool prim[maxn];
int main()
{
    int n;
    int m = sqrt(maxn + 0.5);
    for (int i = ; i <= m; i++)
        if (!prim[i])
            for (int j = i * i; j < maxn; j += i)
                prim[j] = ;
    prim[] = prim[] = ;
    scanf("%d", &n);
    for (int i = ; i <= ; i++)
        if (prim[i*n+])
        {
            printf("%d\n", i);
            return ;
        }
    return ;
}

B题题意是A和B玩游戏，A有n个单词，B有m个单词，每个人轮流说一句单词，但是不能说重复的单词，说到最后没得说的人输，A先说，问最后A是否能赢。比较简单的贪心，找出两个人单词中的相同数量，这一部分是必须先讲的，可以处理出AB可以讲的单词数，比较一下大小就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#define X first
#define Y second
#define clr(u,v); memset(u,v,sizeof(u));
#define in() freopen("data","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e5 + ;
const int INF = 0x3f3f3f3f;
set <string> s;
int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    string str;
    int same = ;
    for (int i = ; i < n; i++)
    {
        cin >> str;
        s.insert(str);
    }
    for (int i = ; i < m; i++)
    {
        cin >> str;
        if (s.count(str))
            same++;
    }
    int num1 = n - same + (same - same / );
    int num2 = m - same + same / ;
    if (num1 > num2) puts("YES");
    else puts("NO");
    return ;
}

C题题意没看懂。。直接看提示大概是给你n个节点，第i个节点和第a[i]个节点在同一棵树上，问有多少棵树。直接并查集或dfs处理出联通块的数量就行了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#define X first
#define Y second
#define clr(u,v); memset(u,v,sizeof(u));
#define in() freopen("data","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e5 + ;
const int INF = 0x3f3f3f3f;
int f[maxn];
int find(int x)
{
    return f[x] == x ? x : f[x] = find(f[x]);
}
int mix(int x, int y)
{
    int fx = find(x), fy = find(y);
    if (fx == fy) return ;
    f[fx] = fy;
    return ;
}
int main()
{
    int n, x;
    scanf("%d", &n);
    int sum = n;
    for (int i = ; i <= n; i++) f[i] = i;
    for (int i = ; i <= n; i++)
    {
        scanf("%d", &x);
        if(mix(i, x)) sum--;
    }
    printf("%d\n", sum);
    return ;
}

D题是给你n边形，然后每次相隔k个点连上一条边，问每次连边后该n边形一共有多少个面。可以考虑，每连一条边，多出的面数是两点直接线段的数量，所以直接树状数组/线段树统计下走过的点，每次增加的时候直接查询他们之间有多少个点就行了。注意要long long，还有要k=min(k,n-k);不然会多算。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#define X first
#define Y second
#define clr(u,v); memset(u,v,sizeof(u));
#define in() freopen("data","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e6 + ;
const int INF = 0x3f3f3f3f;
ll c[maxn];
int n;
int lowbit(int x)
{
    return x & (-x);
}
ll getsum(int x)
{
    ll ans = ;
    while (x > )
    {
        ans += c[x];
        x -= lowbit(x);
    }
    return ans;
}
void updata(int x)
{
    while (x <= n)
    {
        c[x]++;
        x += lowbit(x);
    }
}
int main()
{
    int k;
    scanf("%d%d", &n, &k);
    k = min(n - k, k);
    ll add = , ans = , cur = ;
    for (int i = ; i < n; i++)
    {
        updata(cur);
        if (cur + k <= n) add = getsum(cur + k) - getsum(cur) + ;
        else add = getsum(n) - getsum(cur) + getsum((cur + k - ) % n + ) + ;
        ans += add;
        if (i == n - ) cout << ans - ;
        else cout << ans << " ";
        cur = (cur + k - ) % n + ;
        updata(cur);
//        for (int i=1;i<=n;i++)
//        cout<<getsum(i)-getsum(i-1)<<" ";
//        cout<<endl;
    }
    return ;
}

这题也可以找规律做。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#define X first
#define Y second
#define clr(u,v); memset(u,v,sizeof(u));
#define in() freopen("data","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e6 + ;
const int INF = 0x3f3f3f3f;
int main()
{
    ll n, k;
    cin >> n >> k;
    k = min(k, n - k);
    ll c = , add = , ans = ;
    for (ll i = ; i <= n; i++)
    {
        c += k;
        if (c > n)
        {
            add++;
            ans += add;
            add++;
            c %= n;
        }
        else ans += add;
        if (i == n) cout << ans -  << " ";
        else cout << ans << " ";
    }
    return ;
}

总结：这次前面的题比较水，虽然是摸黑打的，手速慢了点，好在D没fst掉，升了171分，直接蓝名了，不枉熬夜到4点。可惜没把房间里十几个D题hack掉，终测完发现自己变成房间第一。总之再接再厉。

2017-01-16 15:34:14