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The 14th Jilin Provincial Collegiate Programming Contest

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The 14th Jilin Provincial Collegiate Programming Contest

目录

Problem A. Chord

Problem B. Problem Select

题解

签到题，sscanf从字符串中读入数字，排个序取前k个。

代码

点击查看代码 
#include <bits/stdc++.h>

#define rep(i, l, r) for (int i = l, iss = r; i <= iss; ++i)
#define per(i, l, r) for (int i = r, iss = l; i >= iss; --i)
#define pb emplace_back
#define endl '\n'
typedef long long ll;

using namespace std;

const int N = 1050;

int main()
{
    int tc;
    int n, m, k, t, in;
    char s[N];
    vector<int> ans;
    for (scanf("%d", &tc); tc; --tc)
    {
        scanf("%d%d", &n, &k);
        rep(i, 1, n)
        {
            scanf("%s", s);
            m = strlen(s);
            rep(i, 0, m - 1) if (s[i] == '/') in = i;
            sscanf(s + in + 1, "%d", &t);
            ans.push_back(t);
        }
        sort(ans.begin(), ans.end());
        rep(i, 0, k - 1) printf(i + 1 == k ? "%d\n" : "%d ", ans[i]);
        ans.clear();
    }

    return 0;
}
/*
2
3 2
http://acm.hit.edu.cn/problemset/1003
http://acm.hit.edu.cn/problemset/1002
http://acm.hit.edu.cn/problemset/1001
4 1
http://acm.hit.edu.cn/problemset/1001
http://acm.hit.edu.cn/problemset/2001
http://acm.hit.edu.cn/problemset/3001
http://acm.hit.edu.cn/problemset/501

*/

Problem C. String Game

题解

求一个串在另一个串中作为子串的出现次数

记 a[1:n] 为母串, b[1:m] 为子串;
设 f[i][j] = k : b[1:j]在a[1:i]中作为子串 且 a[i]=b[j] 出现的次数, 有转移方程$ f[i][j]= \Sigma_{x=1}^{i-1}{f[x][j-1]}$

代码

点击查看代码 
#include<bits/stdc++.h>

#define rep(i,l,r) for(int i=l,iss=r;i<=iss;++i)
#define per(i,l,r) for(int i=r,iss=l;i>=iss;--i)
#define pb emplace_back
#define endl '\n'
typedef long long ll;

using namespace std;

const int N=5050,M=1050,P=1e9+7;

int n,m;
char a[N],b[M];
ll f[N];

int main ()
{
    while(scanf("%s%s",a+1,b+1)!=EOF)
    {
        n=strlen(a+1),m=strlen(b+1);
        rep(i,1,n)f[i]=(a[i]==b[1]?1:0);
        rep(j,2,m)
        {
            ll prefix_sum=0;
            rep(i,1,n)prefix_sum+=f[i];
            per(i,1,n)
            {
                prefix_sum-=f[i];
                if(a[i]==b[j])f[i]=prefix_sum%P;
                else f[i]=0;
            }
        }
        ll ans=0;
        rep(i,1,n)ans+=f[i];
        printf("%lld\n",ans%P);

    }

    return 0;
}

/*
eeettt
et
eeettt
te
*/

Problem D. Trie

Problem E. Shorten the Array

题解

结论题

什么情况下会是1?: \(\exist s[i],s.t.\ s[i]\ \%\ min_x{s[x]}!=0\)
其它情况: \(ans=cnt(min_x{s[x]})+1>>1\)

代码

点击查看代码 
#include <bits/stdc++.h>

#define rep(i, l, r) for (int i = l, iss = r; i <= iss; ++i)
#define per(i, l, r) for (int i = r, iss = l; i >= iss; --i)
#define pb emplace_back
#define endl '\n'
typedef long long ll;

using namespace std;

const int N = 1e6 + 500;

int n, a[N];

int main()
{
    int tc, ele, cnt;
    bool flag;
    for (scanf("%d", &tc); tc; --tc)
    {
        scanf("%d", &n);
        rep(i, 1, n) scanf("%d", a + i);
        ele = a[1], flag = false;
        rep(i, 1, n) if (a[i] < ele) ele = a[i];
        rep(i, 1, n) if (a[i] % ele != 0)
        {
            flag = true;
            break;
        }
        if (flag)
        {
            printf("1\n");
        }
        else
        {
            cnt = 0;
            rep(i, 1, n) if (a[i] == ele) cnt++;
            printf("%d\n", cnt + 1 >> 1);
        }
    }

    return 0;
}

Problem F. Queue

题解

可以是一道较难的题目, 动态逆序对可持久化线段树可做, 出题人仁慈地放过了暴力;

考察交换两个数字对逆序对数量的贡献, 仅有 \(p_1<i<p_2, s[i] \in \left[s[p_1],s[p_2]\right]\) 中的元素会产生贡献;

代码

点击查看代码 
#include <bits/stdc++.h>

#define rep(i, l, r) for (int i = l, iss = r; i <= iss; ++i)
#define per(i, l, r) for (int i = r, iss = l; i >= iss; --i)
#define pb emplace_back
#define endl '\n'
typedef long long ll;

using namespace std;

const int N = 1e5 + 500;

int n, s[N], m;
ll ori, ans;

struct BIT
{
#define Low_Bit(x) (x & (-x))

private:
    int ran, bit[N];

public:
    BIT(int ran) : ran(ran)
    {
        memset(bit, 0, sizeof(bit));
    }
    void add(int p, int x)
    {
        for (; p <= ran; p += Low_Bit(p))
            bit[p] += x;
    }
    int sum(int p)
    {
        int rec = 0;
        for (; p > 0; p -= Low_Bit(p))
            rec += bit[p];
        return rec;
    }

#undef Low_Bit
};

void Init();
void Get_inversions();
void Process();

int main()
{
    int tc;
    for (scanf("%d", &tc); tc; --tc)
    {
        Init();
        Get_inversions();
        Process();
        printf("%lld\n", ans);
    }

    return 0;
}

void Process()
{
    scanf("%d", &m);
    int l, r,f,lv,rv, cnt;
    rep(i,1,m)
    {
        scanf("%d%d", &l, &r);
        if (s[l] == s[r])
            continue;

        cnt = 1;
        if (s[l] < s[r])f=1,lv=s[l],rv=s[r];
        else f=-1,lv=s[r],rv=s[l];

        rep(j,l+1,r-1)
        {
            if (s[j] >= lv && s[j] <= rv)cnt++;
            if (s[j] >  lv && s[j] <  rv)cnt++;
        }

        ori=ori+f*cnt;

        if(ori<ans)ans=ori;
        swap(s[l], s[r]);
    }
}

void Init()
{
    n = m = 0;
    ori = ans = 0ll;
    memset(s, 0, sizeof(s));
    return;
}

void Get_inversions()
{
    scanf("%d", &n);
    rep(i, 1, n) scanf("%d", s + i);
    rep(i, 1, n)++ s[i];

    int max_s = s[1];
    rep(i, 1, n) if (s[i] > max_s) max_s = s[i];

    BIT *bit = new BIT(max_s);
    per(i, 1, n)
    {
        ori += bit->sum(s[i] - 1);
        bit->add(s[i], 1);
    }
    ans = ori;

    delete bit;
}

Problem G. Matrix

题解

当横纵坐标的乘积为奇数时, $ a(i,j)=1 $ ;
即横纵坐标值应分别为奇数;
$ans= \left\lfloor \sqrt{n} \right\rfloor* \left\lfloor \sqrt{m} \right\rfloor $

Problem H. Curious

题解

莫比乌斯反演 unfinished

Problem I. World Tree

Problem J. Situation

Problem K. Forager

Problem L. Swimmer

Problem M. Upanishad

Problem N. Expressway

标签：Provincial,Contest,int,题解,rep,Programming,ans,Problem,define
来源： https://www.cnblogs.com/-CXF-/p/15971052.html