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2021 Jiangsu Collegiate Programming Contest题解

佃成成成成 2022-05-02 阅读 120
图论算法团队训练赛

A - Spring Couplets

题目描述：

思路：

#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 1000000007
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d %d",&n,&m)
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
typedef pair <int,int> pii;

#define MAX 300000 + 50
int n, m, k, op;
string s;
int ar[MAX];
int br[MAX];
void work(){
    cin >> n;
    for(int i = 1; i <= n; ++i){
        cin >> s;
        ar[i] = s.back() - '0';
    }
    for(int i = 1; i <= n; ++i){
        cin >> s;
        br[i] = s.back() - '0';
    }
    for(int i = 1; i < n; ++i){
        if(ar[i] == 1 || ar[i] == 2){
            if(br[i] == 3 || br[i] == 4)continue;
            else {
                cout << "NO\n";
                return;
            }
        }
        else if(ar[i] == 3 || ar[i] == 4){
            if(br[i] == 1 || br[i] == 2)continue;
            else {
                cout << "NO\n";
                return;
            }
        }
    }
    if(ar[n] == 3 || ar[n] == 4){
        if(br[n] == 1 || br[n] == 2){
            cout << "YES\n";
            return;
        }
    }
    cout << "NO\n";
}


int main(){
    io;
    int tt;cin>>tt;
    for(int _t = 1; _t <= tt; ++_t){
        work();
    }
    return 0;
}

C - Magical Rearrangement

题目描述：

思路：

//Work by: Chelsea
#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 1000000007
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d %d",&n,&m)
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
typedef pair <int,int> pii;

#define MAX 300000 + 50
int n, m, k, op;
int tr[MAX];

int update(int x){
    for(int i = 0; i < 10; ++i){
        if(!tr[i] || i == x)continue;
        --tr[i];
        int p = 0, maxn = 0, sum = 0;
        for(int j = 0; j <= 9; ++j){
            sum += tr[j];
            if(maxn < tr[j]){
                maxn = tr[j];
                p = j;
            }
        }
        sum -= maxn;
        if(p == i && maxn > sum){
            ++tr[i];
            continue;
        }
        if(p != i && maxn > sum + 1){
            ++tr[i];
            continue;
        }
        return i;
    }
    return -1;
}

void work(){
    int sum = 0;
    int p = 0, maxn = 0;
    for(int i = 0; i <= 9; ++i){
        cin >> tr[i];
        if(maxn < tr[i]){
            maxn = tr[i];
            p = i;
        }
        sum += tr[i];
    }
    sum -= maxn;
    if(sum == 0 && tr[0] == 1){
        cout << 0 << endl;
        return;
    }
    if(p == 0 && sum < maxn){
        cout << -1 << endl;
        return;
    }
    if(p != 0 && sum < maxn - 1){
        cout << -1 << endl;
        return;
    }
    string ans = "";
    int num = update(0);
    while(num >= 0){
        ans += (char)(num + '0');
        num = update(num);
    }
    cout << ans << endl;
}


int main(){
    int tt;cin>>tt;
    for(int _t = 1; _t <= tt; ++_t){
        work();
    }
    return 0;
}

D - Pattern Lock

题目描述：

思路：

#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod7 1000000007
#define mod9 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
typedef pair <int,int> pii;
#define y1 y114514
#define MAX 300000 + 50
int n, m, k, x;

vector<pii>ans;

void calleft_rigth(int x1, int y1, int x2, int y2){
    ans.push_back(m_p(x1, y1));
    for(int i = 1; i <= y2 - y1; ++i){
        ans.push_back(m_p(x1, y1 + i));
        ans.push_back(m_p(x2, y1 + i - 1));
    }
    ans.push_back(m_p(x2, y2));
}

void calup_down(int x1, int y1, int x2, int y2){
    ans.push_back(m_p(x1, y1));
    for(int i = 1; i <= x2 - x1; ++i){
        ans.push_back(m_p(x1 + i, y1));
        ans.push_back(m_p(x1 + i - 1, y2));
    }
    ans.push_back(m_p(x2, y2));
}

void cal_3(int x, int y){
    ans.push_back(m_p(x, y));
    ans.push_back(m_p(x, y + 1));
    ans.push_back(m_p(x + 2, y));
    ans.push_back(m_p(x + 1, y + 2));
    ans.push_back(m_p(x + 2, y + 2));
    ans.push_back(m_p(x + 1, y + 1));
    ans.push_back(m_p(x + 2, y + 1));
    ans.push_back(m_p(x, y + 2));
    ans.push_back(m_p(x + 1, y));
}

void work(){
    cin >> n >> m;
    if(n % 2 == 0){
        for(int i = 1; i <= n; i += 2){
            calleft_rigth(i, 1, i + 1, m);
        }
    }
    else {
        if(m % 2 == 0){
            for(int i = 1; i <= m; i += 2){
                calup_down(1, i, n, i + 1);
            }
        }
        else{
            for(int i = 1; i <= n - 3; i += 2){
                calleft_rigth(i, 1, i + 1, m);
            }
            cal_3(n - 2, 1);
            for(int i = 4; i <= m; i += 2){
                calup_down(n - 2, i, n, i + 1);
            }
        }
    }
    for(auto [x, y] : ans)cout << x << ' ' << y << endl;

}


int main(){
    io;
    work();
    return 0;
}

I - Fake Walsh Transform

题目描述：

思路：

//Work by: Chelsea
#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 1000000007
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d %d",&n,&m)
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
typedef pair <int,int> pii;

#define MAX 300000 + 50
ll n, m;

void work(){
    cin >> n >> m;
    if(n == 1){
        if(m == 0)cout << 1 << endl;
        else cout << 2 << endl;
    }
    else if(m == 0){
        cout << (1ll << n) << endl;
    }
    else{
        cout << (1ll << n) - 1 << endl;
    }
}


int main(){
    int tt;cin>>tt;
    for(int _t = 1; _t <= tt; ++_t){
        work();
    }
    return 0;
}

J - Anti-merge

题目描述：

思路：

#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod7 1000000007
#define mod9 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
typedef pair <int,int> pii;

#define MAX 300000 + 50
int n, m, k, x;
int tr[505][505];
bool vis[505][505];
vector<pii>ans;
vector<pii>num0, num1;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};

bool judge(int x, int y){
    if(x > n || x < 1 || y > m || y < 1)return false;
    else if(vis[x][y])return false;
    return true;
}

void dfs(int x, int y, int c){
    vis[x][y] = 1;
    if(c == 0)num0.push_back(m_p(x, y));
    else num1.push_back(m_p(x, y));
    for(int i = 0; i < 4; ++i){
        int xx = x + dx[i];
        int yy = y + dy[i];
        if(judge(xx, yy) && tr[x][y] == tr[xx][yy]){
            dfs(xx, yy, c ^ 1);
        }
    }
}

void work(){
    cin >> n >> m;
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <= m; ++j){
            cin >> tr[i][j];
        }
    }
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <= m; ++j){
            if(!vis[i][j]){
                num0.clear();
                num1.clear();
                dfs(i, j, 0);
                if(num0.size() >= num1.size()){
                    for(auto [x, y] : num1){
                        ans.push_back(m_p(x, y));
                    }
                }
                else{
                    for(auto [x, y] : num0){
                        ans.push_back(m_p(x, y));
                    }
                }
            }
        }
    }
    if(ans.size() == 0)cout << 0 << ' ';
    else cout << 1 << ' ';
    cout << ans.size() << endl;
    for(auto [x, y] : ans)cout << x << ' ' << y << ' ' << 1 << endl;
    
}

int main(){
    io;
    work();
    return 0;
}

K - Longest Continuous 1

好像是个打表题，队友写的，这里懒得补了，贴一下他的代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<vector>
using namespace std;
int t;
long long n;
long long sum=0;
int ans;
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld",&n);
		if(n==1)
		{
			printf("0\n");
			continue;
		}
		sum=1;
		ans=1;
		for(int i=1;i<=50;++i)
		{
			sum+=(1ll<<(i-1))*i;
			if(n>=sum+1) ans=i+1;
			else break;
		}
		printf("%d\n",ans);
	}
	return 0;
}

日常赛后小作文

还是因为疫情，没法在集训室统一实行3人1机，就还是打的三人三机。

我们队伍签到挺快的，我和djk一起开A题，yj去开K题，我上来没怎么看题面，直接看样例猜题意，发现是个春联的平仄问题，问了djk一句谁是平谁是仄以后就直接写了，7分钟的时候就过了

然后我和djk就去看I题，这个题当时太急了，没考虑仔细，wa了3发，后来才改出来，我的锅，都怪我没事瞎试，在比赛最开始贡献了60发的罚时，下次比赛在前期一定会多检查检查再交，尽量1a

我和djk又开C，我发现这个题有0的存在不好写，感觉有点模拟的意思，就不想写，就去看J题

这个时候yj过了K题，我们就让他去写C

我和djk开J，读完题想了一会就想到了01染色，把数量小的记为答案，然后我火速写了就过了，这个时候大概是1小时17分

我写J的时候，djk去读D了，我过了J后我俩就开始讨论怎么D，开始他糊了个假做法给我，我直接去敲了，敲了个头的时候，yj过了C，djk就把题意和思路讲个yj，yj一眼就发现这个思路的问题，得亏发现的早，不然就白写了

我们三就开始疯狂画图讨论，讨论了有快一个小时，画了写写了画，终于讨论出来四种情况改怎么写，然后两个人都写了，都wa了，我和djk讨论觉得一致没问题，甚至怀疑到数据有问题，笑死，后来我改了一下3 * 3的矩阵构造方法，就过了，然后就发现了我们的一个小问题，由于我们俩都是用的我画出来的图写的，所以俩人都写挂了，这个时候大概是赛后的3点45，打完就下班了

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