Codeforces Round #683 (Div. 2, by Meet IT)

最新推荐文章于 2024-07-31 17:49:42 发布

原创最新推荐文章于 2024-07-31 17:49:42 发布 · 135 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#templates #powershell #smalltalk #libgdx #云安全

本文分享了Codeforces Round #683 (Div. 2, by Meet IT)中各题的解题思路。A题进行n次增加，第i次选第i个数；B题答案与奇偶性有关；C题先判断元素范围，再排序累加；D题考虑dp推出状态转移；E题将数二进制表示挂字典树跑dfs。

Codeforces Round #683 (Div. 2, by Meet IT)

A - Add Candies solved

发现只要进行n次增加，第i次增加时选择第i个数即可

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int _, n;

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}


int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    _ = read();
    while (_--) {
        n = read();
        cout << n << '\n';
        rep(i, 1, n)
            printf("%d%c", i, i == n ? '\n' : ' ');
    }

    // #ifndef ONLINE_JUDGE
    //     system("pause");
    // #endif
}

View Code

B - Numbers Box solved

观察一下可以发现最终答案与奇偶性有关

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 210, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int n, m, _, Min;
int sum, cnt;

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}


int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    _ = read();
    while (_--) {
        n = read(); m = read();
        Min = INF;
        int x;
        sum = 0; cnt = 0;
        rep(i, 1, n) {
            rep(j, 1, m) {
                x = read();
                sum += abs(x);
                Min = min(Min, abs(x));
                if (x <= 0) {
                    cnt++;
                }
            }
        }
        if (cnt & 1) {
            cout << sum - 2 * Min << '\n';
        } else {
            cout << sum << '\n';
        }
    }


    // #ifndef ONLINE_JUDGE
    //     system("pause");
    // #endif
}

View Code

C - Knapsack solved

先判断一下有没有元素在[w / 2, w]之间，如果有的话直接输出这一个数即可

如果没有，那么剩下的数值一定全是小于w / 2或者大于w的。我们不考虑大于w的数字。

将小于w/2的数字排序后，从大到小累加，当累加的值满足在区间[w / 2，w]就符合条件，若加完后值小于w/2就不符合。

证明：因为剩下的数字都小于w / 2，就不可能存在加上某个数后答案突然从一个小于w/2的值跃迁到超过w的值的情况

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
ll _, n, W, up;

struct node {
    int id;
    ll val;
}a[N];

bool cmp(node a, node b) {
    if (a.val != b.val) return a.val < b.val;
    return a.id < b.id;
}

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}


int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    _ = read();
    while (_--) {
        n = read(); W = read();
        up = (W + 1) / 2;
        rep(i, 1, n) {
            a[i].val = read();
            a[i].id = i;
        }

        int flag = 0;
        rep(i, 1, n) {
            if (a[i].val >= up && a[i].val <= W) {
                flag = i;
                break;
            }
        }
        if (flag) {
            cout << 1 << '\n';
            cout << flag << '\n';
            continue;
        }
        sort(a + 1, a + 1 + n, cmp);
        
        int l = 1;
        int r = -1;
        for (int i = 1; i <= n; ++i) {
            if (a[i].val > W) {
                r = i - 1;
                break;
            }
        }
        if (!r) {
            puts("-1");
            continue;
        }
        flag = 0;
        ll sum = 0;
        if (r == -1) r = n;
        for (int i = r; i >= 1; --i) {
            sum += a[i].val;
            if (sum >= up && sum <= W) {
                flag = i;
                break;
            }
        }
        if (!flag) puts("-1");
        else {
            cout << r - flag + 1 << '\n';
            for (int i = flag; i <= r; ++i) {
                cout << a[i].id << " ";
            }
            puts("");
        }
    }

    // #ifndef ONLINE_JUDGE
    //     system("pause");
    // #endif
}

View Code

D - Catching Cheaters unsolved

考虑dp

可以推出状态转移为

a[i] == b[j] dp[i][j] = dp[i - 1][j - 1] + 2; (lcm值加一，同时距离加二，总贡献加2)

a[i] != b[j] dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]) - 1 (lcm不变但是距离加以，总贡献减一)

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 5050, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int n, m, ans;
char a[N], b[N];
int dp[N][N];

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}


int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    n = read(); m = read();
    scanf("%s", a + 1);
    scanf("%s", b + 1);

    rep(i, 1, n)
        rep(j, 1, m) {
            if (a[i] == b[j]) {
                dp[i][j] = dp[i - 1][j - 1] + 2;
                // dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 2);
            } else {
                dp[i][j] = max(0, max(dp[i - 1][j], dp[i][j - 1]) - 1);
                // dp[i][j] = max(dp[i - 1][j] - 1, 0);
                // dp[i][j] = max(dp[i][j - 1] - 1, 0);
            }
            ans = max(ans, dp[i][j]);
        }
    
    cout << ans << '\n';

    // #ifndef ONLINE_JUDGE
    //     system("pause");
    // #endif
}

View Code

E - Xor Tree unsolved

把每个数的二进制表示挂在字典树上找规律后直接跑dfs即可

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 6e6 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int n, idx, ans;
int a[N], tr[N][2], ed[N];

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}

void add(int x)
{
    int now = 0;
    for (int i = 30; i >= 0; i--)
    {
        int t = x>>i&1;
        if (!tr[now][t]) tr[now][t] = ++idx;
        now = tr[now][t];
    }
    ed[now] = 1;
}

void dfs(int u)
{
    int ls = tr[u][0];
    int rs = tr[u][1];
    if (ls) dfs(ls);
    if (rs) dfs(rs);
            
    if (ls && rs && ed[ls] >= 2 && ed[rs] >= 2) 
    {
        int t = min(ed[ls], ed[rs]) - 1;
        ans += t;
        ed[u] += ed[ls] + ed[rs] - t;
    } 
    else
        ed[u] += ed[ls] + ed[rs];
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    
    n = read();
    rep(i, 1, n) {
        a[i] = read();
        add(a[i]);
    }

    dfs(0);

    cout << ans << '\n';

    // #ifndef ONLINE_JUDGE
    //     system("pause");
    // #endif
}

View Code