• 【AtCoder】ARC080


    C - 4-adjacent

    我们挑出来4的倍数和不是4的倍数而是2的倍数,和奇数

    然后就是放一个奇数,放一个4,如果一个奇数之后无法放4,然后它又不是最后一个,那么就不合法

    #include <bits/stdc++.h>
    #define fi first
    #define se second
    #define pii pair<int,int>
    #define mp make_pair
    #define pb push_back
    #define space putchar(' ')
    #define enter putchar('
    ')
    #define MAXN 100005
    #define eps 1e-12
    //#define ivorysi
    using namespace std;
    typedef long long int64;
    typedef unsigned int u32;
    typedef double db;
    template<class T>
    void read(T &res) {
        res = 0;T f = 1;char c = getchar();
        while(c < '0' || c > '9') {
            if(c == '-') f = -1;
            c = getchar();
        }
        while(c >= '0' && c <= '9') {
            res = res * 10 + c - '0';
            c = getchar();
        }
        res *= f;
    }
    template<class T>
    void out(T x) {
        if(x < 0) {x = -x;putchar('-');}
        if(x >= 10) {
            out(x / 10);
        }
        putchar('0' + x % 10);
    }
    int N,a[MAXN];
    vector<int> v[5],ans;
    void Solve() {
        read(N);
        for(int i = 1 ; i <= N ; ++i) read(a[i]);
        for(int i = 1 ; i <= N ; ++i) {
            if(a[i] % 4 == 0) v[4].pb(a[i]);
            else if(a[i] % 2 == 0) v[2].pb(a[i]);
            else v[1].pb(a[i]);
        }
        while(v[1].size()) {
            ans.pb(v[1].back());
            v[1].pop_back();
            if(ans.size() == N) break;
            if(v[4].size()) {
                ans.pb(v[4].back());
                v[4].pop_back();
            }
            else {
                puts("No");return;
            }
        }
        puts("Yes");
    }
    int main() {
    #ifdef ivorysi
    	freopen("f1.in","r",stdin);
    #endif
        Solve();
    }
    

    D - Grid Coloring

    就是每行填,如果填到末尾,就下一行从末尾开始填

    如果填到开头,下一行从开始填

    #include <bits/stdc++.h>
    #define fi first
    #define se second
    #define pii pair<int,int>
    #define mp make_pair
    #define pb push_back
    #define space putchar(' ')
    #define enter putchar('
    ')
    #define MAXN 100005
    #define eps 1e-12
    //#define ivorysi
    using namespace std;
    typedef long long int64;
    typedef unsigned int u32;
    typedef double db;
    template<class T>
    void read(T &res) {
        res = 0;T f = 1;char c = getchar();
        while(c < '0' || c > '9') {
            if(c == '-') f = -1;
            c = getchar();
        }
        while(c >= '0' && c <= '9') {
            res = res * 10 + c - '0';
            c = getchar();
        }
        res *= f;
    }
    template<class T>
    void out(T x) {
        if(x < 0) {x = -x;putchar('-');}
        if(x >= 10) {
            out(x / 10);
        }
        putchar('0' + x % 10);
    }
    int H,W,N;
    int a[100005],c[105][105],g[105];
    void Solve() {
        read(H);read(W);
        read(N);
        for(int i = 1 ; i <= N ; ++i) read(a[i]);
        int f = 0;
        int t = 1;
        for(int i = 1 ; i <= N ; ++i) {
            if(g[t] + a[i] < W) {
                if(!f) {
                    for(int j = g[t] + 1 ; j <= g[t] + a[i] ; ++j) c[t][j] = i;
                }
                else {
                    for(int j = W - g[t] ; j >= W - g[t] - a[i] + 1 ; --j) c[t][j] = i;
                }
                g[t] += a[i];
            }
            else {
                for(int j = 1 ; j <= W ; ++j) {
                    if(!c[t][j]) c[t][j] = i;
                }
                a[i] -= W - g[t];
                if(c[t][W] == i) f = 1;
                else f = 0;
                ++t;
                while(a[i]) {
                    if(!f) c[t][g[t] + 1] = i;
                    else c[t][W - g[t]] = i;
                    ++g[t];--a[i];
                    if(g[t] == W) ++t;
                }
            }
        }
        for(int i = 1 ; i <= H ; ++i) {
            for(int j = 1 ; j <= W ; ++j) {
                out(c[i][j]);space;
            }
            enter;
        }
    }
    int main() {
    #ifdef ivorysi
    	freopen("f1.in","r",stdin);
    #endif
        Solve();
    }
    

    E - Young Maids

    对于一个序列,我们取一个最小的奇数位置,然后取这个最小的奇数后面的偶数位置,用st表实现,序列会被分成三份,然后递归会形成树,我们求一个最小的dfs序即可

    #include <bits/stdc++.h>
    #define fi first
    #define se second
    #define pii pair<int,int>
    #define mp make_pair
    #define pb push_back
    #define space putchar(' ')
    #define enter putchar('
    ')
    #define MAXN 200005
    #define eps 1e-12
    //#define ivorysi
    using namespace std;
    typedef long long int64;
    typedef unsigned int u32;
    typedef double db;
    template<class T>
    void read(T &res) {
        res = 0;T f = 1;char c = getchar();
        while(c < '0' || c > '9') {
            if(c == '-') f = -1;
            c = getchar();
        }
        while(c >= '0' && c <= '9') {
            res = res * 10 + c - '0';
            c = getchar();
        }
        res *= f;
    }
    template<class T>
    void out(T x) {
        if(x < 0) {x = -x;putchar('-');}
        if(x >= 10) {
            out(x / 10);
        }
        putchar('0' + x % 10);
    }
    int N,p[MAXN],tot;
    int st[2][MAXN][19],len[MAXN],pos[MAXN];
    pii t[MAXN];
    vector<int> son[MAXN];
    set<pii > S;
    int query_min(int k,int l,int r) {
        int s = len[r - l + 1];
        return min(st[k][l][s],st[k][r - (1 << s) + 1][s]);
    }
    int build_tree(int l,int r) {
        if(r < l) return 0;
        int k = (l & 1);
        int a = query_min(k,l,r);
        int b = query_min(k ^ 1,pos[a] + 1,r);
        t[++tot] = mp(a,b);
        int res = tot;
        son[res].pb(build_tree(l,pos[a] - 1));
        son[res].pb(build_tree(pos[a] + 1,pos[b] - 1));
        son[res].pb(build_tree(pos[b] + 1,r));
        return res;
    }
    void Solve() {
        read(N);
        for(int i = 1 ; i <= N ; ++i) {
            read(p[i]);
            st[i & 1][i][0] = p[i];
            st[(i & 1) ^ 1][i][0] = 0x7fffffff;
            pos[p[i]] = i;
        }
        for(int k = 0 ; k <= 1 ; ++k) {
            for(int j = 1 ; j <= 18 ; ++j) {
                for(int i = 1 ; i <= N ; ++i) {
                    if(i + (1 << j) - 1 > N) break;
                    st[k][i][j] = min(st[k][i][j - 1],st[k][i + (1 << j - 1)][j - 1]);
                }
            }
        }
        for(int i = 2 ; i <= N ; ++i) {
            len[i] = len[i / 2] + 1;
        }
        int rt = build_tree(1,N);
        S.insert(mp(t[rt].fi,rt));
        while(!S.empty()) {
            auto b = *S.begin();S.erase(S.begin());
            out(t[b.se].fi);space;out(t[b.se].se);space;
            for(auto k : son[b.se]) {
                if(k) S.insert(mp(t[k].fi,k));
            }
        }
    }
    int main() {
    #ifdef ivorysi
    	freopen("f1.in","r",stdin);
    #endif
        Solve();
    }
    

    F - Prime Flip

    把每个数改成异或差分

    然后修改一个区间相当于修改两个点,不需要加额外的点,我们需要把所有异或差分值为1的地方两两配对

    如果相差为质数,那么花费为1

    相差为偶数 花费为2

    相差为奇数且非质数,花费为3(一个偶数可以拆成两个奇素数的和,怎么地这么小的范围总得成立吧,不然哥德巴赫猜想早证伪了,然后只需要挑一个大一点的奇素数减掉这个偶数就行了)

    之后我们把这些点分为奇数点和偶数点,差值为质数则连边,做一个最大匹配,之后每个点集两两匹配,如果两个点集还剩一个,就连一条长度为3的边

    #include <bits/stdc++.h>
    #define fi first
    #define se second
    #define pii pair<int,int>
    #define mp make_pair
    #define pb push_back
    #define space putchar(' ')
    #define enter putchar('
    ')
    #define MAXN 200005
    #define eps 1e-12
    //#define ivorysi
    using namespace std;
    typedef long long int64;
    typedef unsigned int u32;
    typedef double db;
    template<class T>
    void read(T &res) {
        res = 0;T f = 1;char c = getchar();
        while(c < '0' || c > '9') {
            if(c == '-') f = -1;
            c = getchar();
        }
        while(c >= '0' && c <= '9') {
            res = res * 10 + c - '0';
            c = getchar();
        }
        res *= f;
    }
    template<class T>
    void out(T x) {
        if(x < 0) {x = -x;putchar('-');}
        if(x >= 10) {
            out(x / 10);
        }
        putchar('0' + x % 10);
    }
    int N;
    bool nonprime[10000005],vis[10000005];
    int prime[10000005],tot,x[205],b[205],M[2],matc[205];
    vector<int> to[505];
    bool used[205];
    bool match(int u) {
        for(auto t : to[u]) {
            if(!used[t]) {
                used[t] = 1;
                if(!matc[t] || match(matc[t])) {
                    matc[t] = u;
                    return true;
                }
            }
        }
        return false;
    }
    void Solve() {
        read(N);
        for(int i = 1 ; i <= N ; ++i) {
            read(x[i]);
            vis[x[i]] = 1;
        }
        for(int i = 2 ; i <= 10000000 ; ++i) {
            if(!nonprime[i]) {
                prime[++tot] = i;
            }
            for(int j = 1 ; j <= tot ; ++j) {
                if(prime[j] > 10000000 / i) break;
                nonprime[i * prime[j]] = 1;
                if(i % prime[j] == 0) break;
            }
        }
        tot = 0;
        for(int i = 1 ; i <= 10000001 ; ++i) {
            if(vis[i] != vis[i - 1]) b[++tot] = i;
        }
        for(int i = 1 ; i <= tot ; ++i) {
            M[b[i] & 1]++;
            for(int j = 1 ; j <= tot ; ++j) {
                if(i == j) continue;
                if(abs(b[i] - b[j]) < 3) continue;
                if(!nonprime[abs(b[i] - b[j])]) to[i].pb(j);
            }
        }
        int ans = 0;
        for(int i = 1 ; i <= tot ; ++i) {
            if(b[i] & 1) {
                memset(used,0,sizeof(used));
                if(match(i)) ++ans;
            }
        }
        out(ans + ((M[0] - ans) / 2 + (M[1] - ans) / 2) * 2 + ((M[0] - ans) & 1) * 3);enter;
    }
    int main() {
    #ifdef ivorysi
    	freopen("f1.in","r",stdin);
    #endif
        Solve();
    }
    
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  • 原文地址:https://www.cnblogs.com/ivorysi/p/10428387.html
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