【AtCoder】AGC025題解

時間  2019-12-14
標籤 AtCoder agc025 agc 題解 简体版
原文   原文鏈接

A - Digits Sum

枚舉便可node

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar('\n')
#define enter putchar(' ')
//#define ivorysi
using namespace std;
typedef long long int64;
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){putchar('-');x = -x;}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
int N;
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(N);
    int ans = 0x7fffffff;
    for(int i = 1 ; i < N ; ++i) {
        int A = i,B =  N - i;
        int tmp = 0;
        while(A) {tmp += A % 10;A /= 10;}
        while(B) {tmp += B % 10;B /= 10;}
        ans = min(ans,tmp);
    }
    out(ans);enter;
}

B - RGB Coloring

枚舉有幾個A，能夠算出有幾個B
能夠看做給序列染兩次色（每一個塊能夠有兩種顏色），被同時染了紅和藍的就是綠色算就行了c++

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar('\n')
#define enter putchar(' ')
#define MAXN 300005
//#define ivorysi
using namespace std;
typedef long long int64;
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){putchar('-');x = -x;}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
const int MOD = 998244353;
int N,fac[MAXN],invfac[MAXN],inv[MAXN];
int64 A,B,K;
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int C(int n,int m) {
    if(m > n) return 0;
    return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(N);read(A);read(B);read(K);
    inv[1] = 1;
    for(int i = 2 ; i <= N ; ++i) {
        inv[i] = mul(inv[MOD % i],MOD - MOD / i);
    }
    fac[0] = invfac[0] = 1;
    for(int i = 1 ; i <= N ; ++i) {
        fac[i] = mul(fac[i - 1],i);
        invfac[i] = mul(invfac[i - 1],inv[i]);
    }
    int ans = 0;
    for(int i = 0 ; i <= N ; ++i) {
        if(K < i * A) break;
        if((K - i * A) % B != 0) continue;
        int s = i,t = (K - i * A) / B;
        ans = inc(ans,mul(C(N,s),C(N,t)));
    }
    out(ans);enter;
}

C - Interval Game

每次都往最左和最右拉
枚舉第一次是往左仍是往右git

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar('\n')
#define enter putchar(' ')
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
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){putchar('-');x = -x;}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
const int MOD = 998244353;
int N,ans;
pii S[MAXN];
int id[2][MAXN];
bool vis[MAXN];
bool cmp1(int a,int b) {
    return S[a].fi < S[b].fi;
}
bool cmp2(int a,int b) {
    return S[a].se < S[b].se;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(N);
    for(int i = 1 ; i <= N ; ++i) {
        read(S[i].fi);read(S[i].se);
        id[0][i] = i;id[1][i] = i;
    }
    sort(id[0] + 1,id[0] + N + 1,cmp1);
    sort(id[1] + 1,id[1] + N + 1,cmp2);
    int64 ans = 0;
    memset(vis,0,sizeof(vis));
    int p[2];p[0] = N,p[1] = 1;
    int pos = 0;int64 tmp = 0;
    if(S[id[0][N]].fi > 0) {
        vis[id[0][p[0]]] = 1;--p[0];
        tmp += abs(S[id[0][N]].fi - pos);
        pos = S[id[0][N]].fi;
        int cur = 1;
        while(1) {
            if(cur == 1) {
                while(p[cur] <= N && vis[id[cur][p[cur]]]) ++p[cur];
                if(p[cur] > N) break;
                if(S[id[cur][p[cur]]].se >= pos) break;
                tmp += abs(S[id[cur][p[cur]]].se - pos);
                pos = S[id[cur][p[cur]]].se;
                vis[id[cur][p[cur]]] = 1;
            }
            else {
                while(p[cur] >= 1 && vis[id[cur][p[cur]]]) --p[cur];
                if(p[cur] < 1) break;
                if(S[id[cur][p[cur]]].fi <= pos) break;
                tmp += abs(S[id[cur][p[cur]]].fi - pos);
                pos = S[id[cur][p[cur]]].fi;
                vis[id[cur][p[cur]]] = 1;
            }
            cur ^= 1;
        }
        tmp += abs(pos);
    }
    ans = max(ans,tmp);
    memset(vis,0,sizeof(vis));
    pos = 0;tmp = 0;p[0] = N;p[1] = 1;
    if(S[id[1][1]].se < 0) {
        vis[id[1][1]] = 1;++p[1];
        tmp += abs(S[id[1][1]].se - pos);
        pos = S[id[1][1]].se;
        int cur = 0;
        while(1) {
            if(cur == 1) {
                while(p[cur] <= N && vis[id[cur][p[cur]]]) ++p[cur];
                if(p[cur] > N) break;
                if(S[id[cur][p[cur]]].se >= pos) break;
                tmp += abs(S[id[cur][p[cur]]].se - pos);
                pos = S[id[cur][p[cur]]].se;
                vis[id[cur][p[cur]]] = 1;
            }
            else {
                while(p[cur] >= 1 && vis[id[cur][p[cur]]]) --p[cur];
                if(p[cur] < 1) break;
                if(S[id[cur][p[cur]]].fi <= pos) break;
                tmp += abs(S[id[cur][p[cur]]].fi - pos);
                pos = S[id[cur][p[cur]]].fi;
                vis[id[cur][p[cur]]] = 1;
            }
            cur ^= 1;
        }
        tmp += abs(pos);
    }
    ans = max(ans,tmp);
    out(ans);enter;
}

D - Choosing Points

咱們根據\(D_1\)和\(D_2\)連邊，會獲得兩個二分圖
爲何是二分圖呢
對於\(D\)是奇數來講，\((x,y)\)和\((x + s,y + t)\)連邊，咱們很容易發現這兩個點橫縱座標加和奇偶性不一樣
對於\(D\)是偶數來講，仍是根據奇偶性染色，咱們發現邊只在顏色相同的塊裏有，咱們把\(\sqrt{2}\)當單位長度，再一次對每一個聯通塊二分圖染色，這樣不斷消去2，最後就會獲得一個二分圖spa

那麼每一個點在兩個圖裏一共四種搭配，那麼根據鴿巢原理，最多的那個顏色必然有超過\(N^2\)個點code

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define pb push_back
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
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){putchar('-');x = -x;}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
int N,D[2];
struct node {
    int to,next;
}E[10000005];
int head[400005],sumE,dx[] = {-1,-1,1,1},dy[] = {1,-1,-1,1};
int col[400005],cr[400005],cnt[5];
int to_int(pii p) {
    return p.fi * 2 * N + p.se;
}
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void dfs(int u) {
    if(col[u] == -1) col[u] = 0;
    for(int i = head[u] ; i ; i = E[i].next) {
        int v = E[i].to;
        if(col[v] == -1) {col[v] = col[u] ^ 1;dfs(v);}
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(N);read(D[0]);read(D[1]);
    for(int t = 0 ; t <= 1 ; ++t) {
        memset(head,0,sizeof(head));sumE = 0;
        for(int x = 0 ; x < 2 * N ; ++x) {
            if(D[t] < x * x) break;
            int y = sqrt(D[t] - x * x);
            if(y * y != D[t] - x * x) continue;
            for(int i = 0 ; i < 2 * N ; ++i) {
                for(int j = 0 ; j < 2 * N ; ++j) {
                    for(int k = 0 ; k < 4 ; ++k) {
                        int tx = i + dx[k] * x,ty = j + dy[k] * y;
                        if(tx >= 0 && tx < 2 * N && ty >= 0 && ty < 2 * N) {
                            add(to_int(mp(i,j)),to_int(mp(tx,ty)));
                        }
                    }
                }
            }
        }
        memset(col,-1,sizeof(col));
        for(int i = 0 ; i < 4 * N * N ; ++i) {
            if(col[i] == -1) dfs(i);
            cr[i] = cr[i] << 1 | col[i];
        }
    }
    for(int i = 0 ; i < 4 * N * N ; ++i) ++cnt[cr[i]];
    int t = 0;
    for(int i = 1 ; i < 4 ; ++i) {
        if(cnt[i] > cnt[t]) t = i;
    }
    int a = N * N;
    for(int i = 0 ; i < 4 * N * N ; ++i) {
        if(!a) break;
        if(cr[i] == t) {
            out(i / (2 * N));space;out(i % (2 * N));enter;
            --a;
        }
    }
}

E - Walking on a Tree

答案最大是\(min(c_i,2)\)的和，\(c_i\)是每條邊被覆蓋了幾回get

構造方法是每次選一個葉子u
找到連着u的惟一一條邊v，\((u,v)\)若是這條邊覆蓋的鏈有1條或0條，那麼就刪掉u
不然將全部覆蓋着u的鏈兩兩組合\((u,a)\)和\((u,b)\)變成\((a,b)\)經過\((a,b)\)的方向來判斷\((u,a)\)和\((u,b)\)的方向it

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 4005
#define mp make_pair
#define pb push_back
//#define ivorysi
using namespace std;
typedef long long int64;
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,M;
struct node {
    int to,next;
}E[MAXN * 2];
int head[MAXN],sumE,d[MAXN],dep[MAXN],fa[MAXN],cnt[MAXN];
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void dfs(int u) {
    dep[u] = dep[fa[u]] + 1;
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa[u]) {
        fa[v] = u;
        dfs(v);
    }
    }
}
int col(int u,int v) {
    if(dep[u] < dep[v]) swap(u,v);
    while(dep[u] != dep[v]) {
    ++cnt[u];
    u = fa[u];
    }
    while(u != v) {
    ++cnt[u];++cnt[v];
    u = fa[u];v = fa[v];
    }
}
vector<pii > Process(vector<pii> pth) {
    int x = 0;
    for(int i = 1 ; i <= N ; ++i) {
    if(d[i] == 1) {
        x = i;break;
    }
    }
    d[x]--;int u = 0;
    for(int i = head[x] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(d[v]) {
        --d[v];u = v;break;
    }
    }
    if(!u) return pth;
    vector<pii > pass,npth;
    vector<int> id;
    id.clear();pass.clear();npth.clear();
    int pos[2005];
    int t,h;
    memset(pos,-1,sizeof(pos));
    for(int i = 0 ; i < pth.size() ; ++i){
    if(pth[i].fi == x && pth[i].se != x) {
        pass.pb(pth[i]);id.pb(i);
    }
    else if(pth[i].fi != x && pth[i].se == x) {
        swap(pth[i].fi,pth[i].se);
        pass.pb(pth[i]);id.pb(i);
    }
    else {
        npth.pb(pth[i]);
        pos[i] = npth.size() - 1;
    }
    }
    t = npth.size();
    if(!pass.size()) return Process(pth);
    bool flag = 0;
    for(int i = 0 ; i < pass.size() ; i += 2) {
    if(i + 1 >= pass.size()) break;
    if(pass[i].se == pass[i + 1].se) swap(pth[id[i]].fi,pth[id[i]].se);
    npth.pb(mp(pass[i].se,pass[i + 1].se));
    pos[id[i]] = npth.size() - 1;
    pos[id[i + 1]] = npth.size() - 1;
    }
    if(pass.size() & 1) {
    flag = 1;
    h = pass.size() - 1;
    npth.pb(mp(u,pass[h].se));
    pos[id[h]] = npth.size() - 1;
    }
    npth = Process(npth);
    for(int i = 0 ; i < pth.size() ; ++i) {
    if(pos[i] < t) {pth[i] = npth[pos[i]];}
    else {
        if(flag && i == id[h]) {
        if(npth[pos[i]].fi != u) swap(pth[i].fi,pth[i].se);
        }
        else {
        if(npth[pos[i]].fi != npth[pos[i]].se && npth[pos[i]].fi == pth[i].se) swap(pth[i].fi,pth[i].se);
        }
    }
    }
    return pth;
}
void Solve() {
    read(N);read(M);
    int a,b;
    for(int i = 1 ; i < N ; ++i) {
    read(a);read(b);
    add(a,b);add(b,a);
    ++d[a];++d[b];
    }
    dfs(1);
    vector<pii > v;
    for(int i = 1 ; i <= M ; ++i) {
    read(a);read(b);
    v.pb(mp(a,b));col(a,b);
    }
    v = Process(v);
    int ans = 0;
    for(int i = 1 ; i <= N ; ++i) ans += min(2,cnt[i]);
    out(ans);enter;
    for(int i = 0 ; i < v.size() ; ++i) {
    out(v[i].fi);space;out(v[i].se);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Addition and Andition

咱們能夠從高位到低位處理每一位的移動io

若是兩位是(1,1)，咱們用一個棧從棧頂到棧底下標遞增維護\((0,1)\)和\((1,0)\)的形狀
進位分紅三種
11，s和t都進位
10，s進位，t不進位
01，s不進位，t進位class

若是有11，咱們看看走的步數能不能走到一個\((1,0)\)或\((0,1)\)，再計算下一種進位方式
若是有10，或01，咱們看看下一個形狀是否是就在它的下一位，而後再計算下一種計算方式原理

若是都不成立，那麼若是進位是11，移動到最後的位置並把這一位都寫成1
若是進位是10和01，那麼放進棧裏

每次走兩步必然會少一個元素，因此複雜度是\(O(n)\)的

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 2000005
#define mp make_pair
#define pb push_back
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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);
}
struct node {
    int id,par;
}sta[MAXN],tmp[MAXN];
int N,M,K,top,cnt,ns[MAXN],nt[MAXN],as[MAXN],at[MAXN];
char s[MAXN],t[MAXN];
void Solve() {
    read(N);read(M);read(K);
    scanf("%s",s + 1);scanf("%s",t + 1);
    reverse(s + 1,s + N + 1);reverse(t + 1,t + M + 1);
    for(int i = 1 ; i <= N ; ++i) ns[i] = s[i] - '0';
    for(int i = 1 ; i <= M ; ++i) nt[i] = t[i] - '0';
    int T = max(N,M);
    for(int i = T ; i >= 1 ; --i) {
    if(ns[i] == 1 && nt[i] == 1) {
        int op = K,pos = i,x = 3;cnt = 0;
        while(top) {
        if(x == 3) {
            if(op >= sta[top].id - pos) {
            op -= sta[top].id - pos;
            pos = sta[top].id;
            tmp[++cnt] = (node){pos,x ^ sta[top].par};
            x = x & sta[top].par;
            }
            else break;
        }
        else if(x != 0){
            if(sta[top].id == pos + 1) {
            pos = sta[top].id;
            if(x ^ sta[top].par) x = 3;
            else x = x & sta[top].par;
            }
            else break;
        }
        else break;
        --top;
        }
        if(x != 0 && x != 3) {
        ++pos;
        tmp[++cnt] = (node){pos,x};
        }
        else if(x) {
        pos += op;
        as[pos] = at[pos] = 1;
        }
        for(int j = cnt ; j >= 1 ; --j) sta[++top] = tmp[j];
    }
    else {
        if(ns[i] || nt[i]) {
        sta[++top] = (node){i,ns[i] << 1 | nt[i]};
        }
    }
    }
    for(int i = 1 ; i <= top ; ++i) {
    as[sta[i].id] = (sta[i].par >> 1) & 1;
    at[sta[i].id] = sta[i].par & 1;
    }
    int c = T + K;
    while(as[c] == 0) --c;
    for(int i = c ; i >= 1 ; --i) {out(as[i]);}
    enter;
    c = T + K;
    while(at[c] == 0) --c;
    for(int i = c ; i >= 1 ; --i) {out(at[i]);}
    enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}
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