牛客小白月赛4

A.三角形

经典暴力：由于不能构成三角形边长至少乘二，只用考虑最大的log个暴力

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
typedef long long LL;
LL a[maxn], id[maxn], vis[maxn], ans[maxn];
 
bool cmp(LL i, LL j) {
    return a[i] > a[j];
}
 
int main() {
    int n, q;
    scanf("%d %d", &n, &q);
    for(int i = 1; i <= n; ++i) scanf("%lld", a + i), id[i] = i;
    sort(id + 1, id + 1 + n, cmp);
    LL M = -1;
    for(int i = 1; i <= min(35, n); ++i) {
        vis[id[i]] = 1;
        for(int j = i + 1; j <= min(35, n); ++j) {
            for(int k = j + 1; k <= min(35, n); ++k) {
                LL I = id[i], J = id[j], K = id[k];
                if(a[I] + a[J] > a[K] && a[J] + a[K] > a[I] && a[I] + a[K] > a[J])
                    M = max(M, a[I] + a[J] + a[K]);
            }
        }
    }
    for(int o = 1; o <= n; ++o) {
        if(!vis[o]) ans[o] = M;
        else {
            LL MM = -1;
            for(int i = 1; i <= min(35, n); ++i) {
                if(id[i] == o) continue;
                for(int j = i + 1; j <= min(35, n); ++j) {
                    if(id[j] == o) continue;
                    for(int k = j + 1; k <= min(35, n); ++k) {
                        if(id[k] == o) continue;
                        LL I = id[i], J = id[j], K = id[k];
                        if(a[I] + a[J] > a[K] && a[J] + a[K] > a[I] && a[I] + a[K] > a[J])
                            MM = max(MM, a[I] + a[J] + a[K]);
                    }
                }
            }
            ans[o] = MM;
        }
    }
    while(q--) {
        int x;
        scanf("%d", &x);
        printf("%lld
", ans[x]);
    }
    return 0;
}

B.博弈论

三位数不到1000个，直接暴力

#include <bits/stdc++.h>
using namespace std;
int a[1111];
set<int> S;
 
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%d", a + i);
    for(int l = 1; l <= 4; ++l) {
        for(int st = 1; st + l - 1 <= n; ++st) {
            int t = 0;
            for(int i = st; i <= st + l - 1; ++i) t = t * 10 + a[i];
            S.insert(t);
        }
    }
    for(int i = 0; ; ++i) {
        if(S.find(i) == S.end()) {
            printf("%d
", i);
            break;
        }
    }
    return 0;
}

C.病菌感染

BFS

#include <bits/stdc++.h>
using namespace std;
int n, cnt[1111][1111], vis[1111][1111];
int dx[] = {0, 1, 0, -1};
int dy[] = {-1, 0, 1, 0};
 
typedef pair<int, int> pii;
 
 
bool in(int i, int j) {
    return i >= 1 && i <= n && j >= 1 && j <= n;
}
 
int main() {
    int m;
    scanf("%d %d", &n, &m);
    int tot = m;
    queue<pii> q;
    for(int i = 1; i <= m; ++i) {
        int x, y;
        scanf("%d %d", &x, &y);
        q.push(pii(x, y));
        vis[x][y] = 1;
    }
    while(!q.empty()) {
        pii tmp = q.front(); q.pop();
        int x = tmp.first, y = tmp.second;
        for(int d = 0; d < 4; ++d) {
            int nx = x + dx[d], ny = y + dy[d];
            if(!in(nx, ny)) continue;
            if(vis[nx][ny]) continue;
            cnt[nx][ny]++;
            if(cnt[nx][ny] >= 2) {
                q.push(pii(nx, ny));
                tot++;
                vis[nx][ny] = 1;
            }
        }
    }
    puts(tot == n * n ? "YES" : "NO");
    return 0;
}

D.郊区春游

题意感人害我中奖，Floyd状压

#include <bits/stdc++.h>
using namespace std;
int G[222][222], id[22], f[1<<15][16];
const int INF = 1e9;
 
int main() {
    int n, m, R;
    scanf("%d %d %d", &n, &m, &R);
    for(int i = 1; i <= R; ++i) scanf("%d", id + i);
    for(int i = 1; i <= n; ++i)
        for(int j = i + 1; j <= n; ++j)
                G[i][j] = G[j][i] = INF;
    for(int i = 1; i <= m; ++i) {
        int A, B, C;
        scanf("%d %d %d", &A, &B, &C);
        G[A][B] = G[B][A] = C;
    }
    for(int k = 1; k <= n; ++k)
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
                G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
    for(int i = 0; i < (1 << R); ++i)
        for(int j = 1; j <= R; ++j)
            f[i][j] = INF;
    for(int i = 1; i <= R; ++i) f[1<<(i - 1)][i] = 0;
    for(int i = 0; i < (1 << R); ++i){
        for(int j = 1; j <= R; ++j) {
            if(f[i][j] == INF) continue;
            for(int k = 1; k <= R; ++k) {
                if(i & (1 << (k - 1))) continue;
                f[i ^ (1 << (k - 1))][k] = min(f[i ^ (1 << (k - 1))][k], f[i][j] + G[id[j]][id[k]]);
            }
        }
    }
    int ans = INF;
    for(int i = 1; i <= R; ++i) ans = min(ans, f[(1<<R)-1][i]);
    printf("%d
", ans);
    return 0;
}

E.浮点数输出

直接字符串

#include <bits/stdc++.h>
using namespace std;
char s[1111111];
 
int main() {
    cin >> s;
    cout << s << endl;
    return 0;
}

F.等价串

把操作2看成对B串增加111或者000，最后只要判一下模3余数

#include <bits/stdc++.h>
using namespace std;
char a[111], b[111];
 
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int n, m, x = 0, y = 0;
        scanf("%d %d %s %s", &n, &m, a + 1, b + 1);
        for(int i = 1; i <= n; ++i)
            if(a[i] == '0') y = (y + 1) % 3;
            else y = (y + 2) % 3;
        for(int i = 1; i <= m; ++i) {
            if(b[i] - '0' != x) x ^= 1, y = y * 2 % 3;
            y = (y + 2) % 3;
        }
        puts(y ? "NO" : "YES");
    }
    return 0;
}

G.黑白棋

很好写的模拟？

#include <bits/stdc++.h>
using namespace std;
int dx[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int dy[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int G[11][11];
 
int flip(int x, int y, int o) {
    int ret = 0;
    for(int d = 0; d < 8; ++d) {
        int nx = x + dx[d], ny = y + dy[d];
        if(G[nx][ny] != 1 - o) continue;
        for(int xx = nx + dx[d], yy = ny + dy[d]; ; xx += dx[d], yy += dy[d]) {
            if(G[xx][yy] == -1) break;
            if(G[xx][yy] == o) {
                for(int tx = nx, ty = ny; tx != xx || ty != yy; tx += dx[d], ty += dy[d]) {
                    G[tx][ty] ^= 1;
                    ret++;
                }
                break;
            }
        }
    }
    return ret;
}
 
int main() {
    memset(G, -1, sizeof(G));
    G[4][5] = G[5][4] = 1;
    G[4][4] = G[5][5] = 0;
    int x, y, o = 1;
    while(~scanf("%d %d", &x, &y)) {
        if(!flip(x, y, o)) flip(x, y, o ^ 1), G[x][y] = o ^ 1;
        else G[x][y] = o, o ^= 1;
    }
    int w = 0, b = 0;
    for(int i = 1; i <= 8; ++i)
        for(int j = 1; j <= 8; ++j)
            if(G[i][j] == 1) b++;
            else if(G[i][j] == 0) w++;
    printf("%d:%d
", b, w);
    return 0;
}

H.相邻的糖果

贪心从后往前吃，可以用一个栈维护没吃完的

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e6 + 10;
stack<int> st;
LL a[maxn];
 
int main() {
    int n, m, x;
    scanf("%d %d %d", &n, &m, &x);
    for(int i = 1; i <= n; ++i) scanf("%lld", a + i);
    LL sum = 0, ans = 0;
    for(int i = 1; i < m; ++i) st.push(i), sum += a[i];
    for(int i = m; i <= n; ++i) {
        sum -= a[i-m];
        sum += a[i];
        st.push(i);
        while(sum > x) {
            int id = st.top();
            LL y = min(sum - x, a[id]);
            a[id] -= y;
            sum -= y;
            ans += y;
            if(a[id] == 0) st.pop();
        }
    }
    printf("%lld
", ans);
    return 0;
}

I.合唱队形

枚举一下连接点

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int L[maxn], R[maxn];
char s[maxn];
 
int main() {
    int n, tot = 0;
    scanf("%d %s", &n, s + 1);
    for(int i = 1; i <= n; ++i) {
        if(s[i] == '0') L[i] = L[i-1] + 1, tot++;
        else L[i] = 0;
    }
    for(int i = n; i >= 1; --i) {
        if(s[i] == '0') R[i] = R[i+1] + 1;
        else R[i] = 0;
    }
    int ans = 0;
    for(int i = 1; i <= n; ++i) {
        if(L[i] == tot) ans = tot;
        else ans = max(ans, L[i] + 1);
    }
    for(int i = 2; i < n; ++i) {
        int x = L[i-1] + R[i+1];
        if(x != tot) ans = max(ans, x + 1);
    }
    printf("%d
", ans);
    return 0;
}

J.强迫症

每次把目前最大的数加到一个重复的数上

#include <bits/stdc++.h>
using namespace std;
set<int> S;
 
int main() {
    int n, x, ans = 0;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &x);
        if(S.find(x) != S.end()) ans++;
        S.insert(x);
    }
    printf("%d
", ans);
    return 0;
}