Codeforces Round #695 (Div. 2)

标签：bool   make   路径   计算   --   标记   const   code   std   

A

9，98,989,9890......

#include <bits/stdc++.h>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define dow(i,j,k) for (int i = j; i >= k; i--)
#define fi first
#define se second
#define mp make_pair
#define pb push_back
using namespace std;
typedef pair<int, int> pi;
typedef long long ll;

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int n; 
        scanf("%d", &n);
        if (n == 1) puts("9");
        else if (n == 2) puts("98");
        else {
            printf("98");
            n -= 2;
            int i = 9;
            while (n--) {
                putchar(‘0‘ + i);
                i = (i + 1) % 10;
            }
            puts("");
       }
    }
}

B

考虑中间那个数，它如果要变相当于只有变成左边这个数或者右边这个数这两种情况，然后求一下最多能减少几个就好。

#include <bits/stdc++.h>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define dow(i,j,k) for (int i = j; i >= k; i--)
#define fi first
#define se second
#define mp make_pair
#define pb push_back
using namespace std;
typedef pair<int, int> pi;
typedef long long ll;
const int N = 3e5 + 100;
int a[N];
int n;
inline bool pd(int x, int y, int z){
    return (x > 0 && z <= n && ((a[x] < a[y] && a[y] > a[z]) || (a[x] > a[y] && a[y] < a[z])));
}
void solve() {
    int cnt = 0, ans = 0;
    scanf("%d", &n);
    rep(i,1,n) scanf("%d", &a[i]);
    rep(i,2,n-1) {
        if (a[i] > a[i-1] && a[i] > a[i+1]) cnt++;
        else if (a[i] < a[i-1] && a[i] < a[i+1]) cnt++;
        else continue;
        int tmp1 = pd(i-2, i-1, i) + pd(i, i+1, i+2) - pd(i-1, i+1, i+2);
        int tmp2 = pd(i-2, i-1, i) - pd(i-2, i-1, i+1) + pd(i, i+1, i+2);
        ans = max(ans, 1 + max(tmp1, tmp2));
    }
    printf("%d\n", cnt - ans);
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--) solve();
}

C

两种情况，牺牲元素和最小的背包，牺牲两个背包里的最小元素。然后去最大就好了。

#include <bits/stdc++.h>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define dow(i,j,k) for (int i = j; i >= k; i--)
#define fi first
#define se second
#define mp make_pair
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<long long, long long> pi;

pi b[4];
bool cmp(pi x, pi y) {
    if (x.se == y.se) return x.fi < y.fi;
    return x.se < y.se;
}
int main() {
    int n[4];
    rep(i,1,3) scanf("%d", &n[i]);
    
    
    rep(j,1,3)
    rep(i,1,n[j]) {
        ll x;
        scanf("%lld", &x);
        b[j].fi+= x;
        if (!b[j].se) b[j].se = x;
        else b[j].se = min(b[j].se, x);
    }
    ll ans = 0;
    sort(b + 1, b + 4);
    ans = b[3].fi + b[2].fi - b[1].fi;
    sort(b + 1, b + 4, cmp);
    ans = max(ans, b[1].fi + b[2].fi + b[3].fi - 2 * b[1].se - 2 * b[2].se);
    cout << ans << endl;
}

D

计数问题，考虑每个位置会被计算多少次。

\(dp[i][j]\)表示长度为i的路径j为结尾的路有多少种。\(dp[0][i] = 1\)作为初始值。那么位置i出现在第j个的路径数是\(dp[i][j] * dp[i][k-j]\)，这里是将第j个位置到开头和到结尾分成两条长度为\(j\)和\(k-j\)的分别以\(i\)作为结尾和开头的路径。

#include <bits/stdc++.h>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define dow(i,j,k) for (int i = j; i >= k; i--)
#define fi first
#define se second
#define mp make_pair
#define pb push_back
using namespace std;
typedef pair<int, int> pi;
typedef long long ll;
const int N = 5005;
const int mod = 1e9 + 7;
ll f[N][N];
ll cnt[N];
int a[N];
int n, k, q;
int main() {
    scanf("%d%d%d", &n, &k, &q);
    rep(i,1,n) f[0][i] = 1;
    rep(i,1,k) rep(j,1,n) {
        f[i][j] = (f[i-1][j-1] + f[i-1][j+1]) % mod;
    }
    rep(i,0,k) rep(j,1,n) cnt[j] =  (cnt[j] + (f[i][j] * f[k-i][j]) % mod) % mod;
    rep(i,1,n) scanf("%d", &a[i]);
    ll ans = 0;
    rep(i,1,n) ans = (ans + cnt[i] * a[i] % mod) % mod;
    rep(i,1,q) {
        int x;
        scanf("%d", &x);
        ans = (ans - a[x] * cnt[x] % mod) % mod;
        scanf("%d", &a[x]);
        ans = (ans + a[x] * cnt[x] % mod + mod) % mod;
        printf("%lld\n", ans);
    }
}
    
    

E

考虑把\(x\)当做树根，如果在某个儿子为根的子树中有\(u\)满足\(a[x] = a[u]\)，那么满足条件的点一定在\(x\)的这个子树中。

用上面这个结论，用树上差分来打标记，就能找到所有的符合条件的点。

#include <bits/stdc++.h>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define dow(i,j,k) for (int i = j; i >= k; i--)
#define fi first
#define se second
#define mp make_pair
#define pb push_back
using namespace std;
typedef pair<int, int> pi;
typedef long long ll;
const int N = 2e5 + 100;
vector<int>G[N];
int cnt[N], num[N], a[N], b[N];
int st[N], ed[N], n, sum[N];
int tot;

void dfs(int x, int fa) {
    st[x] = ++tot;
    int tmp = num[a[x]];
    num[a[x]]++;
    
    for (auto v:G[x]) {
        if (v == fa) continue;
        int tmp1 = num[a[x]];
        dfs(v, x);
        if (num[a[x]] > tmp1) {
            sum[1]++;
            sum[st[v]]--;
            sum[ed[v]+1]++;
        }
    }
    ed[x] = tot;
    if (num[a[x]] - tmp != cnt[a[x]]) {
        sum[st[x]]++;
        sum[ed[x]+1]--;
    }
}



int main() {                                                                          
    scanf("%d", &n);
    rep(i,1,n) scanf("%d", &a[i]), b[i] = a[i];
    sort(b + 1, b + n + 1);
    rep(i,1,n) a[i] = lower_bound(b + 1, b + n + 1, a[i]) - b, cnt[a[i]]++;
    rep(i,1,n-1) {
        int u, v;
        scanf("%d%d", &u, &v);
        G[u].pb(v);
        G[v].pb(u);
    }
    dfs(1,-1);
    int ans = 0;
    rep(i,1,n) {
        sum[i]+=sum[i-1];
        if (sum[i] == 0) ans++;
    }
    printf("%d\n", ans);

}

Codeforces Round #695 (Div. 2)

标签：bool   make   路径   计算   --   标记   const   code   std   

原文地址：https://www.cnblogs.com/Fermat/p/14310184.html