CodeForces Round 521 div3

A:Frog Jumping

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+;
const int N = 1e5 + ;
int main(){
    int T, a, b, c;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d%d", &a, &b, &c);
        LL t = a - b;
        t = t * (c/);
        if(c&) t += a;
        printf("%I64d\n", t);
    }
    return ;
}

B：Disturbed People

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+;
const int N = 1e5 + ;
int a[N];
int main(){
    int n;
    scanf("%d", &n);
    a[] = ; a[n+] = ;
    for(int i = ; i <= n; ++i)
        scanf("%d", &a[i]);
    int ans = ;
    for(int i = ; i <= n; ++i){
        if(a[i-] ==  && a[i+] ==  && a[i] == ){
            a[i+] = ;
            ans++;
        }
    }
    printf("%d\n", ans);
    return ;
}

C：Good Array

题解：模拟， 判断的时候注意 如果是用数组可能会下标越界。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+;
const int N = 1e6 + ;
int a[N];
int vis[N];
vector<int> vc;
int main(){
    int n;
    scanf("%d", &n);
    LL sum = ;
    for(int i = ; i <= n; ++i){
        scanf("%d", &a[i]);
        ++vis[a[i]];
        sum += a[i];
    }
    for(int i = ; i <= n; ++i){
        sum -= a[i];
        --vis[a[i]];
        if(sum% ==   && sum/ < N){
            int t = sum/;
            //cout << i <<"  "<< t << endl;
            if(vis[t]) vc.pb(i);
        }
        sum += a[i];
        ++vis[a[i]];
    }
    printf("%d\n", vc.size());
    for(auto i : vc){
        printf("%d ", i);
    }
    return ;
}

D：Cutting Out

题解：二分次数，然后输出。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+;
const int N = 1e6 + ;
int a[N];
int b[N];
int n, k, m = ;
bool check(int x){
    int ret = ;
    for(int i = ; i <= m; ++i){
         ret += a[b[i]]/x;
    }
    return ret >= k;
}
int main(){
    scanf("%d%d", &n, &k);
    for(int i = , t; i <= n; ++i){
        scanf("%d", &t);
        ++a[t];
    }
    for(int i = ; i < N; ++i){
        if(a[i]){
            b[++m] = i;
        }
    }
    int l = , r = n;
    while(l <= r){
        int mid = l+r >> ;
        if(check(mid)) l = mid+;
        else r = mid-;
    }
    l--;
    for(int i = , c = ; c <= k; ){
        if(a[b[i]] >= l){
            a[b[i]] -= l;
            c++;
            printf("%d ", b[i]);
        }
        else i++;
    }
    return ;
}

E：Thematic Contests

题解：将每种类型的话题存在一起， 然后sort一下，把小的排前面，然后跑一下背包就好了。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+;
const int N = 2e5 + ;
int a[N];
int b[N];
int dp[N];
int main(){
    int n;
    scanf("%d", &n);
    for(int i = ; i <= n; ++i) scanf("%d", &a[i]);
    sort(a+, a++n);
    int m = ;
    for(int i = ; i <= n; ++i){
        if(a[i] == a[i-]) b[m]++;
        else b[++m] = ;
    }
    sort(b+, b++m);
    memset(dp, -inf, sizeof(dp));
    dp[] = ;
    for(int i = ; i <= m; ++i){
        for(int j = b[i]; j > ; --j){
            dp[j] = max(dp[j], j);
            if(j% == ) dp[j] = max(dp[j], dp[j/] + j);
        }
    }
    int ans = ;
    for(int i = ; i < N; ++i)
        ans = max(ans, dp[i]);
    printf("%d\n", ans);
    return ;
}

F：Pictures with Kittens

题解：dp[i][u] 表示 处理到i 之后选了u个点， 他的最大价值是多少。

画画图之后就发现发现，  dp[i][u]  可以从 dp[i - x][u-1] 到 dp[i-1][u-1] 转移过来。

所以对于 dp[i][u] 来说我们需要找到 dp[i-x][u-1] 到 dp[i-1][u-1] 里面的最大值。

我一开始是想用线段树搞，写好了之后MLE了......

后来也发现线段树有太多浪费的点了，然后用set， 可能操作太多了， 然后TLE了。。。。。

然后把set改成优先队列就过了，但是跑的太慢了。。。。

最后改成了单调栈， 这个东西不带log 就跑到200ms内了，我一开始是想用这个东西，然后忘了怎么写，就搞了这么多奇奇怪怪的东西。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<LL,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+;
const int N = ;
int a[N];
int n, k, x;
deque<pll> dq[N];
int main(){
    scanf("%d%d%d", &n, &k, &x);
    for(int i = ; i <= n; ++i)
        scanf("%d", &a[i]);
    dq[].push_back({,});
    LL ans = -;
    for(int i = ; i <= n; ++i){
        for(int j = x-; j >= ; --j){
            while(!dq[j].empty() && dq[j].front().se < i-k) dq[j].pop_front();
            if(dq[j].empty()) continue;
            LL val = dq[j].front().fi;
            val += a[i];
            while(!dq[j+].empty() && dq[j+].back().fi <= val) dq[j+].pop_back();
            dq[j+].push_back({val,i});
            if(j+ == x && i+k > n) ans = max(ans, val);
        }
    }
    printf("%lld\n", ans);
    return ;
}