暑假集训csp提高模拟3

赛时 rank 20，T1 0，T2 45，T3 20,T4 95

T1粘了两遍(因为OJ卡第一次没有显示出来)CE了，挂了100，T4没卡常挂了5

汤碗了！！！！！！！！！！！！！！！

T1 abc猜想

对于任意整数c都有

⌊ab⌋+c=⌊ab+c⌋=⌊a+bcb⌋

所以

⌊abc⌋=kc+ans

ans=⌊ab−kc2c⌋=⌊abmodc2c⌋

点此查看代码 
#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;
#ifdef LOCAL
    FILE *InFile = infile("in.in"),*OutFile = outfile("out.out");
    // FILE *ErrFile=errfile("err.err");
#else
    FILE *Infile = stdin,*OutFile = stdout;
    //FILE *ErrFile = stderr;
#endif
ll a,b,c;
inline ll power(ll a,ll b,ll mod){
    ll res = 1;
    while(b){
        if(b & 1) res = res * a % mod;
        b >>= 1;
        a = a*a%mod;
    }
    return res;
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    cin>>a>>b>>c;
    cout<<power(a,b,c*c)/c<<'\n';
}

死亡回放

image

image

T2 简单的排列最优化题

O(n)做法

将排列中的数分为两类，πi>i，πi≤i

记第一类的总数为zcnt，贡献为ztot；

记第二类的总数为fcnt，贡献为ftot。

除了放到队首的数，第一类数做出的贡献和减掉zcnt，第二类数做出的贡献和加上fcnt

当一个数从第一类数变为第二类数时，那么有πi−i=0，预处理每个数变化的时间即可。

第二类数变成第一类数时，只能是是队尾变队首，特判。

点此查看代码 
#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;
#ifdef LOCAL
    FILE *InFile = infile("in.in"),*OutFile = outfile("out.out");
    // FILE *ErrFile=errfile("err.err");
#else
    FILE *Infile = stdin,*OutFile = stdout;
    //FILE *ErrFile = stderr;
#endif
const int N = 1e7 + 10;
int n,zcnt,fcnt,a[N],Z[N];
ll ztot,ftot;
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    cin>>n;
    for(int i = 1;i <= n; ++i) cin>>a[i];
    for(int i = 1;i <= n; ++i){
        if(a[i] <= i) fcnt++,ftot += i-a[i];
        else zcnt++,ztot+=a[i]-i,Z[a[i]-i]++;
    }
    ll ans = ztot+ftot,id = 0;
    for(int i = 1;i < n; ++i){
        ztot -= zcnt;
        zcnt -= Z[i];
        ftot += fcnt;
        fcnt += Z[i];
        int x = a[n - i + 1];
        ftot -= n+1-x,--fcnt;
        if(x > 1) Z[x-1+i]++,ztot+= x-1,++zcnt;
        else ++fcnt;
        if(ans > ftot + ztot) ans = ftot+ztot,id = i;
    }
    cout<<ans<<' '<<id<<'\n';
}

T3 简单的线性做法题

小清新的分治做法

区间[l∼r]的绝对众数x(即出现次数大于len2的众数)满足对于任意的l≤k≤r,x一定为区间[l∼k]或(k∼r]的绝对众数。利用这一性质分治，O(n)从mid出发扫描能够成为众数的数。

枚举众数x，求当前区间中的满足绝对众数为x的子区间个数。

先从mid开始向左扫描左端点b，记录mid−b+1−cntx不同取值的个数。
再向右扫描右端点e，求满足e−b+1−cntx>⌊e−b+12⌋的[b∼e]个数。

时间复杂度O(nlog2⁡n)

点此查看代码 
#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;
#ifdef LOCAL
    FILE *InFile = infile("in.in"),*OutFile = outfile("out.out");
    // FILE *ErrFile=errfile("err.err");
#else
    FILE *Infile = stdin,*OutFile = stdout;
    //FILE *ErrFile = stderr;
#endif
const int N = 5e5 + 10;
int n,a[N],pos[N],num[N],cnt[N*2];
ll ans;
void solve(int l,int r){
    if(l == r) {ans++;return;}
    int mid = (l+r)>>1;
    solve(l,mid);solve(mid+1,r);
    for(int i = mid;i >= l;--i){
        if(++cnt[a[i]] > (mid - i + 1)/2){
            if(!pos[a[i]]) num[pos[a[i]] = ++num[0]] = a[i];
        }
    }
    for(int i = mid+1;i <= r;++i){
        if(++cnt[a[i]] > (i - mid)/2){
            if(!pos[a[i]]) num[pos[a[i]] = ++num[0]] = a[i];
        }
    }
    for(int i = l;i <= r; ++i) pos[a[i]] = cnt[a[i]] = 0;
    for(int i = 1;i <= num[0]; ++i){
        int sum = r-l+1,mx = sum,mn = sum;
        cnt[sum] = 1;
        for(int j = l;j < mid; ++j){
            if(a[j] == num[i]) sum++;
            else sum--;
            mx = max(mx,sum);
            mn = min(mn,sum);
            cnt[sum]++;
        }
        if(a[mid] == num[i]) sum++;
        else sum--;
        for(int j = mn;j <= mx; ++j) cnt[j] += cnt[j - 1];
        for(int j = mid+1;j <= r; ++j){
            if(a[j] == num[i]) sum++;
            else sum--;
            ans += cnt[min(mx,sum-1)];
        }
        for(int i = mn;i <= mx; ++i) cnt[i] = 0;
    }
    num[0] = 0;
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    cin>>n;int x;cin>>x;
    for(int i = 1;i <= n; ++i) cin>>a[i];
    solve(1,n);
    cout<<ans;
}

T4 简单的线段树题

n很大，分块无法过，线段树维护就可以了，其他的和分块一样，卡卡常数。

点此查看代码 
#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;
#ifdef LOCAL
    FILE *InFile = infile("in.in"),*OutFile = outfile("out.out");
    // FILE *ErrFile=errfile("err.err");
#else
    FILE *Infile = stdin,*OutFile = stdout;
    //FILE *ErrFile = stderr;
#endif
const int N = 1e6 + 10;
inline void read(ll &x){
    x = 0;
    char s = getchar();
    for(;s<'0'||s >'9';s=getchar());
    for(;'0' <= s && s<='9';s=getchar()) x = (x<<1)+(x<<3)+(s^48);
}
ll n,m;ll a[N];
void write(ll x){
    if(x > 9) write(x/10);
    putchar(x%10+48);
}
struct Segment_Tree{
    struct segment_tree{
        int l,r,lazy,tot;ll sum;
        #define l(x) tree[x].l
        #define r(x) tree[x].r
        #define sum(x) tree[x].sum
        #define lazy(x) tree[x].lazy
        #define tot(x) tree[x].tot
    }tree[N<<2];
    inline void pushup(int k){
        sum(k) = sum(k<<1) + sum(k<<1|1);
        tot(k) = min(tot(k<<1),tot(k<<1|1));
    }
    inline void pushdown(int k){
        if(lazy(k)){
            int ls = k<<1,rs = k<<1|1;
            lazy(ls) += lazy(k);
            lazy(rs) += lazy(k);
            if(tot(ls) >= 6) tot(ls) = 6;
            else{
                tot(ls) += lazy(k);
                sum(ls) = 0;
                for(int i = l(ls);i <= r(ls); ++i) sum(ls) += a[i];
            }
            if(tot(rs) >= 6) tot(rs) = 6;
            else{
                tot(rs) += lazy(k);
                sum(rs) = 0;
                for(int i = l(rs);i <= r(rs); ++i) sum(rs) += a[i];
            }
            lazy(k) = 0;
        }
    }
    void build(int k,int l,int r){
        l(k) = l,r(k) = r;
        if(l == r)return sum(k) = a[l],void();
        int mid = (l+r)>>1;
        build(k<<1,l,mid);build(k<<1|1,mid+1,r);
        pushup(k);
    }
    void update(int k,int l,int r){
        if(tot(k) == 6) return;
        if(l <= l(k) && r(k) <= r){
            if(tot(k) == 6) return;
            tot(k)++;lazy(k)++;
            sum(k) = 0;
            for(int i = l(k);i <= r(k); ++i) sum(k) += (a[i] = sqrt(a[i]));
            return;
        }
        pushdown(k);
        int mid = (l(k) + r(k))>>1;
        if(l <= mid) update(k<<1,l,r);
        if(r > mid) update(k<<1|1,l,r);
        pushup(k);
    }
    ll query(int k,int l,int r){
        if(l <= l(k) && r(k) <= r) return sum(k);
        pushdown(k);
        int mid = (l(k) + r(k))>>1;
        ll res = 0;
        if(l <= mid) res += query(k<<1,l,r);
        if(r > mid) res += query(k<<1|1,l,r);
        return res;
    }
}T;
signed main(){
    read(n);
    for(int i = 1;i <= n; ++i) read(a[i]);
    T.build(1,1,n);
    read(m);
    int ttt = 0;
    while(m--){
        ll op,l,r;
        read(op),read(l),read(r);
        if(op == 0) T.update(1,l,r);
        else write(T.query(1,l,r)),puts("");
        // if(op) ttt++;
        // if(ttt == 141 && op) cerr<<l<<' '<<r<<'\n';
    }
}
posted @   CuFeO4  阅读(13)  评论(0)  编辑  收藏  举报
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