【AtCoder】ARC077

C - pushpush

若是是按下標說的話

若是是偶數個

那麼是

\(N,N - 2,N - 4...1,3,5...N - 1\)

若是是奇數個

\(N,N - 2,N - 4...2,4,6...N - 1\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
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;
int a[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    if(N & 1) {
    for(int i = N ; i >= 1 ; i -= 2) {out(a[i]);space;}
    for(int i = i = 2 ; i <= N ; i += 2) {out(a[i]);space;}
    enter;
    }
    else {
    for(int i = N ; i >= 1 ; i -= 2) {out(a[i]);space;}
    for(int i = 1 ; i <= N ; i += 2) {out(a[i]);space;}
    enter;
    }

}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - 11

就至關於只有兩個位置相同，那麼即爲

\(...1 .... 1....\)

算全部位置不一樣的子序列

若是這個序列只包括其中一個1，剩下的數不在兩個1之間，那麼這個序列就被重複計算了，很簡單的計數

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
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);
}
const int MOD = 1000000007;
int N,a[MAXN],fac[MAXN],invfac[MAXN];
int pos[MAXN],p;
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int fpow(int x,int c) {
    int t = x,res = 1;
    while(c) {
    if(c & 1) res = mul(res,t);
    t = mul(t,t);
    c >>= 1;
    }
    return res;
}
int C(int n,int m) {
    if(n < m) return 0;
    return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
void Solve() {
    read(N);
    for(int i = 1 ; i <= N + 1; ++i) read(a[i]);
    for(int i = 1 ; i <= N + 1; ++i) {
        if(pos[a[i]]) p = i;
        else pos[a[i]] = i;
    }
    fac[0] = 1;
    for(int i = 1 ; i <= N + 1 ; ++i) {
        fac[i] = mul(fac[i - 1],i);
    }
    invfac[N + 1] = fpow(fac[N + 1],MOD - 2);
    for(int i = N ; i >= 0 ; --i) {
    invfac[i] = mul(invfac[i + 1],i + 1);
    }
    for(int i = 1 ; i <= N + 1 ; ++i) {
    int res = C(N + 1,i);
    res = inc(res,MOD - C(pos[a[p]] - 1 + N + 1 - p,i - 1));
    out(res);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - guruguru

顯然若是不按第二個按鈕，只用第一種狀況增長，增長的方式不能改變，把每次增長的段畫在數軸上，會發現每次增長對於一段區間至關於若是選了x當喜好按鈕，那麼費用會減小y，這個y在數軸上是個等差數列，用線段樹維護一下就好

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
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;
int a[MAXN];
int64 ans,res;
struct node {
    int L,R;
    int64 a,d;
}tr[MAXN * 4];
void build(int u,int L,int R) {
    tr[u].L = L;tr[u].R = R;
    tr[u].a = tr[u].d = 0;
    if(L == R) return;
    int mid = (L + R) >> 1;
    build(u << 1,L,mid);
    build(u << 1 | 1,mid + 1,R);
}
void addlz(int u,int64 a,int64 d) {
    tr[u].a += a;tr[u].d += d;
}
void pushdown(int u) {
    int l = tr[u].L,m = (tr[u].L + tr[u].R) >> 1,r = tr[u].R;
    addlz(u << 1,tr[u].a,tr[u].d);
    addlz(u << 1 | 1,tr[u].a + (m + 1 - l) * tr[u].d,tr[u].d);
    tr[u].a = tr[u].d = 0;
}
void Add(int u,int ql,int qr,int64 a,int64 d) {
    if(ql > qr) return;
    if(ql == tr[u].L && qr == tr[u].R) {addlz(u,a,d);return;}
    int mid = (tr[u].L + tr[u].R) >> 1;
    pushdown(u);
    if(qr <= mid) Add(u << 1,ql,qr,a,d);
    else if(ql > mid) Add(u << 1 | 1,ql,qr,a,d);
    else {Add(u << 1,ql,mid,a,d),Add(u << 1 | 1,mid + 1,qr,a + (mid + 1 - ql) * d,d);}
}
void Getans(int u) {
    if(tr[u].L == tr[u].R) {res = max(tr[u].a,res);return;}
    pushdown(u);
    Getans(u << 1);Getans(u << 1 | 1);
}
void Solve() {
    read(N);read(M);
    build(1,1,M);
    for(int i = 1 ; i <= N ; ++i) {
    read(a[i]);
    }
    for(int i = 2 ; i <= N ; ++i) {
    if(a[i] > a[i - 1]) {
        ans += a[i] - a[i - 1];
        Add(1,a[i - 1] + 1,a[i],0,1);
    }
    else {
        ans += M - a[i - 1] + a[i];
        Add(1,a[i - 1] + 1,M,0,1);
        Add(1,1,a[i],M - a[i - 1],1);
    }
    }
    Getans(1);
    out(ans - res);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - SS

先把原始的字符串變成兩個相同的字符串的形式

設爲\(SS\)，若是\(S\)最後一位的next是\(k\),那麼顯然要在後邊加\(S\)的後N-k和\(S\)的前N-k位

設\(S\)的後\(N - k\)位爲\(T\)，且\(k\)最大，那麼

\(f(SS) = STST\)

設\(g(S) = ST\)

顯然\(f(SS) = g(S) + g(S)\)

咱們實際上只要求一個足夠長的\(g(S)\)就行

\(g(S) = ST\)

當\(|T|\)是\(|S|\)的約數，能夠獲得\(g(ST) = STT\)

若是\(|T|\)不是\(|S|\)的約數，那麼\(g(ST) = STS\)

而後第一種狀況\(g^{\infty}(S) = STTTTTTT....\)

第二種狀況\(g^{i + 2}(S) = g^{i + 1}(S) + g^{i}(S)\)

而後能夠直接模擬，從第二種狀況開始遞推，若是出現第一種狀況就能夠記錄退出，第二種狀況增加速度是指數級的，遞推100個左右就行

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 200005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
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 {
    int64 c[26],len;
    node() {memset(c,0,sizeof(c));len = 0;}
    friend node operator + (const node &a,const node &b) {
        node r;
        r.len = a.len + b.len;
        for(int i = 0 ; i < 26 ; ++i) r.c[i] = a.c[i] + b.c[i];
        return r;
    }
    friend node operator - (const node &a,const node &b) {
        node r;
        r.len = a.len - b.len;
        for(int i = 0 ; i < 26 ; ++i) r.c[i] = a.c[i] - b.c[i];
        return r;
    }
    friend node operator * (const node &a,const int64 &d) {
        node r = a;
        for(int i = 0 ; i < 26 ; ++i) {
            r.c[i] = r.c[i] * d;
        }
        return r;
    }
}g[MAXN];
char s[MAXN * 2],t[MAXN * 2];
int nxt[MAXN],N;
int st,all;
node Calc(int64 R) {
    if(R > g[all].len && st) {
        node res;
        res = g[st];
        int64 t = (R - g[st].len) / g[st - 1].len;
        res = res + g[st - 1] * t;
        return res + Calc(R - g[st].len - g[st - 1].len * t);
    }
    else if(R <= g[1].len){
        node res;res.len = R;
        for(int i = 1 ; i <= R ; ++i) {
            res.c[s[i] - 'a']++;
        }
        return res;
    }
    else {
        for(int i = all ; i >= 0 ; --i) {
            if(g[i].len <= R) {
                return g[i] + Calc(R - g[i].len);
            }
        }
    }
}
void Solve() {
    scanf("%s",s + 1);
    int64 r,l;
    read(l);read(r);
    N = strlen(s + 1);
    for(int i = 2 ; i <= N ; ++i) {
        int p = nxt[i - 1];
        while(p && s[i] != s[p + 1]) p = nxt[p];
        if(s[i] == s[p + 1]) nxt[i] = p + 1;
        else nxt[i] = 0;
    }
    int p = nxt[N];
    while(p > (N - 1) / 2) p = nxt[p];
    int L = N;
    for(int i = p + 1 ; i <= N - p ; ++i) {
        s[++L] = s[i];
    }
    for(int i = 1 ; i <= L / 2; ++i) {
        g[1].c[s[i] - 'a']++;
    }
    g[1].len = L / 2;
    p = nxt[L / 2];
    g[0].len = L / 2 - p;
    for(int i = 1 ; i <= L / 2 - p ; ++i) {
        t[i] = s[i];
        g[0].c[t[i] - 'a']++;
    }
    for(int i = 2 ; i <= 100 ; ++i) {
        if(g[i - 1].len % (g[i - 1].len - g[i - 2].len) == 0) {all = i - 1;st = i - 1;break;}
        if(g[i - 1].len + g[i - 2].len <= 1e18) g[i] = g[i - 1] + g[i - 2];
        else {all = i - 1;break;}
    }
    if(!all) all = 100;
    node ans = Calc(r) - Calc(l - 1);
    for(int i = 0 ; i < 26 ; ++i) {
        out(ans.c[i]);space;
    }
    enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}