【AtCoder】ARC084

C - Snuke Festival

對於每一個B二分求出幾個A比它小記爲sum
而後對於每一個C就是比它小的B的sum的和

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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],B[MAXN],C[MAXN];
int64 sum[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(A[i]);
    for(int i = 1 ; i <= N ; ++i) read(B[i]);
    for(int i = 1 ; i <= N ; ++i) read(C[i]);
    sort(A + 1,A + N + 1);sort(B + 1,B + N + 1);sort(C + 1,C + N + 1);
    for(int i = 1 ; i <= N ; ++i) {
    sum[i] = lower_bound(A + 1, A + N + 1, B[i]) - A - 1;
    sum[i] += sum[i - 1];
    }
    int64 ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
    ans += sum[lower_bound(B + 1,B + N + 1,C[i]) - B - 1];
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - Small Multiple

一開始記了一個位數，結果開到了2000多位都WA
其實發現答案不會超過5 * 9
直接設\(dp[i][j]\)爲如今答案是\(i\),\(%K\)意義下是\(j\)
每次轉移日後加一位就行，就是若是加了一位\(h\)，第二維是\((10j + h) % K\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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 K;
bool dp[55][100005];
queue<int> Q;
void Solve() {
    read(K);
    for(int i = 1 ; i <= 9 ; ++i) dp[i][i % K] = 1;
    for(int i = 1 ; i <= 50 ; ++i) {
    for(int a = 0 ; a < K ; ++a) {
        if(dp[i][a]) Q.push(a);
    }
    while(!Q.empty()) {
        int v = Q.front();Q.pop();
        if(!dp[i][v * 10 % K]) {
        dp[i][v * 10 % K] = 1;
        Q.push(v * 10 % K);
        }
    }
    for(int a = 0 ; a < K ; ++a) {
        if(!dp[i][a]) continue;
        for(int j = 1 ; j <= 9 ; ++j) {
        if(i + j > 50) break;
        int h = (a * 10 + j) % K;
        dp[i + j][h] = 1;
        }
    }
    }
    for(int i = 1 ; i <= 50 ; ++i) {
    if(dp[i][0]) {
        out(i);enter;return;
    }
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - Finite Encyclopedia of Integer Sequences

若是K是偶數的話，答案是K / 2，K，K，K，K.....
若是K是奇數構造一個數列
K / 2 +1，K / 2 + 1，K / 2 + 1，K / 2 + 1.....
長度爲N
發現一個序列
\(a_i\)映射到\(K - a_i + 1\)若是\(a_i\)不是構造出的這個序列的一個前綴，那麼確定這兩個序列中，一個在這個序列前，一個在這個序列後
因此這個序列的排名是\((S + N) / 2\)
往前倒退\(N / 2\)個便可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 300005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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,K;
int a[MAXN];
void Solve() {
    read(K);read(N);
    if(K % 2 == 0) {
        out(K / 2);space;
        for(int i = 2 ; i <= N ; ++i) {
            out(K);space;
        }
        enter;
    }
    else {
        for(int i = 1 ; i <= N ; ++i) a[i] = (K + 1) / 2;
        int t = N / 2;
        int p = N;
        while(t--) {
            --a[p];
            if(a[p] != 0) {
                for(int i = p + 1 ; i <= N ; i++) a[i] = K;
                p = N;
            }
            if(a[p] == 0) --p;
        }
        for(int i = 1 ; i <= N ; ++i) {
            if(a[i] != 0) {
                out(a[i]);space;
            }
        }
        enter;
    }

}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - XorShift

就是考慮兩個多項式，它們互相消，確定會消成0，這個消成0以前的，就是他們的gcd
全部的數確定都是這個gcd的倍數
因此咱們設gcd長度爲a，x長度爲b
這個數前\(b - a +1\)隨便填，後面的位數能夠惟一肯定，因此答案就是\(x\)的前\(b - a +1\)位二進制構成的數+1
可是有個特殊的，就是若是前面剛好等於\(x\)的狀況，可能後面肯定好的數就超過了x，這個特判一下減掉就好

#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 MAXN 100005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
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 = 998244353;
vector<int> num[7],x,g,t;
int N;
char s[4005];
vector<int> gcd(vector<int> a,vector<int> b) {
    if(a.size() < b.size()) swap(a,b);
    if(b.size() == 1 && b[0] == 0) return a;
    vector<int> d(a.size());
    int t = a.size() - 1;
    for(int i = b.size() - 1 ; i >= 0 ; --i) {
        d[t] = a[t] ^ b[i];
        --t;
    }
    for(int i = t ; i >= 0 ; --i) d[i] = a[i];
    while(d.size() > 1) {
        if(d.back() == 0) d.pop_back();
        else break;
    }
    return gcd(b,d);
}
void Init() {
    read(N);
    scanf("%s",s + 1);
    int len = strlen(s + 1);
    for(int i = len ; i >= 1 ; --i) {
        x.pb(s[i] - '0');
    }
    for(int i = 1 ; i <= N ; ++i) {
        scanf("%s",s + 1);
        len = strlen(s + 1);
        for(int j = len ; j >= 1 ; --j) num[i].pb(s[j] - '0');
    }
    g = num[1];
    for(int i = 2 ; i <= N ; ++i) g = gcd(g,num[i]);
}
void Solve() {
    t = x;
    if(g.size() > x.size()) {out(1);enter;return;}
    int l = x.size() - g.size() + 1;
    int d = x.size() - 1;
    int ans = 0;
    for(int i = 0 ; i < l ; ++i) {
        ans = (1LL * ans * 2 + x[d - i]) % MOD;
    }
    ans = (ans + 1) % MOD;
    for(int i = l ; i < x.size() ; ++i) t[d - i] = 0;
    for(int i = 0 ; i < l ; ++i) {
        if(t[d - i] == 1) {
            int k = g.size() - 1;
            for(int j = 0 ; j < g.size() ; ++j) {
                t[d - i - j] = t[d - i - j] ^ g[k - j];
            }
        }
    }
    for(int i = l ; i < x.size() ; ++i) {
        if(t[d - i] != x[d - i]) {
            if(t[d - i] > x[d - i]) ans = (ans + MOD - 1) % MOD;
            break;
        }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}