[CF1228] 简要题解

标签：define   val   stderr   time   namespace   clock   char   push   ems   

A

题意

求\(l \le x \le r\)的所有数位不同的数\(x\), 任意输出一个.

\(1 \leq l \leq r \leq 10 ^5\)

Solution

按照题意模拟即可.

#include <bits/stdc++.h>
 
using namespace std;
#define rep(i, a, b) for (int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for (int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
template <typename T> bool chkmax(T &a, const T &b) { return a < b ? a = b, true : false; }
template <typename T> bool chkmin(T &a, const T &b) { return a > b ? a = b, true : false; }
template <typename T> T getSign(const T&a) { return (a > T(0)) - (a < T(0)); }
typedef long long LL;
typedef long double LD;
const double pi = acos(-1);
void procStatus() {
    ifstream t("/proc/self/status");
    cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}
int read() {
    int x = 0, flag = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return x * flag;
}
void write(LL x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
} 

int l, r;
void Init() {
    l = read(), r = read();
}

bool check(int val) {
    static int tmp[20], cnt; cnt = 0;
    while (val) 
        tmp[++cnt] = val % 10, val /= 10;
    sort(tmp + 1, tmp + cnt + 1);
    return cnt == unique(tmp + 1, tmp + cnt + 1) - (tmp + 1);
}

void Solve() {
    rep (i, l, r) 
        if (check(i)) {
            printf("%d\n", i);
            exit(0);
        }
    puts("-1");
}

int main() {
//  freopen("bosky.in", "r", stdin);
//  freopen("bosky.out", "w", stdout);

    Init();
    Solve();

#ifdef Qrsikno
    debug("\nRunning time: %.3lf(s)\n", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
    return 0;
}

B

题意

有一个被黑白染色的\(h \times w\)的网格, 定义\(r_i\)表示从上到下第i行\([1, r_i] \cap N^+\)全部是黑色, 并且第i行\(r_i + 1\)为白色(如果该位置在网格内), 定义\(c_i\)表示从左到右第i列\([1, c_i] \cap N^+\)全部是黑色, 并且第i列\(c_i + 1\)为白色(如果该位置在网格内), 其他位置的情况不清楚.

给定\(h, w, r_i, c_i\)求满足条件的网格方案数, 答案对\(10^9 + 7\)取模的答案.

Solution

在矩阵上打标记, 没有被打标记的地方随便填, 这部分对答案的贡献是\(2 ^{cnt}\)

注意可能给出的矩阵本身不合法.

#include <bits/stdc++.h>
 
using namespace std;
#define rep(i, a, b) for (int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for (int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
template <typename T> bool chkmax(T &a, const T &b) { return a < b ? a = b, true : false; }
template <typename T> bool chkmin(T &a, const T &b) { return a > b ? a = b, true : false; }
template <typename T> T getSign(const T&a) { return (a > T(0)) - (a < T(0)); }
typedef long long LL;
typedef long double LD;
const double pi = acos(-1);
void procStatus() {
    ifstream t("/proc/self/status");
    cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}
int read() {
    int x = 0, flag = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return x * flag;
}
void write(LL x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
} 

const int Maxn = 3009, Mod = 1e9 + 7;
int h, w, r[Maxn], c[Maxn];

void Init() {
    h = read(), w = read();
    rep (i, 1, h) r[i] = read();
    rep (i, 1, w) c[i] = read();
}

int vis[Maxn][Maxn], col[Maxn][Maxn];

bool judge() {
    int flag = 1;
    rep (i, 1, h) {
        if (r[i] == 0) 
            flag &= (!col[i][1]);
        else 
        if (r[i] != w) flag &= (col[i][r[i]] && !col[i][r[i] + 1]);
    }
    if (!flag) return 0;
    rep (i, 1, w) {
        if (c[i] == 0) 
            flag &= (!col[1][i]);
        else if (c[i] != h) flag &= (col[c[i]][i] && !col[c[i] + 1][i]);
    }
    return flag;    
}

void Solve() {
    rep (i, 1, h) 
        rep (j, 1, r[i] + 1) vis[i][j] = 1;
    rep (i, 1, w)
        rep (j, 1, c[i] + 1) vis[j][i] = 1;
    rep (i, 1, h) 
        rep (j, 1, r[i]) col[i][j] = 1;
    rep (i, 1, w)
        rep (j, 1, c[i]) col[j][i] = 1;

    if (!judge()) {
        cout << 0 << endl;
        return ;
    }
    int cnt = h * w;
    rep (i, 1, h)
        rep (j, 1, w) cnt -= vis[i][j];
    int ans = 1;
    rep (i, 1, cnt) ans = (ans << 1) % Mod;
    cout << ans << endl;
}

int main() {
//  freopen("bosky.in", "r", stdin);
//  freopen("bosky.out", "w", stdout);

    Init();
    Solve();

#ifdef Qrsikno
    debug("\nRunning time: %.3lf(s)\n", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
    return 0;
}

C

题意

定义\(prime(x)\)表示x因子中的质数形成的集合(\(prime(140)=\{2,7,5\}\)), \(g(x, p)\)表示整除\(x\)的最大的质数\(p\)的次幂(\(g(45, 3) = 3^2, g(100, 3) = 3^0\)), $f(x, y) = \prod_{p \in prime(x)} g(y, p) $.

现在给定\(x, n\), 计算\(\prod_{i = 1}^{n} g(x, i) \pmod {1e9 + 7}\)

\(x \leq 10^9, n \leq 10^{18}\)

Solution

考虑每个x的质因子的贡献, 对于一个质因子\(p\), \(p\)的倍数会因为\(p\)被算一次, \(p^2\)的倍数会因为\(p^2\)被算一次.

将\(x\)质因数分解, 暴力枚举每个质数的幂计算, 注意可能最后一次计算质数的幂会爆\(LL\)

#include <bits/stdc++.h>
 
using namespace std;
#define rep(i, a, b) for (int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for (int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
template <typename T> bool chkmax(T &a, const T &b) { return a < b ? a = b, true : false; }
template <typename T> bool chkmin(T &a, const T &b) { return a > b ? a = b, true : false; }
template <typename T> T getSign(const T&a) { return (a > T(0)) - (a < T(0)); }
typedef long long LL;
typedef long double LD;
const double pi = acos(-1);
void procStatus() {
    ifstream t("/proc/self/status");
    cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}
LL read() {
    LL x = 0, flag = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return x * flag;
}
void write(LL x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
} 

const int Mod = 1e9 + 7;

LL x, n;
void Init() {
    x = read(), n = read();
}

vector <LL> divs;

int fpm(int base, LL tims) {
    tims %= (Mod - 1);
    int r = 1;
    while (tims) {
        if (tims & 1) r = 1ll * base * r % Mod;
        base = 1ll * base * base % Mod;
        tims >>= 1;
    }
    return r;
}

void Solve() {
    rep (i, 2, sqrt(x)) 
        if (x % i == 0) {
            divs.push_back(i);
            while (x % i == 0) x /= i;
        }
    if (x != 1) divs.push_back(x);
    LL ans = 1;
    rep (i, 0, divs.size() - 1) {
        LL val = divs[i], cnt = 0;
        while (val <= n) {
            cnt += n / val;
            if (n / divs[i] < val) break;
            val = val * divs[i];
        }
        ans = 1ll * ans * fpm(divs[i], cnt) % Mod;
    }
    cout << ans << endl;
}

int main() {
//  freopen("bosky.in", "r", stdin);
//  freopen("bosky.out", "w", stdout);

    Init();
    Solve();

#ifdef Qrsikno
    debug("\nRunning time: %.3lf(s)\n", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
    return 0;
}

D

题意

给你一个\(n\)个点\(m\)条边的图, 要求进行三分图染色, 要求三种颜色每种颜色的点都向另外两种颜色的所有点连边, 颜色内部没有边, 不能没有点不被染色, 必须要有三种颜色出现.

给出一种合法方案或判定无解.

\(n \leq 10^5\), \(m \leq 3e5\)

Solution

随便选取一个点, 与之相邻的必定是2/3色, 与之不相邻的必定是1色, 调整与之相邻的点的状态, 如果无法调整判定无解, 还要判定一些别的部分, 详见代码.

#include <bits/stdc++.h>
 
using namespace std;
#define rep(i, a, b) for (int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for (int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
template <typename T> bool chkmax(T &a, const T &b) { return a < b ? a = b, true : false; }
template <typename T> bool chkmin(T &a, const T &b) { return a > b ? a = b, true : false; }
template <typename T> T getSign(const T&a) { return (a > T(0)) - (a < T(0)); }
typedef long long LL;
typedef long double LD;
const double pi = acos(-1);
void procStatus() {
    ifstream t("/proc/self/status");
    cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}
LL read() {
    LL x = 0, flag = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return x * flag;
}
void write(LL x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
} 

const int Mod = 1e9 + 7;

LL x, n;
void Init() {
    x = read(), n = read();
}

vector <LL> divs;

int fpm(int base, LL tims) {
    tims %= (Mod - 1);
    int r = 1;
    while (tims) {
        if (tims & 1) r = 1ll * base * r % Mod;
        base = 1ll * base * base % Mod;
        tims >>= 1;
    }
    return r;
}

void Solve() {
    rep (i, 2, sqrt(x)) 
        if (x % i == 0) {
            divs.push_back(i);
            while (x % i == 0) x /= i;
        }
    if (x != 1) divs.push_back(x);
    LL ans = 1;
    rep (i, 0, divs.size() - 1) {
        LL val = divs[i], cnt = 0;
        while (val <= n) {
            cnt += n / val;
            if (n / divs[i] < val) break;
            val = val * divs[i];
        }
        ans = 1ll * ans * fpm(divs[i], cnt) % Mod;
    }
    cout << ans << endl;
}

int main() {
//  freopen("bosky.in", "r", stdin);
//  freopen("bosky.out", "w", stdout);

    Init();
    Solve();

#ifdef Qrsikno
    debug("\nRunning time: %.3lf(s)\n", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
    return 0;
}

E

题意

给定\(n, k\), 要求给\(n \times n\)的网格填数$ \in [1, k] \cap N^+$,要求每行每列至少出现一个1.

\(n \leq 250, k \leq 10^9\)

Bonus: \(n \leq 10^5\)

Solution

正解给的做法是Dp.

设\(Dp[i][j]\)表示我填完了前\(i\)行, 现在还有j个列没有给\(1\).

转移分两种: 1. 在j个里面选一些填1; 2. 在n - j里面选一些填入1.

然后就是:

1. \[dp[i + 1][l] \leftarrow dp[i][j] {j \choose l} (k - 1)^l k^{n - j}\];

2. \[dp[i + 1][j] \leftarrow dp[i][j] {n - j\choose x} (k - 1)^{n - x} [x \geq 1]\]

预处理组合数和幂就可以做到\(O(n^3)\)

还有一种容斥做法:

设性质\(P_{j}\)表示第\(j\)个位置满足条件(j <= n, 为行, j > n为列):
\[ \begin{aligned} Ans &= \sum_{i = 0}^{n} \sum_{j = 0}^{n} (-1)^{i + j}{n \choose i}{n \choose j}k^{n^2 - (i + j)n + ij}(k - 1)^{(i + j)n-ij} \&= \sum_{i = 0}^{n} \sum_{j = 0}^{n} (-1)^{i}(-1)^{j}{n \choose i}{n \choose j}k^{(n - i)(n - j)}(k - 1)^{j(n - i) + ni}\&= \sum_{i = 0}^{n}(-1)^{i}{n \choose i}(k - 1)^{ni} \sum_{j = 0}^{n} {n \choose j}(k^{n -i})^{n - j}{(-(k - 1)^{n - i})}^{j} \&= \sum_{i = 0}^{n}(-1)^{i}{n \choose i}(k - 1)^{ni} (k^{n-i} -(k - 1)^{n - i})^n \end{aligned} \]
去掉快速幂的复杂度就是O(n)的.

#include <bits/stdc++.h>
 
using namespace std;
#define rep(i, a, b) for (int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for (int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
template <typename T> bool chkmax(T &a, const T &b) { return a < b ? a = b, true : false; }
template <typename T> bool chkmin(T &a, const T &b) { return a > b ? a = b, true : false; }
template <typename T> T getSign(const T&a) { return (a > T(0)) - (a < T(0)); }
typedef long long LL;
typedef long double LD;
const double pi = acos(-1);
void procStatus() {
    ifstream t("/proc/self/status");
    cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}
LL read() {
    LL x = 0, flag = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return x * flag;
}
void write(LL x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
} 

const int Maxn = 300, Mod = 1e9 + 7;
int n, k, power1[Maxn], power2[Maxn];
int fac[Maxn], _inv[Maxn], invFac[Maxn];

void Init() {
    n = read(), k = read();
    power1[0] = power2[0] = 1;
    rep (i, 1, n) {
        power1[i] = power1[i - 1] * (k - 1ll) % Mod;
        power2[i] = power2[i - 1] * 1ll * k % Mod;
    }
    fac[0] = 1;
    rep (i, 1, n) fac[i] = fac[i - 1] * 1ll * i % Mod;
    _inv[1] = 1;
    rep (i, 2, n) _inv[i] = 1ll * _inv[Mod % i] * (Mod - Mod / i) % Mod;
    invFac[0] = 1;
    rep (i, 1, n) 
        invFac[i] = invFac[i - 1] * 1ll * _inv[i] % Mod;
}

inline int C(int _n, int _m) {
    if (_n < _m) return 0;
    return 1ll * fac[_n] * invFac[_m] % Mod * invFac[_n - _m] % Mod;
}

void Solve() {
    static int dp[Maxn][Maxn];

    dp[0][n] = 1;
    rep (i, 0, n)
        rep (j, 0, n) {
            if (dp[i][j] == 0) continue;
            rep (l, 0, j - 1) {
                dp[i + 1][l] += dp[i][j] * 1ll * C(j, l) % Mod * power1[l] % Mod * power2[n - j] % Mod;
                if (dp[i + 1][l] >= Mod) dp[i + 1][l] -= Mod;
            }
            if (j) 
                rep (l, 1, n - j) {
                    dp[i + 1][j] += dp[i][j] * 1ll * C(n - j, l) % Mod * power1[n - l] % Mod;
                    if (dp[i + 1][j] >= Mod) dp[i + 1][j] -= Mod;
                }
            if (j == 0) 
                rep (l, 1, n) {
                    dp[i + 1][0] += dp[i][0] * 1ll * C(n, l) % Mod * power1[n - l] % Mod;
                    if (dp[i + 1][0] >= Mod) dp[i + 1][0] -= Mod;
                }
        }
    cout << dp[n][0] << endl;
}

int main() {
    freopen("bosky.in", "r", stdin);
    freopen("bosky.out", "w", stdout);

    Init();
    Solve();

#ifdef Qrsikno
    debug("\nRunning time: %.3lf(s)\n", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
    return 0;
}

F

??????????????????

[CF1228] 简要题解

标签：define   val   stderr   time   namespace   clock   char   push   ems   

原文地址：https://www.cnblogs.com/qrsikno/p/11619812.html