SPOJ - AMR11H Array Diversity (水题排列组合或容斥)

题意：给定一个序列，让你求两种数，一个是求一个子序列，包含最大值和最小值，再就是求一个子集包含最大值和最小值。

析：求子序列，从前往记录一下最大值和最小值的位置，然后从前往后扫一遍，每个位置求一下数目就好。

求子集可以用排列组合解决，很简单，假设最大值个数是 n，最小值的数是 m，总数是 N，答案就是 (2^n-1) * (2^m-1)*2^(N-m-n)，

当然要特殊判断最大值和最小值相等的时候。

当然也可以用容斥来求，就是总数 - 不是最大值的数目 - 不是最小值的数目 + 不是最大值也不是最小值的数目，其实也差不多

代码如下：

排列组合：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e5 + 10;

const int mod = 1000000007;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

LL fast_pow(int n){

  LL a = 2, ans = 1;

  while(n){

    if(n & 1)  ans = ans * a % mod;

    n >>= 1;

    a = a * a % mod;

  }

  return ans;

}

int a[maxn];

vector<int> v1, v2;

int main(){

  int T;  cin >> T;

  while(T--){

    scanf("%d", &n);

    int mmin = mod, mmax = 0;

    for(int i = 0; i < n; ++i){

      scanf("%d", a+i);

      mmin = min(mmin, a[i]);

      mmax = max(mmax, a[i]);

    }

    v1.clear();  v2.clear();

    for(int i = 0; i < n; ++i)

      if(mmin == a[i])  v1.push_back(i);

      else if(mmax == a[i])  v2.push_back(i);

    if(v1.size() == n){

      LL ans1 = (LL)n * (n+1) / 2 % mod;

      LL ans2 = (fast_pow(n) - 1 % mod) % mod;

      printf("%lld %lld\n", ans1, ans2);

      continue;

    }

    LL ans2 = (fast_pow(v1.size())-1) * (fast_pow(v2.size())-1) % mod * fast_pow(n-v1.size()-v2.size()) % mod;

    ans2 = (ans2 + mod) % mod;

    int i = 0, j = 0, pre = 0;

    LL ans1 = 0;

    while(true){

      int t1 = min(v1[i], v2[j]);

      int t2 = max(v1[i], v2[j]);

      ans1 = (ans1 + (LL)(t1-pre+1) * (n-t2)) % mod;

      v1[i] < v2[j] ? ++i : ++j;

      if(i == v1.size() || v2.size() == j)  break;

      pre = min(t1+1, min(v1[i], v2[j]));

    }

    printf("%lld %lld\n", ans1, ans2);

  }

  return 0;

}

容斥：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e5 + 10;

const int mod = 1000000007;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

LL fast_pow(int n){

  LL a = 2, ans = 1;

  while(n){

    if(n & 1)  ans = ans * a % mod;

    n >>= 1;

    a = a * a % mod;

  }

  return ans;

}

int a[maxn];

vector<int> v1, v2;

int main(){

  int T;  cin >> T;

  while(T--){

    scanf("%d", &n);

    int mmin = mod, mmax = 0;

    for(int i = 0; i < n; ++i){

      scanf("%d", a+i);

      mmin = min(mmin, a[i]);

      mmax = max(mmax, a[i]);

    }

    v1.clear();  v2.clear();

    for(int i = 0; i < n; ++i)

      if(mmin == a[i])  v1.push_back(i);

      else if(mmax == a[i])  v2.push_back(i);

    if(v1.size() == n){

      LL ans1 = (LL)n * (n+1) / 2 % mod;

      LL ans2 = (fast_pow(n) - 1 % mod) % mod;

      printf("%lld %lld\n", ans1, ans2);

      continue;

    }

    LL ans2 = fast_pow(n);

    ans2 = (ans2 - fast_pow(n-v1.size()) - fast_pow(n-v2.size()) + fast_pow(n-v1.size()-v2.size())) % mod;

    ans2 = (ans2 % mod + mod) % mod;

    int i = 0, j = 0, pre = 0;

    LL ans1 = 0;

    while(true){

      int t1 = min(v1[i], v2[j]);

      int t2 = max(v1[i], v2[j]);

      ans1 = (ans1 + (LL)(t1-pre+1) * (n-t2)) % mod;

      v1[i] < v2[j] ? ++i : ++j;

      if(i == v1.size() || v2.size() == j)  break;

      pre = min(t1+1, min(v1[i], v2[j]));

    }

    printf("%lld %lld\n", ans1, ans2);

  }

  return 0;

}

巴特西

SPOJ - AMR11H Array Diversity (水题排列组合或容斥)

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