URAL 1091 Tmutarakan Exams (DP或容斥)

题意

给出一个K和S,求从小于S的数里取出一个K元组的gcd大于1的K元组的数量。

思路

可以dp做，dp的话是经典的计数问题，而且因为数据比较小写起来也比较简单,dp[i][j][k]表示枚举到i时取了j个数此时的gcd为k的个数，则dp[i+1][j+1][gcd(k,i+1)]+=dp[i][j][k]，dp[i+1][j][k]+=dp[i][j][k]。
或者容斥，因为所有合数都能用素数来表示，所以我们先枚举所有素数，然后对各个相同的或者不同的素数和进行容斥，总之就是枚举出来所有素数的组合然后奇数加偶数减就可以了。

DP代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define LL long long
#define Lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 7;
const double eps = 1e-8;
const double PI = acos(-1.0);
LL dp[55][55][55];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int K, S;
    while (scanf("%d%d", &K, &S) != EOF)
    {
        memset(dp, 0, sizeof(dp));
        for (int i = 0; i <= S; i++) dp[i][1][i] = 1;
        for (int i = 1; i < S; i++)
            for (int j = 1; j <= K; j++)
                for (int k = 1; k <= S; k++)
                {
                    dp[i+1][j][k] += dp[i][j][k];
                    dp[i+1][j+1][__gcd(k, i+1)] += dp[i][j][k];
                }
        int ans = 0;
        for (int i = 2; i <= S && ans <= 10000; i++)
            ans += dp[S][K][i];
        if (ans > 10000) ans = 10000;
        printf("%d\n", ans);
    }
    return 0;
}

容斥代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define LL long long
#define Lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 7;
const double eps = 1e-8;
const double PI = acos(-1.0);

vector<int> prime;
vector<int> cnt;  //某个素数的可取倍数的数量
int vis[55];

void init()
{
    memset(vis, 0, sizeof(vis));
    for (int i = 2; i <= 50; i++)
    {
        if (!vis[i]) prime.push_back(i);
        for (int j = i + i; j <= 50; j += i)
            vis[j] = 1;
    }
}

LL C(int n, int m)
{
    if (m > n - m) m = n - m;
    LL res = 1;
    for (int i = 0; i < m; i++)
    {
        res *= n - i;
        res /= i + 1;
    }
    return res;
}

LL ans;
int K, S;

void dfs(int cur, int cnt, int now)
{
    if (now * K > S) return ;
    if (cur >= prime.size())
    {
        if (!cnt) return ;
        if (cnt & 1) ans += C(S / now, K);
        else ans -= C(S / now, K);
        return ;
    }
    dfs(cur + 1, cnt + 1, now * prime[cur]);
    dfs(cur + 1, cnt, now);
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    init();
    while (scanf("%d%d", &K, &S) != EOF)
    {
        ans = 0;
        dfs(0, 0, 1);
        printf("%lld\n", min(ans, 10000LL));
    }
    return 0;
}