2017 Multi-University Training Contest - Team 1

1006(6038)

就是对a，b分别求循环节，先统计一下b中所有长度循环节的出现次数，再对a求循环节时只要满足： a的循环节长度 % b的循环节长度=0，那么这个b的循环节就可以计入答案，尼玛只要是倍数就可以阿，比赛的时候死命想以为只有长度相同或者b的长度为1才能计算贡献，简直弱智。加了一个for就对了

 

 

/** @Date    : 2017-07-25 13:22:13
  * @FileName: 1006.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([e-mail])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL __int64
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;
const LL mod = 1e9 + 7;
LL a[N];
LL b[N];
bool visa[N];
bool visb[N];
LL cnta[N];
LL cntb[N];
int main()
{
    LL n, m;
    LL acnt;
    int icase = 0;
    while(~scanf("%I64d%I64d", &n, &m))
    {
        MMF(visa);
        MMF(visb);
        MMF(cnta);
        MMF(cntb);
        for(int i = 0; i < n; i++)
            scanf("%I64d", a + i);
        for(int j = 0; j < m; j++)
            scanf("%I64d", b + j);

        for(int i = 0; i < m; i++)
        {
            if(!visb[i])
            {
                visb[i] = 1;
                LL np = b[i];
                LL c = 1;
                while(!visb[np])
                {
                    visb[np] = 1;
                    np = b[np];
                    ++c;
                }
                cntb[c]++;
            }
        }

        LL ans = 1;
        for(int i = 0; i < n; i++)
        {
            if(!visa[i])
            {
                visa[i] = 1;
                LL np = a[i];
                LL c = 1;
                while(!visa[np])
                {
                    visa[np] = 1;
                    np = a[np];
                    ++c;
                }
                cnta[c]++;
                ////
                LL tmp = 0;
                for(int j = 1; j <= m; j++)
                {
                	if(c % j == 0)
	                if(cntb[j])
	                    tmp = (tmp + (j * cntb[j]) % mod) % mod;
                }
	            //if(cntb[1] && c != 1)
	            //    tmp = (tmp + cntb[1]) % mod;	
                //cout << c << " ~"<< c*cntb[c] <<"~" << cntb[1]<< endl;
                ans = (ans * tmp + mod) % mod;
            }
        }
        while(ans < 0)
            ans+=mod;
        printf("Case #%d: %I64d\n", ++icase, (ans + mod) % mod);
    }
    return 0;
}

 

1012(6044)

给出n个区间$l_i$,$r_i$,要求每个$\min{(l_i,r_i)} = p_i$，问能够构成合法情况，且其中的$p_i$大小关系不同的方案有几种。首先我们考虑一个区间$[l_i, r_i]$,如果它的左边界$l_i<i$，那么显然意味着$p_i$左边$i-l_i$个数与右边$r_{i}-i$个数的相对大小是不确定的（因为被$p_i$截断，后续区间的左边界必定不会小于i），而且其后的区间可以不再考虑左边的这些数。那么，我们dfs区间，每次把区间分为两部分$(L, i - 1)$,$(i+1, R)$，其中，一个区间的贡献的情况为$C_{r_{i}-l_{i}}^{i - l_{i}}$，注意判断当前区间是否合法(存在)。这题题目提示还要读入优化的外挂...我不会只能网上找个模板了..

 

 

/** @Date    : 2017-07-25 16:24:31
  * @FileName: 1012 读入优化 组合.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([e-mail])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e6+20;
const double eps = 1e-8;
const LL mod = 1e9 + 7;

LL inv[N];
LL fac[N];

void init()
{
	fac[0] = fac[1] = 1;
	inv[0] = inv[1] = 1;
	for(int i = 2; i < N; i++)
	{
		fac[i] = fac[i - 1] * i % mod;
		inv[i] = (mod - mod / i) * inv[mod % i] % mod;
	}
	for(int i = 1; i < N; i++)
		(inv[i] *= inv[i - 1]) %= mod;
}

LL C(LL n, LL k)
{
	LL ans = 0;
	if(k > n)
		return 0;
	ans = fac[n] * inv[k] % mod * inv[n - k] % mod;
	return ans;
}

struct yuu
{
	LL l, r;
	yuu(){}
	yuu(LL _l, LL _r):l(_l),r(_r){}
	bool operator <(const yuu &b) const 
	{
		if(l != b.l)
			return l < b.l;
		return r < b.r;
	}
}a[N];

map<yuu, LL>q;

LL dfs(LL l, LL r)
{
	LL ans = 1;
	if(l > r)
		return 1;
	yuu tmp;
	tmp.l = l, tmp.r = r;
	LL p = q[tmp];
	if(p == 0)
		return 0;
	else if(l == r)
		return 1;
	ans = ans * C(r - l, p - l) % mod;
	LL x = dfs(l, p - 1) % mod;
	LL y = dfs(p + 1, r) % mod;
	if(!x || !y)
		return 0;
	ans = (ans * x % mod * y % mod + mod) % mod;
	return ans;
}
/////
inline char nc(){
    static char buf[1000000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int fre(LL &x){
    char ch=nc();
    if(ch == EOF)
    	return -1;
    x = 0;
    while(!(ch>='0'&&ch<='9')) ch=nc();
    while(ch>='0'&&ch<='9') x = x*10 + ch - 48, ch = nc();
    return 1;
}
/////
int main()
{
	init();
	int icase = 0;
	LL n;
	while(/*~scanf("%lld", &n)*/~fre(n))
	{
		for(int i = 0; i < n; i++)
			/*scanf("%lld", &a[i].l);*/ fre(a[i].l);
		for(int i = 0; i < n; i++)
			/*scanf("%lld", &a[i].r);*/ fre(a[i].r);
		q.clear();
		for(int i = 0; i < n; i++)
		{
			q[a[i]] = i + 1;
		}
		LL ans = dfs(1, n);
		printf("Case #%d: %lld\n", ++icase, ans);
	}
    return 0;
}