Link

题意：给出n个数，其中有一个数会出现两次，其余数只出现一次，问不同长度且不同的子串的数量。取模1e9+7

思路：组合求出所有情况,减去重复情况,注意用逆元即可

/** @Date    : 2017-07-06 09:56:44  * @FileName: atcoder077 D 组合.cpp  * @Platform: Windows  * @Author  : Lweleth (SoungEarlf@gmail.com)  * @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 = 1e5+20;const double eps = 1e-8;const LL mod = 1e9 + 7;LL a[N];LL inv[N];LL fa[N];LL n;void init(){	fa[0] = fa[1] = 1;    inv[1] = 1;    for(LL i = 2; i < N; i++)    {        fa[i] = fa[i-1] * i % mod;        inv[i] = (mod - mod / i) * inv[mod % i] % mod;    }    inv[0] = 1;    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 ans;	ans = ((fa[n] * inv[k] % mod) * inv[n - k]) % mod;	return ans;}int main(){	init();	while(cin >> n)	{		map<LL, int>q;		LL p = 0;		for(int i = 1; i <= n + 1; i++)		{			scanf("%lld", a + i);			if(!q[a[i]])				q[a[i]] = i;			else 				p = i;		}		for(int i = 0; i <= n; i++)		{			LL ans = 0;			ans = (ans + C(n + 1, i + 1)) % mod;			ans = (ans - C(n - p + q[a[p]], i)) % mod;			while(ans < 0)				ans += mod;			printf("%lld\n", ans);		}	}    return 0;}

Atcoder arc077 D - 11 组合