atcoder D - 11（组合数学）

题目链接：http://arc077.contest.atcoder.jp/tasks/arc077_b

 

题解：有n+1个数只有一个数字是有重复出现的，要求一共有多少不同的组合显然和这两个数的位置有关系，具体看一下代码就能理解了

就是组合数学看一下代码就好理解了，这题比较简单不多加解释。

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#define mod 1000000007
using namespace std;
const int M = 1e5 + 10;
typedef long long ll;
int a[M];
bool vis[M];
ll up[M] , down[M] , up2[M] , down2[M];
ll inv(ll a) {
    return a == 1 ? 1 : (ll)(mod - mod / a) * inv(mod % a) % mod;
}
int main() {
    int n;
    scanf("%d" , &n);
    for(int i = 1 ; i <= n + 1 ; i++) scanf("%d" , &a[i]);
    memset(vis , 0 , sizeof(vis));
    int pos1 = 0 , pos2 = 0 , gg;
    for(int i = 1 ; i <= n + 1 ; i++) {
        if(!vis[a[i]]) {
            vis[a[i]] = true;
            continue;
        }
        else {
            pos2 = i;
            gg = a[i];
            break;
        }
    }
    for(int i = 1 ; i <= n + 1 ; i++) {
        if(a[i] == gg) {
            pos1 = i;
            break;
        }
    }
    int num = n - pos2 + 1;
    int num2 = pos1 - 1;
    num += num2;
    up[0] = 1 , down[0] = 1 , up2[0] = 1 , down2[0] = 1;
    n++;
    for(int i = 1 ; i <= n / 2 ; i++) up[i] = up[i - 1] * (n - i + 1) % mod , down[i] = down[i - 1] * i % mod;
    for(int i = n / 2 + 1 ; i <= n ; i++) up[i] = up[n - i] , down[i] = down[n - i];
    for(int i = 1 ; i <= num / 2 ; i++) up2[i] = up2[i - 1] * (num - i + 1) % mod , down2[i] = down2[i - 1] * i % mod;
    for(int i = num / 2 + 1 ; i <= num ; i++) up2[i] = up2[num - i] , down2[i] = down2[num - i];
    for(int i = 1 ; i <= n ; i++) {
        ll sum = 0;
        if(i == 1) {
            printf("%lld\n" , (ll)(n - 1));
        }
        else {
            sum += up[i] * inv(down[i]) % mod;
            if(num >= i - 1 && num > 0) sum -= up2[i - 1] * inv(down2[i - 1]) % mod;
            printf("%lld\n" , (sum + mod) % mod);
        }
    }
    return 0;
}