51nod1453(排列组合)

题目链接: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1453

题意: 中文题诶~

思路: 因为最后一个球总是在编号比他大的球拿完之前拿完, 所以可以先把每种编号的求都拿出一个来, 按照 1, 2, .... n 排列. 然后再把 xi 个 i 号球插入这里的 i 号球之前即可. 然后直接用排列组合计数即可.

代码:

#include <iostream>
#define ll long long
using namespace std;

const int mod = 1e9 + 7;
const int MAXN = 1e3 + 10;
ll a[MAXN], vis[MAXN];

ll get_pow(ll x, int n){
    ll ans = 1;
    while(n){
        if(n & 1) ans = ans * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return ans;
}

int main(void){
    vis[0] = 1;
    for(ll i = 1; i < MAXN; i++){
        vis[i] = vis[i - 1] * i % mod;
    }
    ll n, x;
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> x;
        a[i] = a[i - 1] + x;
    }
    ll cnt1 = 1, cnt2 = 1;
    for(int i = 1; i <= n; i++){
        for(ll j = a[i - 1] + 1; j <= a[i] - 1; j++){
            cnt1 = cnt1 * j % mod;
        }
        cnt2 = cnt2 * vis[a[i] - a[i - 1] - 1] % mod;
    }
    ll sol = cnt1 * get_pow(cnt2, mod - 2) % mod;
    cout << sol << endl;
    return 0;
}