[題目鏈接]

https://www.lydsy.com/JudgeOnline/problem.php?id=2751

[算法]

考慮k = 0的情況 ， 根據乘法原理 :

Ans = (n * (n + 1) / 2) ^ m

那麽 ， 對於k > 0 , 只需將用一棵平衡樹維護每個位置應減小的值即可

詳見代碼

時間復雜度 : O(NlogN)

[代碼]

#include<bits/stdc++.h>
using namespace std;
typedef 
 
long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int P = 1e9 + 7;

int n , m , k;
map<int , int> mp;
map< pair<int , int> , bool> existed;

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void
 
 chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;
    x *= f;
}
inline 
 
int exp_mod(int a , int n)
{
        int b = a , res = 1;
        while (n > 0)
        {
                if (n & 1) res = 1ll * res * b % P;
                b = 1ll * b * b % P;
                n >>= 1;
        }
        return res;
}

int main()
{
        
        read(n); read(m); read(k);
        for (int i = 1; i <= k; ++i)
        {
                int x , y;
                read(x); read(y);
                if (existed[make_pair(x , y)]) continue;
                existed[make_pair(x , y)] = true;
                mp[x] += y;
        }
        int cnt = (1ll * n * (n + 1) >> 1) % P ,
            rest = m - (int)mp.size();
        int ans = exp_mod(cnt , rest);
        for (map<int , int> :: iterator it = mp.begin(); it != mp.end(); it++)
                ans = (1ll * ans * ((cnt - it -> second % P) % P + P) % P) % P;
        printf("%d\n" , ans);
        
        return 0;
    
}

[HAOI 2012] 容易題