題意

題目連結

Sol

把前一半放在左邊，後一半放在右邊

meet in the middle一波

統計答案的時候開始想的是hash，然而MLE了兩個點

實際上只要排序之後雙指標掃一遍就行了

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 7, MAX = 1e7 + 10;
int K[MAXN], P[MAXN], N, M, ans;
int a1[MAX], c1, a2[MAX], c2, cnt[MAX];
int fp(int a, int p) {
    int base = 1;
    while(p) {
        if(p & 1) base = base * a;
        a = a * a; p >>= 1;
    }
    return base;
}
void dfs(int x, int Lim, int opt, int sum) {
    if(x == Lim + 1) {
        if(!opt) a1[++c1] = sum;
        else a2[++c2] = -sum;
        return ;
    }
    for(int i = 1; i <= M; i++) dfs(x + 1, Lim, opt, sum + K[x] * fp(i, P[x]));
}
int main() {
    ios::sync_with_stdio(false);
    cin >> N >> M;
    for(int i = 1; i <= N; i++) cin >> K[i] >> P[i];
    if(N <= 2) {
        a1[++c1] = 0;
        dfs(1, N, 1, 0);
    } else {
        dfs(1, N / 2, 0, 0);
        dfs(N / 2 + 1, N, 1, 0);
    }
    sort(a1 + 1, a1 + c1 + 1);
    sort(a2 + 1, a2 + c2 + 1);
    int j = 1;
    for(int i = 1; i <= c2; i++) {
        if(i != 1 && (a2[i] == a2[i - 1])) {cnt[i] = cnt[i - 1]; continue;}
        while(a1[j] <= a2[i] && j <= c1) {
            if(a1[j] == a2[i]) cnt[i]++;
            j++;
        }
    }
    /*
    for(int i = 1; i <= c1; i++)
        for(int j = 1; j <= c2; j++)    
            ans += (a1[i] == a2[j]);
    */
    for(int i = 1; i <= c2; i++) ans += cnt[i];
    cout << ans;
    return 0;
}