[NOIP2014普及组T1]珠心算测验 - NTT

求数组有多少个数，恰好等于集合中另外两个（不同的）数之和？

注意到数集比较小，而且涉及到下标的加法，可以很自然地想到卷积

注意减去自己加自己的贡献

真是一道NTT练手好题

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 #define Dollar1 998244353
 #define MAXN 70001
 #define int long long
 int lim, dig, rev[MAXN], n, sums[MAXN], A[MAXN];
 const int G = , Ginv = (Dollar1 + ) / ;
 void init(int sz) {
     lim = ; dig = ;
     while (lim <= sz << ) lim <<= , ++dig;
     for (int i = ; i != lim; ++i)
         rev[i] = (rev[i >> ] >> ) | ((i & ) << (dig - ));
 }
 inline int fastpow(int a, int b) {
     int res = ;
     while (b) {
         if (b & ) res = res * a % Dollar1;
         a = a * a % Dollar1;
         b >>= ;
     }
     return res;
 }
 void NTT(int * A, int type) {
     for (int i = ; i != lim; ++i)
         if (i < rev[i])
             swap(A[i], A[rev[i]]);
     for (int mid = ; mid < lim; mid <<= ) {
         int Wn = fastpow(type ==  ? G : Ginv, (Dollar1 + ) / mid / );
         for (int k = ; k < lim; k += (mid << )) {
             int W = ;
             for (int l = ; l < mid; ++l, W = W * Wn % Dollar1) {
                 int X = A[l + k], Y = A[l + k + mid] * W % Dollar1;
                 A[l + k] = (X + Y) % Dollar1;
                 A[l + k + mid] = (X - Y + Dollar1) % Dollar1;
             }
         }
     }
     if (type == -) {
         const int liminv = fastpow(lim, Dollar1 - );
         for (int i = ; i != lim; ++i)
             A[i] = A[i] * liminv % Dollar1;
     }
 }
 signed main() {
     cin >> n;
     init();
     for (int i = ; i <= n; ++i) cin >> sums[i], A[sums[i]] = ;
     NTT(A, );
     for (int i = ; i != lim; ++i) (A[i] *= A[i]) %= Dollar1;
     NTT(A, -);
     for (int i = ; i <= n; ++i) --A[sums[i] << ];
     int ans = ;
     for (int i = ; i <= n; ++i) ans += (bool)A[sums[i]];
     cout << ans << endl;
     return ;
 }

// 这段时间代码风格有变化