[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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 #define Dollar1 998244353
 6 #define MAXN 70001
 7 #define int long long
 8 int lim, dig, rev[MAXN], n, sums[MAXN], A[MAXN];
 9 const int G = 3, Ginv = (Dollar1 + 1) / 3;
10 void init(int sz) { 
11     lim = 1; dig = 0;
12     while (lim <= sz << 1) lim <<= 1, ++dig;
13     for (int i = 0; i != lim; ++i)
14         rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (dig - 1));
15 }
16 inline int fastpow(int a, int b) {
17     int res = 1;
18     while (b) {
19         if (b & 1) res = res * a % Dollar1;
20         a = a * a % Dollar1;
21         b >>= 1;        
22     }
23     return res;
24 }
25 void NTT(int * A, int type) {
26     for (int i = 0; i != lim; ++i)
27         if (i < rev[i])
28             swap(A[i], A[rev[i]]);
29     for (int mid = 1; mid < lim; mid <<= 1) {
30         int Wn = fastpow(type == 1 ? G : Ginv, (Dollar1 + 1) / mid / 2);
31         for (int k = 0; k < lim; k += (mid << 1)) {
32             int W = 1;
33             for (int l = 0; l < mid; ++l, W = W * Wn % Dollar1) {
34                 int X = A[l + k], Y = A[l + k + mid] * W % Dollar1;
35                 A[l + k] = (X + Y) % Dollar1;
36                 A[l + k + mid] = (X - Y + Dollar1) % Dollar1;
37             }
38         }
39     }
40     if (type == -1) {
41         const int liminv = fastpow(lim, Dollar1 - 2);
42         for (int i = 0; i != lim; ++i)
43             A[i] = A[i] * liminv % Dollar1;
44     }
45 }
46 signed main() {
47     cin >> n;
48     init(10000);
49     for (int i = 1; i <= n; ++i) cin >> sums[i], A[sums[i]] = 1;
50     NTT(A, 1);
51     for (int i = 0; i != lim; ++i) (A[i] *= A[i]) %= Dollar1;
52     NTT(A, -1);
53     for (int i = 1; i <= n; ++i) --A[sums[i] << 1];
54     int ans = 0;
55     for (int i = 1; i <= n; ++i) ans += (bool)A[sums[i]];
56     cout << ans << endl;
57     return 0;
58 }

View Code

// 这段时间代码风格有变化