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SPOJ 8372. Triple Sums
8372. Triple SumsProblem code: TSUM |
You‘re given a sequence s of N distinct integers.
Consider all the possible sums of three integers from the sequence at three different indicies.
For each obtainable sum output the number of different triples of indicies that generate it.
Constraints:
N <= 40000, |si| <= 20000
Input
The first line of input contains a single integer N.
Each of the next N lines contain an element of s.
Output
Print the solution for each possible sum in the following format:
sum_value : number_of_triples
Smaller sum values should be printed first.
Example
Input:
5
-1
2
3
0
5
Output:
1 : 1
2 : 1
4 : 2
5 : 1
6 : 1
7 : 2
8 : 1
10 : 1
Explanation:
4 can be obtained using triples ( 0, 1, 2 ) and ( 0, 3, 4 ).
7 can be obtained using triples ( 0, 2, 4 ) and ( 1, 3, 4 ).
Note: a triple is considered the same as any of its permutations.
用一个系数为该位指数在S中出现次数的多项式\(P\)来表示某位上指数的一次表示方式，那么三次表示方式就是\(P^3\)，然后由于要求三元组的元素在S中的序数不同，容斥原理判重即可。
#include <bits/stdc++.h>using namespace std;typedef complex<long double> Complex;const int maxn = 1 << 17;const int base = 20000;void DFT(Complex *a, int N, int flag) { for(int i = (N >> 1), j = 1, k; j < N; i ^= k, ++j) { if(i < j) swap(a[i], a[j]); for(k = (N >> 1); i & k; i ^= k, k >>= 1); } for(int n = 2; n <= N; n <<= 1) { Complex Wn = Complex(cos(flag * M_PI * 2 / n), sin(flag * M_PI * 2 / n)); for(int i = 0; i < N; i += n) { Complex W = Complex(1.0, 0.0); for(int j = i; j < i + (n >> 1); W = W * Wn, ++j) { Complex u = a[j], v = W * a[j + (n >> 1)]; a[j] = u + v; a[j + (n >> 1)] = u - v; } } }}int a[maxn], b[maxn], c[maxn];Complex A[maxn], B[maxn], C[maxn];int main() { int n; scanf("%d", &n); for(int i = 1, x; i <= n; ++i) { scanf("%d", &x); x += base; ++a[x]; ++b[x + x]; ++c[x + x + x]; } int N = maxn; for(int i = 0; i < N; ++i) { A[i] = a[i], B[i] = b[i]; } DFT(A, N, 1); DFT(B, N, 1); for(int i = 0; i < N; ++i) { C[i] = A[i] * (A[i] * A[i] - 3.0l * B[i]); } DFT(C, N, -1); for(int i = 0; i < N; ++i) { long long ans = ((long long)(C[i].real() / N + 0.5) + 2 * c[i]) / 6; if(ans != 0) { printf("%d : %lld\n", i - 60000, ans); } } return 0;}
SPOJ 8372. Triple Sums