标签：number nbsp from return next printf fine image include
Triple Sums
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
The first line of input contains a single integer N.
Each of the next N lines contain an element of s.
Print the solution for each possible sum in the following format:
sum_value : number_of_triples
Smaller sum values should be printed first.
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.
题意：
给你n个数，问你任选三个不同序号的数和为x的方案数有多少
题解：
FFT；
容斥原理要学好
#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define ls i<<1 #define rs ls | 1 #define mid ((ll+rr)>>1) #define pii pair<int,int> #define MP make_pair typedef long long LL; const long long INF = 1e18+1LL; const double pi = acos(-1.0); const int N = 5e5+10, M = 1e3+20,inf = 2e9,mod = 1e9+7; struct Complex { double r , i ; Complex () {} Complex ( double r , double i ) : r ( r ) , i ( i ) {} Complex operator + ( const Complex& t ) const { return Complex ( r + t.r , i + t.i ) ; } Complex operator - ( const Complex& t ) const { return Complex ( r - t.r , i - t.i ) ; } Complex operator * ( const Complex& t ) const { return Complex ( r * t.r - i * t.i , r * t.i + i * t.r ) ; } } ; void FFT ( Complex y[] , int n , int rev ) { for ( int i = 1 , j , t , k ; i < n ; ++ i ) { for ( j = 0 , t = i , k = n >> 1 ; k ; k >>= 1 , t >>= 1 ) j = j << 1 | t & 1 ; if ( i < j ) swap ( y[i] , y[j] ) ; } for ( int s = 2 , ds = 1 ; s <= n ; ds = s , s <<= 1 ) { Complex wn = Complex ( cos ( rev * 2 * pi / s ) , sin ( rev * 2 * pi / s ) ) , w ( 1 , 0 ) , t ; for ( int k = 0 ; k < ds ; ++ k , w = w * wn ) { for ( int i = k ; i < n ; i += s ) { y[i + ds] = y[i] - ( t = w * y[i + ds] ) ; y[i] = y[i] + t ; } } } if ( rev == -1 ) for ( int i = 0 ; i < n ; ++ i ) y[i].r /= n ; } int num[N],n,x,now[N]; Complex s[N*8],t[N*8],tt[N]; int main() { scanf("%d",&n); for(int i = 1; i <= n; ++i) { scanf("%d",&x); num[x + 20000]++; } int n1; for(int i = 1; i <= 20000*6; i<<=1,n1=i); for(int i = 0; i <= 20000*2; ++i) now[i+i] += num[i]; for(int i = 0; i <= 20000*4; ++i) s[i] = Complex(now[i],0); for(int i = 20000*4+1; i < n1; ++i) s[i] = Complex(0,0); for(int i = 0; i <= 2*20000; ++i) t[i] = Complex(num[i],0); for(int i = 2*20000+1; i < n1; ++i) t[i] = Complex(0,0); for(int i = 0; i < n1; ++i) tt[i] = t[i]; FFT(s,n1,1),FFT(t,n1,1);FFT(tt,n1,1); for(int i = 0; i < n1; ++i) t[i] = t[i]*t[i]*t[i]; for(int i = 0; i < n1; ++i) s[i] = s[i]*tt[i]; FFT(s,n1,-1),FFT(t,n1,-1); int cnt = 0; for(int i = 0; i <= 6*20000; ++i) { int x = ((int)(t[i].r + 0.5)) - 3*((int)(s[i].r+0.5)); if(i%3==0) x += 2*num[i/3]; x/=6; if(x) { printf("%d : %d\n",i - 3*20000,x); } } return 0; }
SPOJ - TSUM Triple Sums FFT+容斥
标签：number nbsp from return next printf fine image include
原文地址：http://www.cnblogs.com/zxhl/p/7102937.html
