D Pairs(FFT)

D Pairs Description Given N integers, count the number of pairs of integers whose sum is less than K. And we have M queries.

Input The first behavior is an integer T (1 < = T < = 10), on behalf of the number of sets of data. Each group of data is the first line N, M (1 < = n, m < = 100000), The next line will give N integers. a[i] (0<=a[i]<=100000). The next M lines, each line contains the query K (1 <= K<=200000).

Output For each query, output the number of pairs of integers whose sum is less than K.

Sample Input：
1
5 2
1 5 3 4 2
5
7

Sample Output:
2
6
题意：给你一个长度为n的数组，m次提问，问两个数和小于k的集合的个数。
题解：
比赛时感觉挺简单。然后就蒙了，不就是维护一个和的有序表，判断一下就好了啊，就是超时。。。。。看了标称才知道，得用FFT来加速这个维护的过程。get新东西。
代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const double PI = acos(-1.0);
struct complex
{
    double r,i;
    complex(double _r = 0,double _i = 0)
    {
        r = _r; i = _i;
    }
    complex operator +(const complex &b)
    {
        return complex(r+b.r,i+b.i);
    }
    complex operator -(const complex &b)
    {
        return complex(r-b.r,i-b.i);
    }
    complex operator *(const complex &b)
    {
        return complex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
};
void change(complex y[],int len)
{
    int i,j,k;
    for(i = 1, j = len/2;i < len-1;i++)
    {
        if(i < j)swap(y[i],y[j]);
        k = len/2;
        while( j >= k)
        {
            j -= k;
            k /= 2;
        }
        if(j < k)j += k;
    }
}
void fft(complex y[],int len,int on)
{
    change(y,len);
    for(int h = 2;h <= len;h <<= 1)
    {
        complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
        for(int j = 0;j < len;j += h)
        {
            complex w(1,0);
            for(int k = j;k < j+h/2;k++)
            {
                complex u = y[k];
                complex t = w*y[k+h/2];
                y[k] = u+t;
                y[k+h/2] = u-t;
                w = w*wn;
            }
        }
    }
    if(on == -1)
        for(int i = 0;i < len;i++)
            y[i].r /= len;
}

const int MAXN = 400040;
complex x1[MAXN];
int a[MAXN/4];
long long num[MAXN];//100000*100000会超int
long long sum[MAXN];

int main()
{
    int T;
    int n;
    int m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n, &m);
        memset(num,0,sizeof(num));
        for(int i = 0;i < n;i++)
        {
            scanf("%d",&a[i]);
            num[a[i]]++;
        }
        sort(a,a+n);
        int len1 = a[n-1]+1;
        int len = 1;
        while( len < 2*len1 )len <<= 1;
        for(int i = 0;i < len1;i++)
            x1[i] = complex(num[i],0);
        for(int i = len1;i < len;i++)
            x1[i] = complex(0,0);
        fft(x1,len,1);
        for(int i = 0;i < len;i++)
            x1[i] = x1[i]*x1[i];
        fft(x1,len,-1);
        for(int i = 0;i < len;i++)
            num[i] = (long long)(x1[i].r+0.5);
        len = 2*a[n-1];
        //减掉取两个相同的组合
        for(int i = 0;i < n;i++)
            num[a[i]+a[i]]--;
        //选择的无序，除以2
        for(int i = 1;i <= len;i++)
        {
            num[i]/=2;
            num[i] += num[i-1];
        }
        while(m--) {
            int k;
            scanf("%d", &k);
            printf("%lldn", num[k-1]);
        }
    }
    return 0;
}

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