H Super Fast Fourier Transform

题目描述

Bobo has two sequences of integers {a1,a2,…,an}{a_1, a_2, \dots, a_n}{a1​,a2​,…,an​} and {b1,b2,…,bm}{b_1, b_2, \dots, b_m}{b1​,b2​,…,bm​}.
He would like to find
∑i=1n∑j=1m⌊∣ai−bj∣⌋.\sum_{i = 1}^n\sum_{j = 1}^m \lfloor \sqrt{|a_i - b_j|} \rfloor.∑i=1n​∑j=1m​⌊∣ai​−bj​∣
​⌋.

Note that ⌊x⌋\lfloor x \rfloor⌊x⌋ denotes the maximum integer does not exceed x,
and |x| denotes the absolute value of x.

输入描述:

The input contains at most 30 sets. For each set:
The first line contains 2 integers n, m (1≤n,m≤1051 \leq n, m \leq 10^51≤n,m≤105).
The second line contains n integers a1,a2,…,ana_1, a_2, \dots, a_na1​,a2​,…,an​.
The thrid line contains m integers b1,b2,…,bmb_1, b_2, \dots, b_mb1​,b2​,…,bm​.
(ai,bi≥0,a1+a2+⋯+an,b1+b2+…,bm≤106a_i, b_i \geq 0, a_1 + a_2 + \dots + a_n, b_1 + b_2 + \dots, b_m \leq 10^6ai​,bi​≥0,a1​+a2​+⋯+an​,b1​+b2​+…,bm​≤106)

输出描述:

For each set,
an integer denotes the sum.

示例1
输入
复制

1 2
1
2 3

输出
复制

2

示例2
输入
复制

2 3
1 2
3 4 5

输出
复制

7

1≤n,m≤1e5，，直接暴力不行，，不过(ai,bi≥0,a1+a2+⋯+an,b1+b2+…,bm≤1e6)通过这个条件可知，，a和b序列会有很多重复数字，可以采取去重后暴力的方法来写。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdio>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;

const int maxn = 1e6 +10;
int a[maxn], b[maxn];
int numa[maxn], numb[maxn];
int tota = 0, totb = 0;
int n, m;

void init()
{
	tota = 0, totb = 0;
	mem(numa, 0), mem(numb, 0);
	mem(a, 0), mem(b, 0);
}

int main()
{
	while(~scanf("%d%d", &n, &m))
	{
		init();
		int x;
		for(int i = 1; i <= n; i++)
		{
			scanf("%d", &x);
			if(numa[x] == 0)
				a[tota++] = x;
			numa[x]++;
		}
		for(int i = 1; i <= m; i++)
		{
			scanf("%d", &x);
			if(numb[x] == 0)
				b[totb++] = x;
			numb[x]++;
		}
		long long ans = 0;
		for(int i = 0; i < tota; i++)
			for(int j = 0; j < totb; j++)
			{
				ans += 1ll * numa[a[i]] * numb[b[j]] * (long long)(sqrt(fabs(a[i] - b[j])));
			}
		printf("%lld\n", ans);
	}
	return 0;
}