[hdu 5307] He is Flying (FFT)

题意

给出长度为$n$的数列$a$，记$sum_{i=1}^na_i=s(s<=50000)$
对于任意$kin[0,s]$,求出区间和为$k$的区间的长度之和

题目分析

有了几道fft题的经验,套路就是把可加性的计算化为幂的次数做多项式卷积
有
$large egin{aligned} Ans(k)=&sum_{i=1}^nsum_{j=0}^{i-1}[sum(i)-sum(j)==k](i-j)\ =&sum_{i=1}^nsum_{j=0}^{i-1}[sum(i)-sum(j)==k]i............(1)\ &-sum_{i=1}^nsum_{j=0}^{i-1}[sum(i)-sum(j)==k]j............(2)\ end{aligned}$
分别把$(1),(2)$看成两个多项式，做FFT即可

• 拿$(1)$举例

第一个多项式:
sum(i)的次数对应的系数加上i(1<=i<=n)
第二个多项式:
-sum(i)的次数对应的系数加上1(0<=i<=n-1)

由于负数次幂FFT无法处理,于是加上一个$s$,即

第一个多项式:
sum(i)的次数对应的系数加上i(1<=i<=n)
第二个多项式:
s-sum(i)的次数对应的系数加上1(0<=i<=n-1)

$k$所对应的答案存在$k+s$的次数的系数上

• $(2)$可以类比

• 注意特判一下$k=0$的情况

• 还有这道题卡精度,必须用Long Double,否则会WA(也可以写NTT)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;
typedef long double LD; //卡精度,不然会WA
const int MAXN = 1<<17;
const LD Pi = acos(-1.0);
struct complex
{
	LD r, i;
	complex(LD _r = 0, LD _i = 0):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, i*t.r + t.i*r); }
	
	
}a1[MAXN], a2[MAXN];

inline void change(complex arr[], const int& len)
{
	for(register int i = 1, j = len/2; i < len-1; ++i)
	{
		if(i < j) swap(arr[i], arr[j]);
		int k = len/2;
		while(j >= k) j -= k, k>>=1;
		j += k;
	}
}

inline void fft(complex arr[], const int& len, const int& flg)
{
	change(arr, len);
	for(register int i = 2; i <= len; i<<=1)
	{
		complex wn = complex(cos(2*Pi*flg/i), sin(2*Pi*flg/i));
		for(register int j = 0; j < len; j += i)
		{
			complex w = complex(1, 0);
			for(register int k = j; k < j + i/2; ++k)
			{
				complex A0 = arr[k];
				complex wA1 = w * arr[k + i/2];
				arr[k] = A0 + wA1;
				arr[k + i/2] = A0 - wA1;
				w = w * wn;
			}
		}
	}
	if(flg == -1)
		for(register int i = 0; i < len; ++i)
			arr[i].r /= len;
}

int sum[MAXN], S;

LL ans[MAXN], pres[MAXN];

int main ()
{
	for(LL i = 1; i <= 100000; ++i) //预处理
		pres[i] = pres[i-1] + i*(i+1)/2;
	int T, n;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d", &n);
		int last = 0; LL ans0 = 0;
		for(int i = 1, x; i <= n; ++i)
		{
			scanf("%d", &x), sum[i] = sum[i-1] + x;
			if(!x) ++last;
			else ans0 += pres[last], last = 0;
		}
		printf("%lld
", ans0 += pres[last]); //0特判
	
		S = sum[n];
	
		int len = 1;
		while(len < (S<<1)) len<<=1;
		
		for(int i = 0; i < len; ++i) a1[i] = a2[i] = complex(); //Part 1
		for(int i = 1; i <= n; ++i)
		{
			a1[sum[i]].r += i;
			if(i<n) a2[S-sum[i]].r += 1;
		}
		a2[S-0].r += 1; //i = sum[i] = 0
		fft(a1, len, 1); fft(a2, len, 1);
		for(int i = 0; i < len; ++i) a2[i] = a1[i] * a2[i];
		fft(a2, len, -1);
		for(int i = S+1; i <= (S<<1); ++i) ans[i-S] = LL(a2[i].r + 0.5);
		
		
		for(int i = 0; i < len; ++i) a1[i] = a2[i] = complex(); //Part 2
		for(int i = 1; i <= n; ++i)
		{
			a1[sum[i]].r += 1;
			if(i < n) a2[S-sum[i]].r += i;
		}
		fft(a1, len, 1); fft(a2, len, 1);
		for(int i = 0; i < len; ++i) a2[i] = a1[i] * a2[i];
		fft(a2, len, -1);
	
		for(int i = S+1; i <= (S<<1); ++i)
			printf("%lld
", ans[i-S] - LL(a2[i].r + 0.5));
	}
}