标签：des 0.00 atom cst 3.2 should org out while
Nikita likes tasks on order statistics, for example, he can easily find the k-th number in increasing order on a segment of an array. But now Nikita wonders how many segments of an array there are such that a given number x is the k-th number in increasing order on this segment. In other words, you should find the number of segments of a given array such that there are exactly k numbers of this segment which are less than x
.
Nikita wants to get answer for this question for each k
from 0 to n, where n
is the size of the array.
InputThe first line contains two integers n
and x (1≤n≤2⋅105,−109≤x.
The second line contains n
integers a1,a2,…,an (−109≤ai≤109)— the given array.
OutputPrint n+1
integers, where the i-th number is the answer for Nikita‘s question for k.
Examples5 3
1 2 3 4 5
6 5 4 0 0 0
2 6
-5 9
1 2 0
6 99
-1 -1 -1 -1 -1 -1
0 6 5 4 3 2 1
把所有小于 x 的数记为 1，剩下的数记为 0，则题目要求的是有多少个区间有 k 个 1。对这个 01 数组求前缀和，设为 s，则题目要求的是有多少个 (l 满足 sr
。
现在相当于有一个长度为 n+1
的数组 s，满足 si≤si+1≤si+1，问两个元素的差的取值的情况数。设多项式 f 的 x 项的系数等于 sk 的 k 的个数，设多项式 g 的 x 项的系数等于 sk 的 k 的个数。对 f 和 g 做卷积，则积的 x 的系数就是对应的答案。当 k 时，首先要减去自己卷上自己的 n+1 次，然后再除以二，就是对应的答案。
1 #include"cstdio" 2 #include"iostream" 3 #include"cmath" 4 #include"cstring" 5 #include"algorithm" 6 using namespace std; 7 typedef long long ll; 8 int n,m; 9 const int nn=2e6+20; 10 const double pi = acos(-1); 11 struct complex { 12 double x,y; 13 complex(double xx=0,double yy=0){x=xx,y=yy;} 14 }; 15 16 complex operator-(complex a,complex b) {return complex(a.x-b.x,a.y-b.y);} 17 complex operator+(complex a,complex b){return complex(a.x+b.x,a.y+b.y);} 18 complex operator*(complex a,complex b) 19 {return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);} 20 21 string s1,s2; 22 int n2; int k; 23 complex a[2000000],b[2000000]; 24 int limit=1,l; 25 ll num[2000000]; 26 int sum[2000000]; 27 int r[2000000]; 28 29 inline void fft(complex *t,int type) 30 { 31 for (int i=0;i<limit;i++)if (i<r[i])swap(t[i],t[r[i]]); 32 for (int mid=1;mid<limit;mid<<=1) 33 { 34 complex wn=complex(cos(pi/mid),type*sin(pi/mid)); 35 for (int R=mid<<1,j=0;j<limit;j+=R) 36 { 37 complex w=complex(1,0); 38 for (int k=0;k<mid;k++,w=w*wn) 39 { 40 complex x=t[j+k],y=w*t[j+mid+k]; 41 t[j+k]=x+y; 42 t[j+k+mid]=x-y; 43 } 44 } 45 46 } 47 } 48 int t[300000]; 49 50 int main() 51 { 52 53 cin>>n>>k; 54 for(int i=1;i<=n;i++)scanf("%d",&t[i]),t[i]=t[i-1]+(t[i]<k); 55 for (int i=0;i<=n;i++)a[n+t[i]].x++,b[n-t[i]].x++; 56 int len = 3*n; 57 while (limit<=len)limit<<=1,l++; 58 for (int i=0;i<limit;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1)); 59 fft(a,1); fft(b,1); 60 for (int i=0;i<=limit;i++)a[i]=a[i]*b[i]; 61 fft(a,-1); 62 for (int i=0;i<=len;i++)num[i]=(ll)(a[i].x/limit+0.5); 63 // cout<<endl; 64 // for (int i=1;i<=len;i++)cout<<num[i]<<" "; 65 // cout<<"**************"<<endl; 66 ll ans = num[n*2]; 67 ans=(ans-n-1)>>1; 68 cout<<ans<<" "; 69 for (int i=1;i<=n;i++)printf("%lld ",num[i+n+n]); 70 71 72 }
标签：des 0.00 atom cst 3.2 should org out while
原文地址：https://www.cnblogs.com/zhangbuang/p/10310925.html
