CF86 D. Powerful array
阿新 • • 發佈:2021-07-01
題目傳送門:https://codeforces.com/problemset/problem/86/D
題目大意:
給定一個長度為\(n\)的序列,有\(m\)組詢問,每次詢問\([l,r]\)中,\(\sum\limits_{s}K_s^2\times s\)的值,其中,\(K_s\)表示\(s\)在子串\([l,r]\)中的出現次數
由於這題沒有修改,僅有詢問,故我們可以考慮莫隊演算法
/*program from Wolfycz*/ #include<map> #include<cmath> #include<cstdio> #include<vector> #include<cstring> #include<iostream> #include<algorithm> #define Fi first #define Se second #define ll_inf 1e18 #define MK make_pair #define sqr(x) ((x)*(x)) #define pii pair<int,int> #define int_inf 0x7f7f7f7f using namespace std; typedef long long ll; typedef unsigned int ui; typedef unsigned long long ull; inline char gc(){ static char buf[1000000],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++; } template<typename T>inline T frd(T x){ int f=1; char ch=gc(); for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } template<typename T>inline T read(T x){ int f=1; char ch=getchar(); for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } inline void print(int x){ if (x<0) putchar('-'),x=-x; if (x>9) print(x/10); putchar(x%10+'0'); } const int N=2e5,M=1e6; int A[N+10],pos[N+10]; struct node{ int l,r,ID; void Read(int i){l=read(0),r=read(0),ID=i;} node(int _l=0,int _r=0,int _ID=0){l=_l,r=_r,ID=_ID;} bool operator <(const node &tis)const{return pos[l]!=pos[tis.l]?l<tis.l:r<tis.r;} }B[N+10]; ll All,Ans[N+10],Cnt[M+10]; void Add(int x,int v){ All-=1ll*x*sqr(Cnt[x]); Cnt[x]+=v; All+=1ll*x*sqr(Cnt[x]); } int main(){ // freopen(".in","r",stdin); // freopen(".out","w",stdout); int n=read(0),m=read(0),size=sqrt(n); for (int i=1;i<=n;i++) A[i]=read(0),pos[i]=(i-1)/size+1; for (int i=1;i<=m;i++) B[i].Read(i); sort(B+1,B+1+m); for (int i=1,l=1,r=0;i<=m;i++){ while (r<B[i].r) Add(A[++r], 1); while (r>B[i].r) Add(A[r--],-1); while (l<B[i].l) Add(A[l++],-1); while (l>B[i].l) Add(A[--l], 1); Ans[B[i].ID]=All; // printf("\n"); } for (int i=1;i<=m;i++) printf("%lld\n",Ans[i]); return 0; }