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cnt数组记录每种数字出现的次数

我们只需要考虑add(x),del(x),其他的交给莫队

我们考虑当前数原本的答案是a_x \times cnt_x^2

现在变成a_x \times (cnt_x+1)^2

cnt_x =y

(y+1)^2 = y^2 + 2y +1

所以更新的时候ans += a_x \times (cnt_x \times 2 + 1)

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;rg char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;rg char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;rg char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;rg char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
#define N 200011
int n,Q,a[N],sz;
struct que{
    int l,r,id;
}q[N];
il int bl(int x){return (x-1)/sz+1;}
il bool cmp(que x,que y){
    if(bl(x.l)==bl(y.l))return x.r<y.r;
    else return x.l<y.l;
}
int cnt[1000001];
ll ans=0,ANS[N];
il void add(int x){
    ans+=1ll*a[x]*(cnt[a[x]]<<1|1);
    ++cnt[a[x]];
}
il void del(int x){
    ans-=1ll*a[x]*(cnt[a[x]]*2-1);
    --cnt[a[x]];
}
int main(){
    in(n,Q);
    sz=sqrt(n);
    Fur(i,1,n)in(a[i]);
    Fur(i,1,Q)in(q[i].l,q[i].r),q[i].id=i;
    sort(q+1,q+Q+1,cmp);
    int l=0,r=0;
    Fur(i,1,Q){
        Fur(j,l,q[i].l-1)del(j);
        Fur(j,r+1,q[i].r)add(j);
        Fdr(j,l-1,q[i].l)add(j);
        Fdr(j,r,q[i].r+1)del(j);
        ANS[q[i].id]=ans;
        l=q[i].l;r=q[i].r;
    }
    Fur(i,1,Q)out(ANS[i],ln);
    flush();
}
LG CF86D Powerful-array
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