[Usaco2015 dec]Breed Counting

原题链接https://www.lydsy.com/JudgeOnline/problem.php?id=4397

用线段树维护区间和即可。时间复杂度(O((N+Q)logN))。

#include<iostream>
#include<cstring>
#include<cstdio>
#define maxn 100010
using namespace std;
 
inline int read(){
    register int x(0),f(1); register char c(getchar());
    while(c<'0'||'9'<c){ if(c=='-') f=-1; c=getchar(); }
    while('0'<=c&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
 
int n,q,a[maxn];
 
struct SegmentTree{
    struct node{
        int l,r,sum[4];
    }t[maxn<<2];
    void build(int d,int l,int r){
        t[d].l=l,t[d].r=r;
        if(l==r){
            for(register int i=1;i<=3;i++) t[d].sum[i]=0;
            t[d].sum[a[l]]=1; return;
        }
        int mid=(l+r)>>1;
        build(d<<1,l,mid),build(d<<1|1,mid+1,r);
        for(register int i=1;i<=3;i++)
            t[d].sum[i]=t[d<<1].sum[i]+t[d<<1|1].sum[i];
    }
    void getsum(int d,const int &l,const int &r,int &ans1,int &ans2,int &ans3){
        if(l<=t[d].l&&t[d].r<=r){ ans1=t[d].sum[1],ans2=t[d].sum[2],ans3=t[d].sum[3]; return; }
        int mid=(t[d].l+t[d].r)>>1,sum[4]={0,0};
        if(l<=mid)
            getsum(d<<1,sum[1],sum[2],sum[3]),ans1=sum[1],ans2=sum[2],ans3=sum[3];
        if(r>mid)
            getsum(d<<1|1,ans1+=sum[1],ans2+=sum[2],ans3+=sum[3];
    }
}tree;
 
int main(){
    n=read(),q=read();
    for(register int i=1;i<=n;i++) a[i]=read();
    tree.build(1,1,n);
    for(register int i=1;i<=q;i++){
        int l=read(),r=read(),ans1=0,ans2=0,ans3=0;
        tree.getsum(1,ans1,ans2,ans3);
        printf("%d %d %dn",ans3);
    }
    return 0;
}