CodeForces-438D The Child and Sequence 线段树

题目

给定一个长为n的序列,要求支持区间求和,区间取余,单点修改

题解

首先我们有一个结论:

若(x > p)那么有((x ext{ }mod ext{ }p) <= frac{x}{2})

所以我们在区间取余时利用区间最大值来剪枝后.

由势能分析法易得总体复杂度是(O(nlon^2n))的.

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
    x=0;char ch;bool flag = false;
    while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
    while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline void read(ll &x){
    x=0;char ch;bool flag = false;
    while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
    while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const int maxn = 100010;
ll mx[maxn<<2],sum[maxn<<2],a[maxn];
void build(int rt,int l,int r){
    if(l == r){
	mx[rt] = sum[rt] = a[l];
	return;
    }
    int mid = l+r >> 1;
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
    mx[rt] = max(mx[rt<<1],mx[rt<<1|1]);
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
ll L,R,val;
ll query(int rt,int l,int r){
    if(L <= l && r <= R) return sum[rt];
    int mid = l+r >> 1;
    if(R <= mid) return query(rt<<1,l,mid);
    if(L >  mid) return query(rt<<1|1,mid+1,r);
    return query(rt<<1,l,mid) + query(rt<<1|1,mid+1,r);
}
void mod(int rt,int l,int r){
    if(l == r){
	mx[rt] = sum[rt] = sum[rt] % val;
	return ;
    }
    if(mx[rt] < val) return;
    int mid = l+r >> 1;
    mod(rt<<1,l,mid);mod(rt<<1|1,mid+1,r);
    mx[rt] = max(mx[rt<<1],mx[rt<<1|1]);
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void search(int rt,int l,int r){
    if(L <= l && r <= R){
	mod(rt,l,r);
	return ;
    }
    int mid = l+r >> 1;
    if(L <= mid) search(rt<<1,l,mid);
    if(R >  mid) search(rt<<1|1,mid+1,r);
    mx[rt] = max(mx[rt<<1],mx[rt<<1|1]);
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
int pos;
void modify(int rt,int l,int r){
    if(l == r){
	mx[rt] = sum[rt] = val;
	return ;
    }
    int mid = l+r >> 1;
    if(pos <= mid) modify(rt<<1,l,mid);
    if(pos >  mid) modify(rt<<1|1,mid+1,r);
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
    mx[rt] = max(mx[rt<<1],mx[rt<<1|1]);
}
int main(){
    int n,m;read(n);read(m);
    for(int i=1;i<=n;++i){
	read(a[i]);
    }build(1,1,n);
    int op;
    while(m--){
	read(op);
	if(op == 1){
	    read(L);read(R);
	    printf("%lld
",query(1,1,n));
	}else if(op == 2){
	    read(L);read(R);read(val);
	    search(1,1,n);
	}else if(op == 3){
	    read(pos);read(val);
	    modify(1,1,n);
	}
    }
    return 0;
}