codeforces 914D(线段树)
2018-01-21 11:44
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题意:
1单点更新
2查询[l,r]区间的值最多改变一个能不能使得区间的gcd等于给出的x,但是此改变不改变原来的值。
思路:线段树维护gcd,很简单。查询的时候可以维护一个cnt的值,表示为当一个点的值不整除x的时候,对数进行改变的个数。当cnt>1的时候,直接结束函数。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 10;
typedef long long ll;
#define clr(x,y) memset(x,y,sizeof x)
#define INF 0x3f3f3f3f
typedef pair<int,int> P;
const ll Mod = 1e9 + 7;
int gcd(int x,int y)
{
return y ? gcd(y,x % y) : x;
}
int a[maxn];
int tree[maxn << 2];
void build(int l,int r,int rt)
{
if(l == r)
{
tree[rt] = a[l];return ;
}
int mid = (l + r) >> 1;
build(l,mid,rt << 1);build(mid + 1,r,rt << 1|1);
tree[rt] = gcd(tree[rt << 1],tree[rt << 1|1]);
}
void update(int pos,int x,int l,int r,int rt)
{
if(l == r)
{
tree[rt] = x;return ;
}
int mid = (l + r) >> 1;
if(pos <= mid)
update(pos,x,l,mid,rt << 1);
else update(pos,x,mid + 1,r,rt << 1|1);
tree[rt] = gcd(tree[rt << 1],tree[rt << 1|1]);
}
int cnt;
void fun(int l,int r,int x,int rt)
{
if(cnt > 1)return;
if(l == r)
{
if(tree[rt] % x)
cnt ++;
return ;
}
int mid = (l + r) >> 1;
if(tree[rt << 1] % x)
fun(l,mid,x,rt << 1);
if(tree[rt << 1|1] % x)
fun(mid + 1,r,x,rt << 1|1);
}
int query(int L,int R,int x,int l,int r,int rt)
{
if(cnt > 1)
return x;
if(L <= l && R >= r)
{
if(tree[rt] % x)
{
fun(l,r,x,rt);
return x;
}
return tree[rt];
}
int mid = (l + r) >> 1;
int ret = 0;
if(L <= mid)
ret = gcd(ret,query(L,R,x,l,mid,rt <<1));
if(R >= mid + 1)
ret = gcd(ret,query(L,R,x, mid + 1,r,rt << 1|1));
return ret;
}
int main()
{
int n;
while( ~ scanf("%d",&n))
{
for(int i = 1;i <= n;i ++)
scanf("%d",&a[i]);
build(1,n,1);
int m ;scanf("%d",&m);
for(int i = 1;i <= m;i ++)
{
int type;scanf("%d",&type);
if(type == 1)
{
int l,r,x;scanf("%d%d%d",&l,&r,&x);
cnt = 0;
query(l,r,x,1,n,1);
if(cnt <= 1)
puts("YES");
else puts("NO");
}
else
{
int x,y;scanf("%d%d",&x,&y);
if(a[x] == y)continue;
a[x] = y;
update(x,y,1,n,1);
}
}
}
return 0;
}
1单点更新
2查询[l,r]区间的值最多改变一个能不能使得区间的gcd等于给出的x,但是此改变不改变原来的值。
思路:线段树维护gcd,很简单。查询的时候可以维护一个cnt的值,表示为当一个点的值不整除x的时候,对数进行改变的个数。当cnt>1的时候,直接结束函数。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 10;
typedef long long ll;
#define clr(x,y) memset(x,y,sizeof x)
#define INF 0x3f3f3f3f
typedef pair<int,int> P;
const ll Mod = 1e9 + 7;
int gcd(int x,int y)
{
return y ? gcd(y,x % y) : x;
}
int a[maxn];
int tree[maxn << 2];
void build(int l,int r,int rt)
{
if(l == r)
{
tree[rt] = a[l];return ;
}
int mid = (l + r) >> 1;
build(l,mid,rt << 1);build(mid + 1,r,rt << 1|1);
tree[rt] = gcd(tree[rt << 1],tree[rt << 1|1]);
}
void update(int pos,int x,int l,int r,int rt)
{
if(l == r)
{
tree[rt] = x;return ;
}
int mid = (l + r) >> 1;
if(pos <= mid)
update(pos,x,l,mid,rt << 1);
else update(pos,x,mid + 1,r,rt << 1|1);
tree[rt] = gcd(tree[rt << 1],tree[rt << 1|1]);
}
int cnt;
void fun(int l,int r,int x,int rt)
{
if(cnt > 1)return;
if(l == r)
{
if(tree[rt] % x)
cnt ++;
return ;
}
int mid = (l + r) >> 1;
if(tree[rt << 1] % x)
fun(l,mid,x,rt << 1);
if(tree[rt << 1|1] % x)
fun(mid + 1,r,x,rt << 1|1);
}
int query(int L,int R,int x,int l,int r,int rt)
{
if(cnt > 1)
return x;
if(L <= l && R >= r)
{
if(tree[rt] % x)
{
fun(l,r,x,rt);
return x;
}
return tree[rt];
}
int mid = (l + r) >> 1;
int ret = 0;
if(L <= mid)
ret = gcd(ret,query(L,R,x,l,mid,rt <<1));
if(R >= mid + 1)
ret = gcd(ret,query(L,R,x, mid + 1,r,rt << 1|1));
return ret;
}
int main()
{
int n;
while( ~ scanf("%d",&n))
{
for(int i = 1;i <= n;i ++)
scanf("%d",&a[i]);
build(1,n,1);
int m ;scanf("%d",&m);
for(int i = 1;i <= m;i ++)
{
int type;scanf("%d",&type);
if(type == 1)
{
int l,r,x;scanf("%d%d%d",&l,&r,&x);
cnt = 0;
query(l,r,x,1,n,1);
if(cnt <= 1)
puts("YES");
else puts("NO");
}
else
{
int x,y;scanf("%d%d",&x,&y);
if(a[x] == y)continue;
a[x] = y;
update(x,y,1,n,1);
}
}
}
return 0;
}
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