hdu 5634 Rikka with Phi

题意：完成三种操作

There are three types of queries: 

1lr  

Change  A[i]  into  φ(A[i]) , for all  i∈[l,r] .

2lrx  

Change  A[i]  into  x , for all  i∈[l,r] .

3lr  

Sum up  A[i] , for all  i∈[l,r] .

题解：线段树+欧拉函数

代码

#include<cstdio>
#include<cstring>
#define M 300100
#define N 10000100
#define ll long long

ll sum[M * 4];
int a[M * 4];
int phi[N];
int prime[N];

void build(int node, int l, int r)
{
	if(l == r)
	{
		scanf("%d", &a[node]);
		sum[node] = a[node];
		return;
	}
	int mid = (l + r) / 2;
	build(node * 2 ,l, mid);
	build(node * 2 + 1, mid + 1, r);
	sum[node] = sum[node * 2] + sum[node * 2 + 1];
    if(a[node * 2] == a[node * 2 + 1])
		a[node] = a[node * 2];
    else a[node] = 0;
}

void updata1(int l, int r, int left, int right, int node)
{
	if(left <= l && right >= r && a[node])
	{
		a[node] = phi[a[node]];
		sum[node] = (ll)(r - l + 1) * a[node];
		return;
	}
	if(l == r)
	{
		a[node] = phi[sum[node]];
		sum[node] = phi[sum[node]];
		return;
	}
	if(a[node])
	{
		int mid = (l + r) / 2;
		sum[node * 2] = (ll)(mid - l + 1) * a[node];
		sum[node * 2 + 1] = (ll)(r - mid) * a[node];
		a[node * 2] = a[node];
		a[node * 2 + 1] = a[node];
		a[node] = 0;
	}
	int mid = (l + r) / 2;
	if(left <= mid)updata1(l, mid, left, right, node * 2);
	if(right > mid)updata1(mid + 1, r, left, right, node * 2 + 1);
	sum[node] = sum[node * 2] + sum[node * 2 + 1];
    if(a[node * 2] == a[node * 2 + 1])
		a[node] = a[node * 2];
    else a[node] = 0;
}

void updata2(int l, int r, int left, int right, int node, int x)
{
	if(left <= l && right >= r)
	{
		a[node] = x;
		sum[node] = (ll)x * (r - l + 1);
		return;
	}
	if(a[node])
	{
		int mid = (l + r) / 2;
		sum[node * 2] = (ll)(mid - l + 1) * a[node];
		sum[node * 2 + 1] = (ll)(r - mid) * a[node];
		a[node * 2] = a[node];
		a[node * 2 + 1] = a[node];
		a[node] = 0;
	}
	int mid = (l + r) / 2;
	if(left <= mid)updata2(l, mid, left, right, node * 2, x);
	if(right > mid)updata2(mid + 1, r, left, right, node * 2 + 1, x);
	sum[node] = sum[node * 2] + sum[node * 2 + 1];
    if(a[node * 2] == a[node * 2 + 1])
		a[node] = a[node * 2];
    else a[node] = 0;
}

ll query(int l, int r, int left, int right, int node)
{
	if(left <= l && right >= r)
		return sum[node];
	if(a[node])
	{
		int mid = (l + r) / 2;
		sum[node * 2] = (ll)(mid - l + 1) * a[node];
		sum[node * 2 + 1] = (ll)(r - mid) * a[node];
		a[node * 2] = a[node];
		a[node * 2 + 1] = a[node];
		a[node] = 0;
	} 
	ll ans = 0;
	int mid = (l + r) / 2;
	if(left <= mid)ans = ans + query(l, mid, left, right, node * 2);
	if(right > mid)ans = ans + query(mid + 1, r, left, right, node * 2 + 1);
	return ans;
}

void phi_table()
{
    memset(phi,0,sizeof(phi));
    phi[1]=1;
    int flag = 0;
    for(int i=2;i<=N;i++)
    {
        if(!phi[i])phi[i]=i-1,prime[flag++]=i;
        for(int j=0;j < flag&&prime[j]*i<=N;j++)
        {
            if(i%prime[j]) phi[prime[j]*i] = phi[i]*(prime[j]-1);
            else
            {
                phi[prime[j]*i] = phi[i]*prime[j];
                break;
            }
        }
    }
}

int main()
{
	phi_table();
	int T;
	scanf("%d", &T);
	for(int t = 1; t <= T; t++)
	{
		int n, m;
		scanf("%d %d", &n, &m);
		build(1, 1, n);
		for(int i = 1; i <= m; i++)
		{
			int x, y, z;
			scanf("%d %d %d", &x, &y, &z);
			if(x == 1)updata1(1, n, y, z, 1);
			else if(x == 2)
			{
				int num;
				scanf("%d", &num);
				updata2(1, n, y, z, 1,num);
			}
			else if(x == 3)
				printf("%lld\n", query(1, n, y, z, 1));
		}
	}
	return 0;
}