大家好，又见面了，我是全栈君。
Problem Description
XXX is puzzled with the question below:
1, 2, 3, …, n (1<=n<=400000) are placed in a line. There are m (1<=m<=1000) operations of two kinds.
Operation 1: among the x-th number to the y-th number (inclusive), get the sum of the numbers which are co-prime with p( 1 <=p <= 400000).
Operation 2: change the x-th number to c( 1 <=c <= 400000).
For each operation, XXX will spend a lot of time to treat it. So he wants to ask you to help him.
Input
There are several test cases.
The first line in the input is an integer indicating the number of test cases.
For each case, the first line begins with two integers — the above mentioned n and m.
Each the following m lines contains an operation.
Operation 1 is in this format: “1 x y p”.
Operation 2 is in this format: “2 x c”.
The first line in the input is an integer indicating the number of test cases.
For each case, the first line begins with two integers — the above mentioned n and m.
Each the following m lines contains an operation.
Operation 1 is in this format: “1 x y p”.
Operation 2 is in this format: “2 x c”.
Output
For each operation 1, output a single integer in one line representing the result.
Sample Input
1 3 3 2 2 3 1 1 3 4 1 2 3 6
Sample Output
7 0
Source
思路：m<=1000 和 数列初始状态为1,2,3,..n 是该题的突破口。
对于每一次询问。先不考虑数字被改动了。那我们能够直接用求和公式把x到y之间的数字的和求出来，然后再减去那些不和p互质的数(用容斥原理)，再对改动过的数进行特判就可以。
#include <stdio.h>
int num,idx[1005],val[1005],prime[10],p[40000];
inline bool check(int x)
{
int i;
for(i=0;i<num;i++) if(x%prime[i]==0) return 0;
return 1;
}
int main()
{
int T,n,m,i,j,k,type,cnt,a,b,c,last,lxdcnt,lxdnum,l,r;
long long ans;
//把40W以内的素数预处理出来-----------------
cnt=0;
for(i=2;i<400000;i++)
{
for(j=2;j*j<=i;j++) if(i%j==0) break;
if(j*j>i) p[cnt++]=i;
}
//------------------------------------------
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
cnt=0;
for(i=1;i<=m;i++)
{
scanf("%d",&type);
if(type==1)
{
scanf("%d%d%d",&a,&b,&c);
ans=(long long)(a+b)*(b-a+1)/2;
num=0;//质因数的个数
//获取c的质因数-----------------
last=0;
while(c>1)
{
if(c%p[last]==0)
{
prime[num++]=p[last];
c/=p[last];
while(c%p[last]==0) c/=p[last];
}
last++;
}
//-------------------------------
//容斥原理-------------------------
for(j=1;j<(1<<num);j++)
{
lxdcnt=0;
lxdnum=1;
for(k=0;k<num;k++) if(j&(1<<k))
{
lxdcnt++;
lxdnum*=prime[k];
}
l=a/lxdnum*lxdnum;
if(l<a) l+=lxdnum;
r=b/lxdnum*lxdnum;
if(r<l) continue;
if(lxdcnt&1) ans-=(long long)(l+r)*((r-l)/lxdnum+1)/2;
else ans+=(long long)(l+r)*((r-l)/lxdnum+1)/2;
}
//-----------------------------------
//对改动过的数字特殊推断-----------------------------------
for(j=0;j<cnt;j++)
{
if(idx[j]>=a && idx[j]<=b)
{
if(!check(idx[j]))
{
if(check(val[j])) ans+=val[j];
}
else
{
if(check(val[j])) ans=ans+val[j]-idx[j];
else ans-=idx[j];
}
}
}
//--------------------------------------------------------
printf("%I64d\n",ans);
}
else
{
scanf("%d%d",&a,&b);
for(j=0;j<cnt;j++) if(idx[j]==a)//注意，改动的点可能之前已被改动过
{
val[j]=b;
break;
}
if(j==cnt)
{
idx[cnt]=a;
val[cnt++]=b;
}
}
}
}
}
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