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".
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 2012 ACM/ICPC Asia Regional Jinhua Online
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，先不考虑数字被修改了，那我们可以直接用求和公式把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)，再对修改过的数进行特判即可。
#include
int num,idx[1005],val[1005],prime[10],p[40000];
inline bool check(int x)
{
int i;
for(i=0;i
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<
=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