Hdu 4407 Sum

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 897    Accepted Submission(s): 259

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

 

Recommend

zhoujiaqi2010

参考思路
http://www.cnblogs.com/Lyush/archive/2012/09/22/2698448.html

#include <iostream>
#include <map>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
bool h[700];
int p[200];
void prime()
{
    int cnt=1;
    int i,j;
    p[0]=2;
    for(i=2;i<=700;i+=2)
      h[i]=1;
    for(i=3;i<=700;i+=2)
     if(!h[i])
     {
         p[cnt++]=i;
        for(j=i*i;j<=700;j+=i)
          h[j]=1;
     }
}
int gcd(int a,int b)
{
    int r;
    while(r=a%b){a=b;b=r;}
    return b;
}
__int64 solve(__int64 n,__int64 pp)
{
    if(n==0) return 0;
    int i,j;
    int rc[10],cnt=0;
    int m=(int)sqrt(pp*1.0);
    for(i=0;p[i]<=m;i++)
    {
        if(pp%p[i]==0) rc[cnt++]=p[i];
        while(pp%p[i]==0)pp=pp/p[i];
        if(pp==1) break;
    }
    if(pp>1) rc[cnt++]=pp;
    int w=(1<<cnt);
    int js;
    __int64 k=0;
    __int64 ans=0,cj;
    for(i=1;i<w;i++)//关键点、将容斥原理的关系转换成二进制、这个还是第一次接触、纪念下
    {    cj=1;
        for(js=j=0;j<cnt;j++)
        if(i&(1<<j))
        {
           js++;
           cj*=rc[j];
        }
        k=n/cj;
        if(js&1)
         ans+=((cj+cj*k)*k)/2;
        else
         ans-=((cj+cj*k)*k)/2;
    }
    return ans;
}

map <int,int> mp;
map <int,int>::iterator it;
int main()
{
    prime();
    int T;
    int n,m;
    __int64 op,x,y,p;
    __int64 sum;
    scanf("%d",&T);
    while(T--)
    {
       scanf("%d %d",&n,&m);
       while(m--)
       {
           scanf("%I64d",&op);
           if(op==1)
           {
               scanf("%I64d %I64d %I64d",&x,&y,&p);
               sum=(y-x+1)*(x+y)>>1;
               for(it=mp.begin();it!=mp.end();it++) 此题关键点是m很小、才1000
                if(it->first>=x)
                {   if(it->first>y) break;
                    sum-=gcd(it->first,p)==1?it->first:0;
                    sum+=gcd(it->second,p)==1?it->second:0;
                }
               sum-=(solve(y,p)-solve(x-1,p));
               printf("%I64d\n",sum);
           }
           else
           {
              scanf("%I64d %I64d",&x,&y);
              mp[x]=y;
           }
       }
       mp.clear();
    }
    return 0;
}