[容斥原理] hdu 4407 Sum

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4407

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1817    Accepted Submission(s): 504

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

 

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题目意思：

开始有1~n的n(n<=40000)个数,有两种操作

操作1: 1  x y p    求在区间[x,y]之间和p互质的所有数的和

操作2: 2  x  c      把x位置上的数改成c

解题思路：

简单容斥原理

实际上题目要求的就是求1~x间和p互质的数的总和。可以先把p分解质因数，然后用容斥原理求出与p不互质的总和，用总和减去即可。

至于改变，只需保存改变的数值，然后对每个查询扫一遍，单独处理改变的就行了。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 410000

bool isp[Maxn];
ll pri[Maxn],pp[Maxn],cnt0,cnt;
ll x,y;
ll ans1,ans2;
map<int,int>myp;
int cc,n,m,p;
struct Cha
{
    int old,ne;
}cha[1100];

void init()  //打质数表
{
    cnt=0;
    int N=400000;

    for(int i=1;i<=N;i++)
        isp[i]=true;
    for(int i=2;i<=N;i++)
    {
        if(isp[i])
        {
            pri[++cnt]=i;
            for(int j=i*2;j<=N;j+=i)
                isp[j]=false;
        }
    }
}
void cal(ll cur) //分解质因数
{
    cnt0=0;

    for(int i=1;pri[i]*pri[i]<=cur;i++)
    {
        if(!(cur%pri[i]))
        {
            pp[++cnt0]=pri[i];
            while(!(cur%pri[i]))
                cur/=pri[i];
        }
    }
    if(cur!=1)
        pp[++cnt0]=cur;

}
void dfs(ll hav,int cur,int num) //容斥原理求出不互质的
{
    if((hav>x&&hav>y)||cur>cnt0)
        return ;
    for(int i=cur;i<=cnt0;i++)
    {
        ll next=hav*pp[i];

        ll tt1=y/next,tt2=x/next;

        if(num&1)
        {
            ans1-=(1+tt1)*tt1/2*next;
            ans2-=(1+tt2)*tt2/2*next;
        }
        else
        {
            ans1+=(1+tt1)*tt1/2*next;
            ans2+=(1+tt2)*tt2/2*next;
        }
        dfs(next,i+1,num^1);

    }
}
ll gcd(ll a,ll b)
{
    if(a%b==0)
        return b;
    return gcd(b,a%b);
}
int main()
{

    //printf("%I64d\n",gcd(12,18));
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   init();

   int t;
   scanf("%d",&t);
   while(t--)
   {
       scanf("%d%d",&n,&m);
       myp.clear();
       cc=0;

       while(m--)
       {
           int a;

           scanf("%d",&a);
           if(a==1)
           {
               scanf("%I64d%I64d%I64d",&x,&y,&p);
               if(x>y)
                    swap(x,y);
               x--;   //注意x要减一
               cal(p);
               ll ans;
               ans1=(1+y)*y/2,ans2=(1+x)*x/2;

               for(int i=1;i<=cnt0;i++)
               {
                   ll tt1=y/pp[i],tt2=x/pp[i];

                   ans1-=(1+tt1)*tt1/2*pp[i];
                   ans2-=(1+tt2)*tt2/2*pp[i];

                   dfs(pp[i],i+1,0);
               }
               ans=ans1-ans2;
               for(int i=1;i<=cc;i++)
               {
                   if(cha[i].ne!=cha[i].old&&cha[i].old<=y&&cha[i].old>x) //要在[x,y]范围内找，x已经减了1
                   {
                       if(gcd(cha[i].old,p)==1) //之前已经加了
                            ans-=cha[i].old;
                       if(gcd(cha[i].ne,p)==1) //改了过后是否要加上
                            ans+=cha[i].ne;
                   }
               }
               printf("%I64d\n",ans);
           }
           else
           {
               scanf("%d%d",&x,&y);
               if(myp.find(x)==myp.end())
               {
                    ++cc;
                   myp[x]=cc;
                   cha[cc].ne=y,cha[cc].old=x;
               }
               else
                   cha[myp[x]].ne=y;

           }
       }

   }

   return 0;
}