Sum(hdu4407)

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2974    Accepted Submission(s): 843


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
思路：容斥原理;
因为原数列是[1,n];所以数的和我们可以直接用求和公式；
然后m<1000;所以改变的最多不超过1000；我们用数组存改变，改变的位置会重复，这时后要覆盖，也就是数组中原先已有这个点，我们可以用map标记是否在数组中已有；
没有的话加入，有的话覆盖。
然后询问是[x,y]区间内与z互质的数的和，所以我们考虑分解z，那么素数打表。
然后容斥求在[x,y]与z互质数的和；然后再循环一遍改变，判断替换的数，和原来的数是否与z互质，原来的数互质的话减去原来的，改变的数互质的话，加上改变的数；
每个数的不同的质因数不超过9个；
复杂度（5120*m）；
  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<stdlib.h>
  5 #include<string.h>
  6 #include<queue>
  7 #include<map>
  8 typedef long long LL;
  9 using namespace std;
 10 bool prime[400005];
 11 int ans[400005];
 12 queue<int>que;
 13 int  cm[400005];
 14 int  ma[400005];
 15 int flag[2000];
 16 int gcd(int n,int m);
 17 map<int,int>my;
 18 int main(void)
 19 {
 20         int i,j,k;
 21         int p,q;
 22         int n,m;
 23         int x,y;
 24         int t;
 25         for(i=2; i<=2000; i++)
 26         {
 27                 if(!prime[i])
 28                 {
 29                         for(j=i; i*j<=400000; j++)
 30                         {
 31                                 prime[i*j]=true;
 32                         }
 33                 }
 34         }
 35         int cnt=0;
 36         for(i=2; i<=400000; i++)
 37         {
 38                 if(!prime[i])
 39                 {
 40                         ans[cnt++]=i;
 41                 }
 42         }
 43         scanf("%d",&k);
 44         while(k--)
 45         {      my.clear();
 46                 scanf("%d %d",&p,&q);
 47                 int bt=1;
 48                 for(i=0; i<=p; i++)
 49                         ma[i]=i;
 50                 while(q--)
 51                 {
 52                         scanf("%d",&n);
 53                         if(n==1)
 54                         {
 55                                 scanf("%d %d %d",&x,&y,&t);
 56                                 int r=0;
 57                                 int yy=0;
 58                                 int xy=t;
 59                                 if(x>y)swap(x,y);
 60                                 while(t>1)
 61                                 {
 62                                         if(r==0&&t%ans[yy]==0)
 63                                         {
 64                                                 r=1;
 65                                                 que.push(ans[yy]);
 66                                                 t/=ans[yy];
 67                                         }
 68                                         else if(r==1&&t%ans[yy]==0)
 69                                         {
 70                                                 t/=ans[yy];
 71                                         }
 72                                         else
 73                                         {
 74                                                 r=0;
 75                                                 yy++;
 76                                         }
 77                                 }
 78                                 x-=1;
 79                                 int ak=0;
 80                                 while(!que.empty())
 81                                 {
 82                                         cm[ak++]=que.front();
 83                                         que.pop();
 84                                 }
 85                                 LL sum1=0;
 86                                 LL sum2=0;
 87                                 for(i=1; i<=(1<<ak)-1; i++)
 88                                 {
 89                                         int nn=0;
 90                                         LL dp=1;
 91                                         for(j=0; j<ak; j++)
 92                                         {
 93                                                 if(i&(1<<j))
 94                                                 {
 95                                                         nn++;
 96                                                         dp*=(LL)cm[j];
 97                                                 }
 98                                         }
 99                                         if(nn%2)
100                                         {
101                                                LL ct=y/dp;
102                                                 LL qt=x/dp;
103                                                  LL ct1=ct+1;
104                                                  LL qt1=qt+1;
105                                                 sum1+=dp*ct1*ct/2;
106                                                 sum2+=dp*qt1*qt/2;
107                                         }
108                                         else
109                                         {
110                                                  LL ct=y/dp;
111                                                 LL qt=x/dp;
112                                                  LL ct1=ct+1;
113                                                  LL qt1=qt+1;
114                                                 sum1-=dp*ct1*ct/2;
115                                                 sum2-=dp*qt1*qt/2;
116                                         }
117                                 }  LL xc=1+x;LL yc=1+y;
118                                 LL ap1=(long long )(xc)*(long long)x/2;
119                                 LL ap2=(long long)(yc)*(long long)y/2;
120                                 ap2-=sum1;
121                                 ap1-=sum2;
122                                 ap2-=ap1;
123                                 for(j=1; j<bt; j++)
124                                 {
125                                         if(flag[j]>=x+1&&flag[j]<=y)
126                                         {
127                                                 int cp=gcd(xy,ma[flag[j]]);
128                                                 int cq=gcd(xy,flag[j]);
129                                                 if(cq==1)
130                                                         ap2-=flag[j];
131                                                 if(cp==1)
132                                                         ap2+=ma[flag[j]];
133                                         }
134                                 }
135                                 printf("%lld
",ap2);
136                         }
137                         else
138                         {
139                                 scanf("%d %d",&x,&t);
140                                 if(my[x]==0)
141                                 {my[x]=1;flag[bt]=x;bt++;}
142                                 ma[x]=t;
143                         }
144                 }
145         }
146         return 0;
147 }
148 int gcd(int n,int m)
149 {
150         if(m==0)
151                 return n;
152         else if(n%m==0)
153                 return m;
154         else return gcd(m,n%m);
155 }
 
油！油！you@
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原文地址：https://www.cnblogs.com/zzuli2sjy/p/5459853.html