51nod1678 lyk与gcd

链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1678

题意：中文题。

分析：比较基础的容斥题，我们预处理出每个i的所有素因子的组合，比如6={2,3,6}，那么我们对于a[6]将它加入到sum[2],sum[3],sum[6]中，统计答案时用容斥思想加加减减就行了。O(nlogn)

代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=100010;
const int MAX=1000000100;
const int mod=100000000;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=1000000007;
const int INF=1000000010;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned long long ull;
vector<int>f
;
vector<int>g
;
int a
,q
;
ll sum
;
void deal(int n) {
int i,j,h,w,k=0;
for (i=2;i<=n;i++) {
if (!q[i]) a[++k]=i;
for (j=1;j<=k;j++) {
if (a[j]*i>n) break ;
q[a[j]*i]=1;
if (i%a[j]==0) break ;
}
}
for (i=1;i<=k;i++)
for (j=a[i];j<=n;j+=a[i]) {
w=f[j].size();
for (h=0;h<w;h++) {
f[j].push_back(f[j][h]*a[i]);g[j].push_back(g[j][h]+1);
}
f[j].push_back(a[i]);g[j].push_back(1);
}
}
int main()
{
int i,j,n,q,x,y;
ll ans;
scanf("%d%d", &n, &q);
deal(n);
for (i=1;i<=n;i++) {
scanf("%d", &a[i]);sum[1]+=a[i];
for (j=0;j<f[i].size();j++) sum[f[i][j]]+=a[i];
}
while (q--) {
scanf("%d", &x);
if (x==1) {
scanf("%d%d", &x, &y);
for (i=0;i<f[x].size();i++) sum[f[x][i]]-=a[x];
sum[1]-=a[x];a[x]=y;sum[1]+=a[x];
for (i=0;i<f[x].size();i++) sum[f[x][i]]+=a[x];
} else {
ans=sum[1];
scanf("%d", &x);
for (i=0;i<f[x].size();i++)
if (g[x][i]&1) ans-=sum[f[x][i]];
else ans+=sum[f[x][i]];
printf("%lld\n", ans);
}
}
return 0;
}