https://www.bnuoj.com/v3/contest_show.php?cid=9149#problem/I
【题意】
给定长度为l的一个数组，初始值为0；规定了两种操作：
【思路】
找到了一个讲解很清楚的博客http://www.cnblogs.com/flipped/p/HDU4947.html
【Accepted】
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <stack> 7 #include <queue> 8 #include <string> 9 #include <vector> 10 #include <set> 11 #include <map> 12 #include <cassert> 13 using namespace std; 14 15 #define CLR(a,b) memset(a,b,sizeof(a)) 16 typedef long long LL; 17 const int N = 200000+20; 18 bool check[N]; 19 int mu[N],prime[N]; 20 21 vector<int> fac[N]; 22 LL f[N]; 23 int l,q; 24 25 void Mobius() 26 { 27 CLR(check, 0); 28 mu[1] = 1; 29 int tot = 0; 30 for(int i = 2; i < N ; i++){ 31 if(!check[i]){ 32 prime[tot ++] = i; 33 mu[i] = -1; 34 } 35 for(int j = 0 ;j < tot; j ++){ 36 if(i * prime[j] >= N)break; 37 check[i * prime[j]] = true; 38 if(i % prime[j] == 0){ 39 mu[i * prime[j]] = 0; 40 break; 41 }else{ 42 mu[i * prime[j]] = -mu[i]; 43 } 44 } 45 } 46 for(int i = 1 ;i < N ; i++){ 47 for(int j = i ; j < N ; j += i){ 48 fac[j].push_back(i); 49 } 50 } 51 } 52 53 inline LL sum(int p){ 54 LL s = 0; 55 while(p > 0){ 56 s += f[p]; 57 p -= p & (-p); 58 } 59 return s; 60 } 61 62 inline void add(int p,int v){ 63 while(p <= l){ 64 f[p] += v; 65 p += (p) & (-p); 66 } 67 } 68 69 void update(int n,int d,int v){ 70 if(n % d != 0)return; 71 72 n = n/d; 73 for(int i = 0 ;i < fac[n].size() ; i++){ 74 int q = fac[n][i]; 75 add(q * d, v * mu[q]); 76 } 77 } 78 79 LL query(int p) 80 { 81 LL ans = 0; 82 for(int i = 1,last = i ; i <= p ; i = last + 1){ 83 last = p/(p/i); 84 ans += (LL)(p/i) * (sum(last) - sum(i-1)) ; 85 } 86 return ans; 87 } 88 89 int main() 90 { 91 Mobius(); 92 int cas = 0; 93 while(~scanf("%d%d",&l,&q)){ 94 if(l == 0 && q == 0)break; 95 CLR(f, 0); 96 cas ++; 97 printf("Case #%d: ",cas); 98 while(q--){ 99 int t; 100 scanf("%d",&t); 101 if(t == 1){ 102 int n,d,v; 103 scanf("%d%d%d",&n,&d,&v); 104 update(n, d, v); 105 }else{ 106 int x; 107 scanf("%d",&x); 108 printf("%I64d ",query(x)); 109 } 110 } 111 } 112 return 0; 113 } 114 115 /* 116 117 6 4 118 1 4 1 2 119 2 5 120 1 3 3 3 121 2 3 122 0 0 123 124 */