You have written on a piece of paper an array of n positive integers a[1], a[2], ..., a[n] and m good pairs of integers (i1, j1), (i2, j2), ..., (im, jm). Each good pair (ik, jk) meets the following conditions: ik + jk is an odd number and 1 ≤ ik < jk ≤ n.
In one operation you can perform a sequence of actions:
- take one of the good pairs (ik, jk) and some integer v (v > 1), which divides both numbers a[ik] and a[jk];
- divide both numbers by v, i. e. perform the assignments: and .
Determine the maximum number of operations you can sequentially perform on the given array. Note that one pair may be used several times in the described operations.
Input
The first line contains two space-separated integers n, m (2 ≤ n ≤ 100, 1 ≤ m ≤ 100).
The second line contains n space-separated integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — the description of the array.
The following m lines contain the description of good pairs. The k-th line contains two space-separated integers ik, jk (1 ≤ ik < jk ≤ n, ik + jk is an odd number).
It is guaranteed that all the good pairs are distinct.
Output
Output the answer for the problem.
Sample Input
3 2
8 3 8
1 2
2 3
0
3 2
8 12 8
1 2
2 3
2
将点拆成多个素数，然后套用网络流，还有一种思路是建很多次图，分开处理素数的匹配。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 const int INF=1000000000; 6 const int N=110,M=1010,K=35010; 7 int P[K],cntx;bool check[K]; 8 void Linear_Shaker(){ 9 for(int i=2;i<K;i++){ 10 if(!check[i])P[++cntx]=i; 11 for(int j=1;j<=cntx;j++){ 12 if(i*P[j]>=K)break; 13 check[i*P[j]]=true; 14 if(i%P[j]==0)break; 15 } 16 } 17 } 18 int v[N][12],c[N][12],id[N][12],h[N],idx; 19 int cnt=1,fir[M],to[M*20],nxt[M*20],cap[M*20]; 20 int dis[M],gap[M],path[M],fron[M],q[M],f,b; 21 void add(int a,int b,int c){ 22 nxt[++cnt]=fir[a]; 23 to[fir[a]=cnt]=b; 24 cap[cnt]=c; 25 } 26 27 void addedge(int a,int b,int c){ 28 //printf("%d %d ",a,b); 29 add(a,b,c);add(b,a,0); 30 } 31 32 bool BFS(int S,int T){ 33 q[f=b]=T;dis[T]=1; 34 while(f<=b){ 35 int x=q[f++]; 36 for(int i=fir[x];i;i=nxt[i]) 37 if(!dis[to[i]]){ 38 dis[to[i]]=dis[x]+1; 39 q[++b]=to[i]; 40 } 41 } 42 return dis[S]; 43 } 44 45 int Max_Flow(int S,int T){ 46 if(!BFS(S,T))return 0; 47 for(int i=S;i<=T;i++)fron[i]=fir[i]; 48 for(int i=S;i<=T;i++)gap[dis[i]]+=1; 49 int ret=0,f,p=S; 50 while(dis[S]<=T+1){ 51 if(p==T){ 52 f=INF; 53 while(p!=S){ 54 f=min(f,cap[path[p]]); 55 p=to[path[p]^1]; 56 }ret+=f,p=T; 57 while(p!=S){ 58 cap[path[p]]-=f; 59 cap[path[p]^1]+=f; 60 p=to[path[p]^1]; 61 } 62 } 63 for(int &i=fron[p];i;i=nxt[i]) 64 if(cap[i]&&dis[to[i]]==dis[p]-1){ 65 path[p=to[i]]=i;break; 66 } 67 if(!fron[p]){ 68 if(!--gap[dis[p]])break;int Min=T+1; 69 for(int i=fir[p];i;i=nxt[i]) 70 if(cap[i])Min=min(Min,dis[to[i]]); 71 gap[dis[p]=Min+1]+=1;fron[p]=fir[p]; 72 if(p!=S)p=to[path[p]^1]; 73 } 74 } 75 return ret; 76 } 77 78 int n,m,S,T,num[N]; 79 int G[N][N]; 80 int main(){ 81 Linear_Shaker(); 82 scanf("%d%d",&n,&m); 83 for(int i=1;i<=n;i++){ 84 scanf("%d",&num[i]); 85 for(int j=1;j<=cntx;j++){ 86 if(P[j]*P[j]>num[i])break; 87 if(num[i]%P[j]==0){ 88 v[i][h[i]]=P[j]; 89 while(num[i]%P[j]==0){ 90 c[i][h[i]]+=1; 91 num[i]/=P[j]; 92 }h[i]+=1; 93 } 94 } 95 if(num[i]!=1){ 96 v[i][h[i]]=num[i]; 97 c[i][h[i]++]=1; 98 } 99 } 100 /* 101 for(int i=1;i<=n;i++){ 102 for(int j=0;j<h[i];j++) 103 printf("<%d %d> ",v[i][j],c[i][j]); 104 puts(""); 105 } 106 */ 107 for(int i=1,a,b;i<=m;i++){ 108 scanf("%d%d",&a,&b); 109 if(a%2)swap(a,b); 110 G[a][b]=1; 111 } 112 for(int i=1;i<=n;i++) 113 for(int j=0;j<h[i];j++) 114 id[i][j]=++idx; 115 S=0;T=idx+1; 116 for(int i=1;i<=n;i++) 117 for(int j=0;j<h[i];j++){ 118 if(i%2==0)addedge(S,id[i][j],c[i][j]); 119 else addedge(id[i][j],T,c[i][j]); 120 } 121 for(int i=2;i<=n;i+=2) 122 for(int j=1;j<=n;j+=2)if(G[i][j]) 123 for(int x=0;x<h[i];x++) 124 for(int y=0;y<h[j];y++) 125 if(v[i][x]==v[j][y]) 126 addedge(id[i][x],id[j][y],INF); 127 printf("%d ",Max_Flow(S,T)); 128 return 0; 129 }