• hdu 5726 GCD


    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5726

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 #define cl(a,b) memset(a,b,sizeof(a))
     4 #define LL long long
     5 #define pb push_back
     6 #define gcd __gcd
     7 
     8 const int maxn = 1e5+200;
     9 const int inf  = 1 << 23;
    10 
    11 int n;LL a[maxn];
    12 LL node[maxn*4];
    13 
    14 void push_up(int rt){
    15     node[rt] = gcd(node[rt<<1],node[rt<<1|1]);
    16 }
    17 
    18 void build(int rt,int l,int r){
    19     if(l==r){
    20         node[rt]=a[l];
    21         return ;
    22     }
    23     int mid = l+r>>1;
    24     build(rt<<1, l, mid);
    25     build(rt<<1|1, mid+1, r);
    26     push_up(rt);
    27 }
    28 
    29 LL query(int rt,int l,int r,int x,int y){
    30     //printf("rt = %d, l = %d, r = %d, x = %d,y = %d
    ",rt,l,r,x,y);
    31     if(x<=l&&y>=r){
    32         return node[rt];
    33     }
    34     int mid = l+r>>1;
    35     LL lson=0 ,rson=0;
    36 
    37     if(x<=mid) lson = query(rt<<1,l,mid,x,y);
    38     if(y>mid)  rson = query(rt<<1|1,mid+1,r,x,y);
    39     if(lson==0)return rson;
    40     if(rson==0)return lson;
    41     return gcd(lson,rson);
    42 }
    43 
    44 map<LL,LL> f[maxn];//f[i]记录下第a[i]形成的gcd的集合
    45 map<LL,LL> ans;
    46 
    47 int main(){
    48     int T;scanf("%d",&T);
    49     int cas = 1;
    50     while(T--){
    51         
    52         scanf("%d",&n);
    53         for(int i=1;i<=n;i++){
    54             scanf("%d",&a[i]);
    55         }
    56         printf("Case #%d:
    ",cas++);
    57 
    58         build(1,1,n);
    59         ans.clear();
    60         for(int i=0;i<=n;i++)f[i].clear();
    61 
    62         ans[a[1]]=1;f[1][a[1]]=1;
    63         for(int i=2;i<=n;i++){
    64             ans[a[i]]++;f[i][a[i]]+=1;
    65             for(map<LL,LL>::iterator it = f[i-1].begin();it!=f[i-1].end();it++){
    66                 LL tmp = gcd(a[i],it->first);
    67                 f[i][tmp]+=it->second;
    68                 ans[tmp]+=it->second;
    69             }
    70         }
    71 
    72         int q;scanf("%d",&q);
    73         while(q--){
    74             int x,y;scanf("%d%d",&x,&y);
    75             LL _gcd = query(1,1,n,x,y);
    76             printf("%lld %lld
    ",_gcd,ans[_gcd]);
    77         }
    78     }
    79     return 0;
    80 }
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  • 原文地址:https://www.cnblogs.com/pter/p/5705385.html
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